{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0001","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f86bb5a8148a1fdaef7e77dfda9141f197edca1783dc59122b363d847552d961","family_id":"qf-density-integral-additivity-triple","seed":202610096001,"semantic_template":"qf-density-integral-additivity-triple/[\"all_zero_segments\"]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [0], constant term first. For arbitrary oriented endpoints a=0,b=1,c=2, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(0/1, 0/1, 0/1)","explanation":"Exact signed moments give (0/1, 0/1, 0/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0002","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f3ee65a7597ceb70bb64aaf668cf287b591cc54ef53524befa89bfd1f9b0b9c","family_id":"qf-density-integral-additivity-triple","seed":202610096002,"semantic_template":"qf-density-integral-additivity-triple/[[3],3,-1,2]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [3], constant term first. For arbitrary oriented endpoints a=3,b=-1,c=2, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-12/1, 9/1, -3/1)","explanation":"Exact signed moments give (-12/1, 9/1, -3/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0003","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23b1b6722b79dfe65179c03e98c98ded089e883ccc71539344b0524ecfce4d4c","family_id":"qf-density-integral-additivity-triple","seed":202610096003,"semantic_template":"qf-density-integral-additivity-triple/[[1,2,3],0,2,0]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [1, 2, 3], constant term first. For arbitrary oriented endpoints a=0,b=2,c=0, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(14/1, -14/1, 0/1)","explanation":"Exact signed moments give (14/1, -14/1, 0/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0004","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ca88935ab330a13a7dd29d66dd284f208b979ff4b77a8bac0baf3ddc6378b8a","family_id":"qf-density-integral-additivity-triple","seed":202610096004,"semantic_template":"qf-density-integral-additivity-triple/[[-4,4,5,1],-2,-6,-11]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [-4, 4, 5, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-6,c=-11, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(160/3, 20015/12, 6885/4)","explanation":"Exact signed moments give (160/3, 20015/12, 6885/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0005","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb84e25a88801e60139e3ef82110b8f4896f66acd459c387402b61192d6ffafd","family_id":"qf-density-integral-additivity-triple","seed":202610096005,"semantic_template":"qf-density-integral-additivity-triple/[[-3,5,6,1],-1,0,6]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [-3, 5, 6, 1], constant term first. For arbitrary oriented endpoints a=-1,b=0,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-15/4, 828/1, 3297/4)","explanation":"Exact signed moments give (-15/4, 828/1, 3297/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0006","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cd0f20254a74b20254eef89c790e2ea95eb277094c07d2b1fd9ade19211c057","family_id":"qf-density-integral-additivity-triple","seed":202610096006,"semantic_template":"qf-density-integral-additivity-triple/[[-2,6,7,1],0,-6,-4]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [-2, 6, 7, 1], constant term first. For arbitrary oriented endpoints a=0,b=-6,c=-4, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-60/1, 92/3, -88/3)","explanation":"Exact signed moments give (-60/1, 92/3, -88/3); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0007","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bdc0079d28c6aa831a3cff39c63b514b30bcec913a498244ab9c809f377a1542","family_id":"qf-density-integral-additivity-triple","seed":202610096007,"semantic_template":"qf-density-integral-additivity-triple/[[-1,7,8,1],-2,1,-7]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [-1, 7, 8, 1], constant term first. For arbitrary oriented endpoints a=-2,b=1,c=-7, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(27/4, -424/3, -1615/12)","explanation":"Exact signed moments give (27/4, -424/3, -1615/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0008","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4644d1835e7df37b6be7d6d2495aa1c658108c9c044b7cc35a4ce71c13ffc0c1","family_id":"qf-density-integral-additivity-triple","seed":202610096008,"semantic_template":"qf-density-integral-additivity-triple/[[0,8,9,1],-1,-9,-5]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [0, 8, 9, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-9,c=-5, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-224/1, 104/1, -120/1)","explanation":"Exact signed moments give (-224/1, 104/1, -120/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0009","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"754fc383a0345928ae450ce6ca2aab5f04e0a688d5b0da1598c248002fec44b2","family_id":"qf-density-integral-additivity-triple","seed":202610096009,"semantic_template":"qf-density-integral-additivity-triple/[[1,9,10,1],0,1,6]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [1, 9, 10, 1], constant term first. For arbitrary oriented endpoints a=0,b=1,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(109/12, 14435/12, 1212/1)","explanation":"Exact signed moments give (109/12, 14435/12, 1212/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0010","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9155a7da6d57e8e554aceeaa640e818da08d559dddfab7f7f4e3a82f3350db6","family_id":"qf-density-integral-additivity-triple","seed":202610096010,"semantic_template":"qf-density-integral-additivity-triple/[[2,10,11,1],-2,-12,-23]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [2, 10, 11, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-12,c=-23, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-1340/3, 340835/12, 111825/4)","explanation":"Exact signed moments give (-1340/3, 340835/12, 111825/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0011","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0478604cc0e26e5c46306abd27bd7768d2594290402f17ca63d138935542f723","family_id":"qf-density-integral-additivity-triple","seed":202610096011,"semantic_template":"qf-density-integral-additivity-triple/[[3,11,12,1],-1,2,4]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [3, 11, 12, 1], constant term first. For arbitrary oriented endpoints a=-1,b=2,c=4, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(261/4, 