{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261007-advance-20261008-r07-definite-integral-0001","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf819ef966d6de80238630fc617dc823204f0b650e481645a7403a58e3a506c7","family_id":"qf-integral-constant","seed":202610087001,"semantic_template":"qf-integral-constant/{\"inputs\":[\"zero_integral\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=3. Use the scoped signed polynomial moments to find I(f,2,2), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"The constant moment c(b−a) gives 0/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0002","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58295147b4ff66c6524479be75da965ffd3ecb91a26caa656d8f781b0f578910","family_id":"qf-integral-constant","seed":202610087002,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-1\",\"2\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-1. Use the scoped signed polynomial moments to find I(f,-11,-9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/1","explanation":"The constant moment c(b−a) gives -2/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0003","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2555f222485436486ba6fcae3d064b4fd6c2004b33f96c7ac9b400e6fb9f0369","family_id":"qf-integral-constant","seed":202610087003,"semantic_template":"qf-integral-constant/{\"inputs\":[\"2\",\"3\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=2. Use the scoped signed polynomial moments to find I(f,-10,-7), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"6/1","explanation":"The constant moment c(b−a) gives 6/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0004","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c732a6afafcc8f5271102825c0722eefe3449e20064e7e2de4280304d5c0f2f5","family_id":"qf-integral-constant","seed":202610087004,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-3\",\"-4\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-3. Use the scoped signed polynomial moments to find I(f,-9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"12/1","explanation":"The constant moment c(b−a) gives 12/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0005","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"109d77e8e5aed43de89e14ea01ebe27498ae044b43dca1181d9fe562b54e3b10","family_id":"qf-integral-constant","seed":202610087005,"semantic_template":"qf-integral-constant/{\"inputs\":[\"4\",\"5\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=4. Use the scoped signed polynomial moments to find I(f,-8,-3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"20/1","explanation":"The constant moment c(b−a) gives 20/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0006","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71dc23be01e6b455d58e32f566ccd85d5f825007c9f2faa3168301780b05076b","family_id":"qf-integral-constant","seed":202610087006,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-5\",\"6\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-5. Use the scoped signed polynomial moments to find I(f,-7,-1), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-30/1","explanation":"The constant moment c(b−a) gives -30/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0007","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b64395cb7e6b4577b6149523fda2e355e07dc054c60342b9f0d8f556fa518f6","family_id":"qf-integral-constant","seed":202610087007,"semantic_template":"qf-integral-constant/{\"inputs\":[\"6\",\"-7\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=6. Use the scoped signed polynomial moments to find I(f,-6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-42/1","explanation":"The constant moment c(b−a) gives -42/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0008","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc017529dd6e9250cde7eded9c720a303596b97c762f0d59fead536ed2d8a4e4","family_id":"qf-integral-constant","seed":202610087008,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-7\",\"8\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-7. Use the scoped signed polynomial moments to find I(f,-5,3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-56/1","explanation":"The constant moment c(b−a) gives -56/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0009","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d8751dcf4ef4a97dad84657af38700c18148bd5dc2bf56c26efccf706505394","family_id":"qf-integral-constant","seed":202610087009,"semantic_template":"qf-integral-constant/{\"inputs\":[\"8\",\"9\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=8. Use the scoped signed polynomial moments to find I(f,-4,5), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"72/1","explanation":"The constant moment c(b−a) gives 72/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0010","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"653eb432853ffe7e7f8ed7ff95977db0aea9a91cfaeb12afd0df8625f4cf5869","family_id":"qf-integral-constant","seed":202610087010,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-9\",\"-10\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-9. Use the scoped signed polynomial moments to find I(f,-3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"90/1","explanation":"The constant moment c(b−a) gives 90/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0011","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7d2093af6715e94f6bffec6631d138e094b1cfbd151584902e55912f63c3a0a","family_id":"qf-integral-constant","seed":202610087011,"semantic_template":"qf-integral-constant/{\"inputs\":[\"10\",\"11\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=10. Use the scoped signed polynomial moments to find I(f,-2,9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"110/1","explanation":"The constant moment c(b−a) gives 110/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0012","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a3b40d8927f38d07c7e809c71fd9bca9f758851f810d107fe1903cf9ce9a29c","family_id":"qf-integral-constant","seed":202610087012,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-11\",\"12\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-11. Use the scoped signed polynomial moments to find I(f,-1,11), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-132/1","explanation":"The constant moment c(b−a) gives -132/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0013","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a52fa992d599dd1e0213f90f5156a62b245c63d7e712be982b63d79ff28b5d8","family_id":"qf-integral-constant","seed":202610087013,"semantic_template":"qf-integral-constant/{\"inputs\":[\"12\",\"-13\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=12. Use