356/1, 1685/4)","explanation":"Exact signed moments give (261/4, 356/1, 1685/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0012","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f8fd5f753547c4ae06e994b757eece79020fcb014cb9b91e37b9cbb6152ad10","family_id":"qf-density-integral-additivity-triple","seed":202610096012,"semantic_template":"qf-density-integral-additivity-triple/[[4,12,13,1],0,-12,-9]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [4, 12, 13, 1], constant term first. For arbitrary oriented endpoints a=0,b=-12,c=-9, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-1488/1, 1677/4, -4275/4)","explanation":"Exact signed moments give (-1488/1, 1677/4, -4275/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0013","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"07b0ee26585e22b0799bbba94a218788f2846c2a61f7dbabd4deaf1251531633","family_id":"qf-density-integral-additivity-triple","seed":202610096013,"semantic_template":"qf-density-integral-additivity-triple/[[5,13,14,1],-2,-1,-15]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [5, 13, 14, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-1,c=-15, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(173/12, -5110/3, -20267/12)","explanation":"Exact signed moments give (173/12, -5110/3, -20267/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0014","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b8e5c3c0588acdd87dbfbbfe952e2ba32bafabec37819d05d32681c9d63304d","family_id":"qf-density-integral-additivity-triple","seed":202610096014,"semantic_template":"qf-density-integral-additivity-triple/[[6,14,15,1],-1,-15,-10]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [6, 14, 15, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-15,c=-10, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-2730/1, 3495/4, -7425/4)","explanation":"Exact signed moments give (-2730/1, 3495/4, -7425/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0015","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5164e7f67a61acfe851acf8250fa2d7d776a714203101a33a28edf19bfac943","family_id":"qf-density-integral-additivity-triple","seed":202610096015,"semantic_template":"qf-density-integral-additivity-triple/[[7,15,16,1],0,3,9]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [7, 15, 16, 1], constant term first. For arbitrary oriented endpoints a=0,b=3,c=9, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(1011/4, 5946/1, 24795/4)","explanation":"Exact signed moments give (1011/4, 5946/1, 24795/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0016","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35a75fb5c5cdade6372ddfd92c15aafd3e0e005390d6aafbfbf5f04110f9106d","family_id":"qf-density-integral-additivity-triple","seed":202610096016,"semantic_template":"qf-density-integral-additivity-triple/[[8,16,17,1],-2,-18,-35]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [8, 16, 17, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-18,c=-35, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-12992/3, 1752887/12, 566973/4)","explanation":"Exact signed moments give (-12992/3, 1752887/12, 566973/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0017","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a599e062281341b9c80bb28a7bca6003336832e47ea6596166019767a9ee63c","family_id":"qf-density-integral-additivity-triple","seed":202610096017,"semantic_template":"qf-density-integral-additivity-triple/[[9,17,18,1],-1,0,3]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [9, 17, 18, 1], constant term first. For arbitrary oriented endpoints a=-1,b=0,c=3, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(25/4, 1143/4, 292/1)","explanation":"Exact signed moments give (25/4, 1143/4, 292/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0018","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68593b63961818ce8cc3169185a3e9c6f6dfccd3a4ce948d85f4b0962f13c481","family_id":"qf-density-integral-additivity-triple","seed":202610096018,"semantic_template":"qf-density-integral-additivity-triple/[[10,18,19,1],0,-18,-14]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [10, 18, 19, 1], constant term first. For arbitrary oriented endpoints a=0,b=-18,c=-14, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-7956/1, 5416/3, -18452/3)","explanation":"Exact signed moments give (-7956/1, 5416/3, -18452/3); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0019","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e8901cef99b08808b4cda2f1852e053e809f0879b797ff74ab4602a5b513727","family_id":"qf-density-integral-additivity-triple","seed":202610096019,"semantic_template":"qf-density-integral-additivity-triple/[[11,19,20,1],-2,1,-19]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [11, 19, 20, 1], constant term first. For arbitrary oriented endpoints a=-2,b=1,c=-19, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(243/4, -29860/3, -118711/12)","explanation":"Exact signed moments give (243/4, -29860/3, -118711/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0020","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2341efb974ebcfb07e72bbf7c32fec50964e79ba9adac83a35e9b2f29a45a5cf","family_id":"qf-density-integral-additivity-triple","seed":202610096020,"semantic_template":"qf-density-integral-additivity-triple/[[12,20,21,1],-1,-21,-15]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [12, 20, 21, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-21,c=-15, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-12040/1, 3150/1, -8890/1)","explanation":"Exact signed moments give (-12040/1, 3150/1, -8890/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0021","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff31af18647e0a7cd30f5b979985253c0ce7b46e72a93b075c9617cc6a89c542","family_id":"qf-density-integral-additivity-triple","seed":202610096021,"semantic_template":"qf-density-integral-additivity-triple/[[13,21,22,1],0,1,3]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [13, 21, 22, 1], constant term first. For arbitrary oriented endpoints a=0,b=1,c=3, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(373/12, 962/3, 1407/4)","explanation":"Exact signed moments give (373/12, 962/3, 1407/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0022","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ae5eee2b53807478d2dcdc9c4d77ed09f13cba0ee66597a15ef6eef9631fe9a","family_id":"qf-density-integral-additivity-triple","seed":202610096022,"semantic_template":"qf-density-integral-additivity-triple/[[14,22,23,1],-2,-24,-47]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [14, 22, 23, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-24,c=-47, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-50996/3, 5575499/12, 1790505/4)","explanation":"Exact signed moments give (-50996/3, 5575499/12, 1790505/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0023","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0289c96d559aa393e45b73d4ed308e0619c464e1715b99514736ebde727d59d3","family_id":"qf-density-integral-additivity-triple","seed":202610096023,"semantic_template":"qf-density-integral-additivity-triple/[[15,23,24,1],-1,2,6]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [15, 23, 24, 1], constant term first. For arbitrary oriented endpoints a=-1,b=2,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(621/4, 2412/1, 10269/4)","explanation":"Exact signed moments give (621/4, 2412/1, 10269/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0024","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0df43401979295629e628659e69f31c0c28a58131acf22dae841648b15318812","family_id":"qf-density-integral-additivity-triple","seed":202610096024,"semantic_template":"qf-density-integral-additivity-triple/[[16,24,25,1],0,-24,-19]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [16, 24, 25, 1], constant term first. For arbitrary oriented endpoints a=0,b=-24,c=-19, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(-25728/1, 62135/12, -246601/12)","explanation":"Exact signed moments give (-25728/1, 62135/12, -246601/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0025","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83bad69f97cc8f4047bc262d10834a6628dce057dc7fa31a2924c6ede2325a35","family_id":"qf-density-integral-additivity-triple","seed":202610096025,"semantic_template":"qf-density-integral-additivity-triple/[[17,25,26,1],-2,-1,-27]"},"fields":{"title":"Three Oriented Polynomial-Moment Integrals","stem":"On all R, f has ascending coefficients [17, 25, 26, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-1,c=-27, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(437/12, -87178/3, -348275/12)","explanation":"Exact signed moments give (437/12, -87178/3, -348275/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof."},"skills":["Three Oriented Polynomial-Moment Integrals"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0026","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3bc2f3b8768e13e7bf86683fb24eb24b0ec321ca1512885dd1727bd0a8044d5","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096026,"semantic_template":"qf-density-integral-parity-symmetric-pair/[\"odd_symmetric_zero\"]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [0, 2, 0, 1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(Yes, 0/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (Yes, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0027","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f5d6794260e816cdfd3409b84e4a72eb47bfeb070d1039dc3f0a9f11db949e4","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096027,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-3,0,1],3]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-3, 0, 1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 0/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0028","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81be2e2946c2b42a1ceaa670e31dacd9a59a4b9aecd4a436ab28e0e98683de1b","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096028,"semantic_template":"qf-density-integral-parity-symmetric-pair/[\"nonodd_zero_radius\"]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [1, 0, 2], constant term first. For r=0≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 0/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0029","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4795d019982ed6d12588636b84673ac5abaf347b15ef639e1dd30cdad98b850c","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096029,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-9,0,4],4]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-9, 4, 4, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 296/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 296/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0030","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c21d19590963e0d5a4d39465b0f5efdf0ace87eed9736896848ea26e060ea2f8","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096030,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-8,0,5],5]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-8, 1, 5, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 1010/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1010/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0031","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02bf8d54ede0b4447e8dba79e9b53d0f768b87f92b6f8db79b8ed60ebc412364","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096031,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-7,0,6],1]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-7, 2, 6, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, -10/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, -10/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0032","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fb8399f99eedb0c9f17053aacca3cd7432da4bfbba01d60d0c82ee0b7ccee91","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096032,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-6,0,7],2]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-6, 3, 7, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 40/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 40/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0033","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5b990a0f13b8bcfbbc2b1f88fd322ad7601ecee68473165753e5604f27d798f","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096033,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-5,0,8],3]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-5, 4, 8, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 114/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 114/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0034","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72783673db04647a87f40bd7f95fad3add6a14fce6247179f235667a6a36113c","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096034,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-4,0,9],4]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-4, 1, 9, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 352/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 352/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0035","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3a9d5c4e532ee7ba532f5d9d4bf631db7d8e0c2230f1251b76e4d0c4fcc4473","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096035,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-3,0,10],5]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-3, 2, 10, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 2410/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 2410/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0036","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81fcb09417bc7c029e021de6704f00c9bca0b9e6e48aa3d3091501a98bdcf48b","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096036,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-2,0,11],1]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-2, 3, 11, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 10/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 10/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0037","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38309fb8cddc4325b2a5f92fe4173c8e0a1b73337672ae8754284b8d5ba24019","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096037,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[-1,0,12],2]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [-1, 4, 12, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 60/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 60/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0038","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"867a211f98e61e7db018411b35fe9a179a99ab76adbf4331575750461610abb0","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096038,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[0,0,13],3]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [0, 1, 13, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 234/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 234/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0039","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c81397491837d4c5dab371639c2a72a734315682ab6b1edaaf201a09a41a0300","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096039,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[1,0,14],4]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [1, 2, 14, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 1816/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1816/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0040","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d16f9de48a85de1c363da30cc8601a37a7edc09cf58478a5c798ffd8f834747","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096040,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[2,0,15],5]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [2, 3, 15, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 1270/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1270/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0041","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d62a18f6dd2eef0c952070b46e0064d02812710e15b79153d2d4e5c2a658d5fb","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096041,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[3,0,16],1]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [3, 4, 16, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 50/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 50/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0042","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b40fdb877820a2c4de995dc4c1b37f941e6e0eed5de10cb053ee3a616bbd9d3","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096042,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[4,0,17],2]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [4, 1, 17, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 320/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 320/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0043","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"091ef32c03c739f090790bc4f76fc7c35cddf67ea127d812a5df95663aa43c1a","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096043,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[5,0,18],3]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [5, 2, 18, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 354/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 354/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0044","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1158f0361f32e8015567c4fb1e8ae19c4dbc91d92069a44d18f8f1506c91d742","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096044,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[6,0,19],4]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [6, 3, 19, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 2576/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 2576/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0045","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"106ecccf9f35c90341d3918b5da54b3207885eaf754bb88b0009f09003d6129e","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096045,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[7,0,20],5]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [7, 4, 20, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 5210/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 5210/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0046","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8aaabc45b52c2af1680127cc46d514338b69392fd310264e4e7b0aee62a7bdfe","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096046,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[8,0,21],1]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [8, 1, 21, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 30/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 30/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0047","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"630c59630f23ee9c8ee833f6cd8c7f6b0f026cd01fe2b561024a862117178eda","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096047,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[9,0,22],2]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [9, 2, 22, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 460/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 460/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0048","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef6cbf53476d955fef73d793a6bc922fe5edfc63d23b5796e057fb04c8d6a403","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096048,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[10,0,23],3]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [10, 3, 23, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 474/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 474/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0049","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"815e926eddda67535550481bb65084b5423a41979502d3210396cff4daf9f7ab","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096049,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[11,0,24],4]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [11, 4, 24, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 1112/1)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1112/1). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0050","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f318d1c4643f6c1d43412474aec03a69092819452ab1251331031c29999a6fdf","family_id":"qf-density-integral-parity-symmetric-pair","seed":202610096050,"semantic_template":"qf-density-integral-parity-symmetric-pair/[[12,0,25],5]"},"fields":{"title":"Oddness Eligibility and Symmetric Integral Pair","stem":"On all R, f has ascending coefficients [12, 1, 25, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"(No, 6610/3)","explanation":"Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 6610/3). A zero symmetric integral does not imply oddness, including the r=0 case."},"skills":["Oddness Eligibility and Symmetric Integral Pair"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0051","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"279bf5553802c24a6f492cc056c93e07f1563e0e441d3a74290199980546b042","family_id":"qf-density-integral-left-error","seed":202610096051,"semantic_template":"qf-density-integral-left-error/[\"zero_error\"]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(0)x+(2). On a=0≤b=2, use N=3 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=0/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0052","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8951bd89bae50bcab63895be0e40e51f4e1cebf2deca8ec3206fb67aa39faf65","family_id":"qf-density-integral-left-error","seed":202610096052,"semantic_template":"qf-density-integral-left-error/[-2,\"2\",1]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-2)x+(-11). On a=-1≤b=1, use N=1 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-4/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-4/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0053","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c25c9cac76b29536d5930dc59ea75df1b32b2fb8a36ca7e6610b4949c20ea3ca","family_id":"qf-density-integral-left-error","seed":202610096053,"semantic_template":"qf-density-integral-left-error/[3,\"3\",3]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(3)x+(-10). On a=0≤b=3, use N=3 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0054","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"826644ea16f2d0479fcdf4bda9007e652504f2f8ec27e2ddedd09b60d767f337","family_id":"qf-density-integral-left-error","seed":202610096054,"semantic_template":"qf-density-integral-left-error/[-4,\"4\",4]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-4)x+(-9). On a=1≤b=5, use N=4 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-8/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0055","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5be9a4cceef69efbf1a67382fae74f1de687b08e6da4f207848cc609f97ffad","family_id":"qf-density-integral-left-error","seed":202610096055,"semantic_template":"qf-density-integral-left-error/[5,\"5\",5]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(5)x+(-8). On a=-2≤b=3, use N=5 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"25/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0056","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ccae1ccfc18c85c1ed974b6559d89a6973c86af5454a6381ca4baa286079fa4","family_id":"qf-density-integral-left-error","seed":202610096056,"semantic_template":"qf-density-integral-left-error/[-6,\"6\",6]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-6)x+(-7). On a=-1≤b=5, use N=6 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-18/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0057","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac44eaa832b87c2de39171638d71ff4196ae86b6af5063ea8de7f30f0323d63c","family_id":"qf-density-integral-left-error","seed":202610096057,"semantic_template":"qf-density-integral-left-error/[7,\"1\",7]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(7)x+(-6). On a=0≤b=1, use N=7 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0058","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed7fbc4aa0a41d1c119bff38d0282b23d8611423c8f8d7a88c8f86bc98a03a15","family_id":"qf-density-integral-left-error","seed":202610096058,"semantic_template":"qf-density-integral-left-error/[-8,\"2\",8]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-8)x+(-5). On a=1≤b=3, use N=8 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0059","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"72b129697a908fba313ddf4705e46cf2d2f39ade62f146a759dd1c597c044d36","family_id":"qf-density-integral-left-error","seed":202610096059,"semantic_template":"qf-density-integral-left-error/[9,\"3\",9]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(9)x+(-4). On a=-2≤b=1, use N=9 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0060","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4fcbba20a948408f67b0bae7e2142374bf2d86d55039ff59a03e9f098d70709","family_id":"qf-density-integral-left-error","seed":202610096060,"semantic_template":"qf-density-integral-left-error/[-10,\"4\",10]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-10)x+(-3). On a=-1≤b=3, use N=10 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-8/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0061","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23ea6919486002bbeab3a414d81975452c51e558bb2ed2e37bb71eb29ac2bd11","family_id":"qf-density-integral-left-error","seed":202610096061,"semantic_template":"qf-density-integral-left-error/[11,\"5\",11]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(11)x+(-2). On a=0≤b=5, use N=11 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"25/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0062","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69cbbe87db2ad418c9e0792af83b486e5a4ca699672feaa6f55bc4adff42b31e","family_id":"qf-density-integral-left-error","seed":202610096062,"semantic_template":"qf-density-integral-left-error/[-12,\"6\",12]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-12)x+(-1). On a=1≤b=7, use N=12 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-18/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0063","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cf4030df9616b1539b5fc66b2f2a8db18c2c991601e8bbf79032ec471f24a45","family_id":"qf-density-integral-left-error","seed":202610096063,"semantic_template":"qf-density-integral-left-error/[13,\"1\",13]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(13)x+(0). On a=-2≤b=-1, use N=13 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0064","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ccb0832bb03b61ab58f07b1bbff14e21d0998d72c14565ab2ce0fbcec8d9082","family_id":"qf-density-integral-left-error","seed":202610096064,"semantic_template":"qf-density-integral-left-error/[-14,\"2\",14]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-14)x+(1). On a=-1≤b=1, use N=14 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0065","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15c99559b13a33ac6f5e69570410cf889db492f6598bedd458289a01ce3678a1","family_id":"qf-density-integral-left-error","seed":202610096065,"semantic_template":"qf-density-integral-left-error/[15,\"3\",15]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(15)x+(2). On a=0≤b=3, use N=15 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0066","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f02b1874afc95a1c7013873a6fb308e12ea2460d9ceff596b02c6676c2ab9245","family_id":"qf-density-integral-left-error","seed":202610096066,"semantic_template":"qf-density-integral-left-error/[-16,\"4\",16]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-16)x+(3). On a=1≤b=5, use N=16 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-8/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0067","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3f5777b92545634b0f562ddb755ab18dcd8691ec4599120f7d2f7fac776cf20","family_id":"qf-density-integral-left-error","seed":202610096067,"semantic_template":"qf-density-integral-left-error/[17,\"5\",17]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(17)x+(4). On a=-2≤b=3, use N=17 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"25/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0068","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6815460a2ff27be93a302373faff452fa80ab9bec6f90dfb11a311738c77fe4d","family_id":"qf-density-integral-left-error","seed":202610096068,"semantic_template":"qf-density-integral-left-error/[-18,\"6\",18]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-18)x+(5). On a=-1≤b=5, use N=18 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-18/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0069","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f2e6cdcee874df678569ecab03a4cb674fed5b4713667e5371b20fce985cbc9","family_id":"qf-density-integral-left-error","seed":202610096069,"semantic_template":"qf-density-integral-left-error/[19,\"1\",19]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(19)x+(6). On a=0≤b=1, use N=19 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0070","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d63214579246b52a4e5b8b38a27bd2a91fceec957f10d1989f1c8bb8beea56b1","family_id":"qf-density-integral-left-error","seed":202610096070,"semantic_template":"qf-density-integral-left-error/[-20,\"2\",20]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-20)x+(7). On a=1≤b=3, use N=20 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0071","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f78b389290b4c722060e58f98a03b77a0e5049e17726453210931554cf4c29d6","family_id":"qf-density-integral-left-error","seed":202610096071,"semantic_template":"qf-density-integral-left-error/[21,\"3\",21]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(21)x+(8). On a=-2≤b=1, use N=21 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0072","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"761ea8ae80dff03f021d8c6288d2f32e90f86a60cb87704530c229e7e226767f","family_id":"qf-density-integral-left-error","seed":202610096072,"semantic_template":"qf-density-integral-left-error/[-22,\"4\",22]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-22)x+(9). On a=-1≤b=3, use N=22 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-8/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0073","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9aee17b3258cf5156318a51faafbc2bf7b687a2867ab49ea062dcd71ff9094b9","family_id":"qf-density-integral-left-error","seed":202610096073,"semantic_template":"qf-density-integral-left-error/[23,\"5\",23]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(23)x+(10). On a=0≤b=5, use N=23 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"25/2","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0074","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"533867a2de579f93f453651fee91b24f05befe94d3d7f95b17375917fde7336a","family_id":"qf-density-integral-left-error","seed":202610096074,"semantic_template":"qf-density-integral-left-error/[-24,\"6\",24]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(-24)x+(11). On a=1≤b=7, use N=24 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-18/1","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0075","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9d5dd056a03791cfe650b3b96d8afcde27f070e51be8c93db8ef551efbc068e","family_id":"qf-density-integral-left-error","seed":202610096075,"semantic_template":"qf-density-integral-left-error/[25,\"1\",50]"},"fields":{"title":"Exact Moment Integral Minus a Finite Left Sum","stem":"On all R, f(x)=(25)x+(12). On a=-2≤b=-1, use N=50 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"1/4","explanation":"Compute the polynomial moment and finite left sum separately, giving I−S_N=1/4. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral."},"skills":["Exact Moment Integral Minus a Finite Left Sum"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0076","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a1afddaaa5f469968d713c677f84bc6b440bf1db694c5294c527a2ab2a924fe","family_id":"qf-density-integral-endpoint-recovery","seed":202610096076,"semantic_template":"qf-density-integral-endpoint-recovery/[\"zero_target\",3]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=1≠0. Starting at a=3, its oriented integral has target 0/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"3/1","explanation":"Solve c(b−a)=target to obtain b=3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0077","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4fc36593c4da357b4f7b88f7f17e7f519baf686cdf7c94e32ca05b9af008cde7","family_id":"qf-density-integral-endpoint-recovery","seed":202610096077,"semantic_template":"qf-density-integral-endpoint-recovery/[-2,-11,\"-4/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-2≠0. Starting at a=-11, its oriented integral has target -8/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-9/1","explanation":"Solve c(b−a)=target to obtain b=-9/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0078","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f112ec16d98a9d962acdf07d00f7e2889d6972610e40778b8206544d4ec3f39","family_id":"qf-density-integral-endpoint-recovery","seed":202610096078,"semantic_template":"qf-density-integral-endpoint-recovery/[3,-10,\"9/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=3≠0. Starting at a=-10, its oriented integral has target 27/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-7/1","explanation":"Solve c(b−a)=target to obtain b=-7/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0079","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc3d2c65ff804172c8029c2f497bf3966638ef62e49e7b10690f672b1ec697d2","family_id":"qf-density-integral-endpoint-recovery","seed":202610096079,"semantic_template":"qf-density-integral-endpoint-recovery/[-4,-9,\"16/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-4≠0. Starting at a=-9, its oriented integral has target 16/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0080","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c88c2897305b12de74f9cd3681c7ba81a10bbf646d06c9b7b0e7a18c1aab28f","family_id":"qf-density-integral-endpoint-recovery","seed":202610096080,"semantic_template":"qf-density-integral-endpoint-recovery/[5,-8,\"25/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=5≠0. Starting at a=-8, its oriented integral has target 50/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-3/1","explanation":"Solve c(b−a)=target to obtain b=-3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0081","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00410655fd22254f2a733ee49c8e0da62485f1ef7e6787fbaba1779aacf8d037","family_id":"qf-density-integral-endpoint-recovery","seed":202610096081,"semantic_template":"qf-density-integral-endpoint-recovery/[-6,-7,\"-36/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-6≠0. Starting at a=-7, its oriented integral has target -108/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-1/1","explanation":"Solve c(b−a)=target to obtain b=-1/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0082","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0d0e7e2ec230d4086d540ffbaa1aefb2a2a62a1d398a759847607fd723b3ce2","family_id":"qf-density-integral-endpoint-recovery","seed":202610096082,"semantic_template":"qf-density-integral-endpoint-recovery/[7,-6,\"-49/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=7≠0. Starting at a=-6, its oriented integral has target -49/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0083","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a846efce7fafe2786ec00d70cd84396ea9b29b12045d43428fcbb7cbbedde17","family_id":"qf-density-integral-endpoint-recovery","seed":202610096083,"semantic_template":"qf-density-integral-endpoint-recovery/[-8,-5,\"-64/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-8≠0. Starting at a=-5, its oriented integral has target -128/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"3/1","explanation":"Solve c(b−a)=target to obtain b=3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0084","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"923fc39e1b95085262707f120e240f308bd9be2eddce9dfc7c37e2e73e7ebf7d","family_id":"qf-density-integral-endpoint-recovery","seed":202610096084,"semantic_template":"qf-density-integral-endpoint-recovery/[9,-4,\"81/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=9≠0. Starting at a=-4, its oriented integral has target 243/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"5/1","explanation":"Solve c(b−a)=target to obtain b=5/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0085","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8478376c63a3da976211ff88855a773c4666466273b8611147fd4c98bd47b70c","family_id":"qf-density-integral-endpoint-recovery","seed":202610096085,"semantic_template":"qf-density-integral-endpoint-recovery/[-10,-3,\"100/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-10≠0. Starting at a=-3, its oriented integral has target 100/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0086","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06897b526f29f0d8bd01af9ae985f4aca5eae29fbe6bd2d5e5ecd89975d654af","family_id":"qf-density-integral-endpoint-recovery","seed":202610096086,"semantic_template":"qf-density-integral-endpoint-recovery/[11,-2,\"121/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=11≠0. Starting at a=-2, its oriented integral has target 242/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"9/1","explanation":"Solve c(b−a)=target to obtain b=9/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0087","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9aa96e293d60b4a487909c51d904a1983da31e30ccab71c29a7929629ad4c9af","family_id":"qf-density-integral-endpoint-recovery","seed":202610096087,"semantic_template":"qf-density-integral-endpoint-recovery/[-12,-1,\"-144/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-12≠0. Starting at a=-1, its oriented integral has target -432/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"11/1","explanation":"Solve c(b−a)=target to obtain b=11/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0088","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19d8da2243a6a93addff2f552239e3f11529a0a5b6698fd2a31df9d49da2e185","family_id":"qf-density-integral-endpoint-recovery","seed":202610096088,"semantic_template":"qf-density-integral-endpoint-recovery/[13,0,\"-169/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=13≠0. Starting at a=0, its oriented integral has target -169/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0089","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6147460c500acd64df93e7fede9a34f7ec3e6c8cdcc9a88d2e2c8e20b99338bd","family_id":"qf-density-integral-endpoint-recovery","seed":202610096089,"semantic_template":"qf-density-integral-endpoint-recovery/[-14,1,\"-196/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-14≠0. Starting at a=1, its oriented integral has target -392/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"15/1","explanation":"Solve c(b−a)=target to obtain b=15/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0090","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fcec96b45a037d58339a4ed535ad9144f89d2c56c85a87e95bd27d95be504fa6","family_id":"qf-density-integral-endpoint-recovery","seed":202610096090,"semantic_template":"qf-density-integral-endpoint-recovery/[15,2,\"225/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=15≠0. Starting at a=2, its oriented integral has target 675/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"17/1","explanation":"Solve c(b−a)=target to obtain b=17/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0091","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2064ccc50e1bebfa1ad52b2224b96a934ef1d8f28c318307c3c74cfd3bb279ee","family_id":"qf-density-integral-endpoint-recovery","seed":202610096091,"semantic_template":"qf-density-integral-endpoint-recovery/[-16,3,\"256/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-16≠0. Starting at a=3, its oriented integral has target 256/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0092","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a94b3bd07765a4eb67cec4fdc742255d13f27eafb9e96d2c78dfc91ba7dabc51","family_id":"qf-density-integral-endpoint-recovery","seed":202610096092,"semantic_template":"qf-density-integral-endpoint-recovery/[17,4,\"289/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=17≠0. Starting at a=4, its oriented integral has target 578/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"21/1","explanation":"Solve c(b−a)=target to obtain b=21/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0093","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24be2be51c2457326e66c7078296b47555c49b8ed481fe9f7fb2ab7072dd9093","family_id":"qf-density-integral-endpoint-recovery","seed":202610096093,"semantic_template":"qf-density-integral-endpoint-recovery/[-18,5,\"-324/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-18≠0. Starting at a=5, its oriented integral has target -972/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"23/1","explanation":"Solve c(b−a)=target to obtain b=23/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0094","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"43b31c38e39b78b6f08c71aa2a1901175e08baf35238e101f6478ba87e9e2aea","family_id":"qf-density-integral-endpoint-recovery","seed":202610096094,"semantic_template":"qf-density-integral-endpoint-recovery/[19,6,\"-361/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=19≠0. Starting at a=6, its oriented integral has target -361/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0095","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c558cd291be7afeb03785aae60677496a1c3b96bc78e07afc2b447446b71d77","family_id":"qf-density-integral-endpoint-recovery","seed":202610096095,"semantic_template":"qf-density-integral-endpoint-recovery/[-20,7,\"-400/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-20≠0. Starting at a=7, its oriented integral has target -800/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"27/1","explanation":"Solve c(b−a)=target to obtain b=27/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0096","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93e4e0d3fe5473623a67075b2dd5d2ee65c0e07808c2967137510a371d08ff48","family_id":"qf-density-integral-endpoint-recovery","seed":202610096096,"semantic_template":"qf-density-integral-endpoint-recovery/[21,8,\"441/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=21≠0. Starting at a=8, its oriented integral has target 1323/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"29/1","explanation":"Solve c(b−a)=target to obtain b=29/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0097","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d61f2c7dc7907720135c0a08cebac77776817752321cf8d22d129011c234f86","family_id":"qf-density-integral-endpoint-recovery","seed":202610096097,"semantic_template":"qf-density-integral-endpoint-recovery/[-22,9,\"484/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-22≠0. Starting at a=9, its oriented integral has target 484/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0098","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1433deec31b2411e0bbc941ffa84788f606bcfac9c8ea092c4dae393bfed2b98","family_id":"qf-density-integral-endpoint-recovery","seed":202610096098,"semantic_template":"qf-density-integral-endpoint-recovery/[23,10,\"529/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=23≠0. Starting at a=10, its oriented integral has target 1058/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"33/1","explanation":"Solve c(b−a)=target to obtain b=33/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0099","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"057d5284381419de880103b6562ec3a15c1d7d34d02498df9e3ed26e1fd0c3c7","family_id":"qf-density-integral-endpoint-recovery","seed":202610096099,"semantic_template":"qf-density-integral-endpoint-recovery/[-24,11,\"-576/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=-24≠0. Starting at a=11, its oriented integral has target -1728/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"35/1","explanation":"Solve c(b−a)=target to obtain b=35/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261009-r06-definite-integral-0100","batch_id":"batch-20261007-advance-20261009-r06-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70487747c78c996e72927866d1646c4fc257af7848d43c9c157fec3ea88d98fd","family_id":"qf-density-integral-endpoint-recovery","seed":202610096100,"semantic_template":"qf-density-integral-endpoint-recovery/[25,12,\"-625/1\"]"},"fields":{"title":"Integer Endpoint Recovered from a Nonzero Constant Integral","stem":"On all R, the constant integrand is c=25≠0. Starting at a=12, its oriented integral has target -625/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.","answer":{"kind":"expression","value":"-13/1","explanation":"Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction."},"skills":["Integer Endpoint Recovered from a Nonzero Constant Integral"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}