the scoped signed polynomial moments to find I(f,0,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-156/1","explanation":"The constant moment c(b−a) gives -156/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0014","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f702cec37eaa9c19c8b48315cc49a3468df0db258b95962e5ee976998e82ae9","family_id":"qf-integral-constant","seed":202610087014,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-13\",\"14\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-13. Use the scoped signed polynomial moments to find I(f,1,15), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-182/1","explanation":"The constant moment c(b−a) gives -182/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0015","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"268159fbf4f3e6763f9ec849ad1c885cbff604f279dc6ccdd6fa81b40c8d7bf0","family_id":"qf-integral-constant","seed":202610087015,"semantic_template":"qf-integral-constant/{\"inputs\":[\"14\",\"15\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=14. Use the scoped signed polynomial moments to find I(f,2,17), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"210/1","explanation":"The constant moment c(b−a) gives 210/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0016","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03005b39bb06959874d5feaac81b2c148d32d5becd82b85f60332be0deb2023f","family_id":"qf-integral-constant","seed":202610087016,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-15\",\"-16\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-15. Use the scoped signed polynomial moments to find I(f,3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"240/1","explanation":"The constant moment c(b−a) gives 240/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0017","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f3d9d5b322e57f9113715725b385bc35cfc788ac971e25d09ca70270a0483f9","family_id":"qf-integral-constant","seed":202610087017,"semantic_template":"qf-integral-constant/{\"inputs\":[\"16\",\"17\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=16. Use the scoped signed polynomial moments to find I(f,4,21), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"272/1","explanation":"The constant moment c(b−a) gives 272/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0018","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0f14577e39c4adf41dab05ffcc9e494e007c8f67d9c64ea6eee6ceb63a3b4c8","family_id":"qf-integral-constant","seed":202610087018,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-17\",\"18\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-17. Use the scoped signed polynomial moments to find I(f,5,23), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-306/1","explanation":"The constant moment c(b−a) gives -306/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0019","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d18caf1273289d417eaee836ee69684dba999584deba747474101fdc7bb5eed8","family_id":"qf-integral-constant","seed":202610087019,"semantic_template":"qf-integral-constant/{\"inputs\":[\"18\",\"-19\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=18. Use the scoped signed polynomial moments to find I(f,6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-342/1","explanation":"The constant moment c(b−a) gives -342/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0020","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e6a3d08465abb5a9d3a2ff5be84adee53cbba2edc87d6629931e09785dd788b","family_id":"qf-integral-constant","seed":202610087020,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-19\",\"20\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-19. Use the scoped signed polynomial moments to find I(f,7,27), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-380/1","explanation":"The constant moment c(b−a) gives -380/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0021","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"898ea2f9ebe2ce629cae3ffcff9bfcc7042cc16bd994afb4793c2e76d8c58007","family_id":"qf-integral-constant","seed":202610087021,"semantic_template":"qf-integral-constant/{\"inputs\":[\"20\",\"21\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=20. Use the scoped signed polynomial moments to find I(f,8,29), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"420/1","explanation":"The constant moment c(b−a) gives 420/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0022","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d03cfefec0887204e2988cbe7cf8540a5bda732cff5f15e35461a758b4fd6371","family_id":"qf-integral-constant","seed":202610087022,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-21\",\"-22\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-21. Use the scoped signed polynomial moments to find I(f,9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"462/1","explanation":"The constant moment c(b−a) gives 462/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0023","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a709f0a266bdaf5b2e888da1e9661665eba027e787b7a1864d56ca4242ff0627","family_id":"qf-integral-constant","seed":202610087023,"semantic_template":"qf-integral-constant/{\"inputs\":[\"22\",\"23\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=22. Use the scoped signed polynomial moments to find I(f,10,33), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"506/1","explanation":"The constant moment c(b−a) gives 506/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0024","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9d8a0621e827da2d3846e0bf673b18a874edd9c6336342872684db38abfd93a","family_id":"qf-integral-constant","seed":202610087024,"semantic_template":"qf-integral-constant/{\"inputs\":[\"-23\",\"24\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=-23. Use the scoped signed polynomial moments to find I(f,11,35), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-552/1","explanation":"The constant moment c(b−a) gives -552/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0025","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"18ebbfc0bb780d67ff3b63ae547fac6a290d083e3b8c27cbcfa88d1bf514faaf","family_id":"qf-integral-constant","seed":202610087025,"semantic_template":"qf-integral-constant/{\"inputs\":[\"24\",\"-25\"],\"operation\":\"constant_integral\"}"},"fields":{"title":"Signed Moment Integral of a Constant","stem":"On all R, f(x)=24. Use the scoped signed polynomial moments to find I(f,12,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-600/1","explanation":"The constant moment c(b−a) gives -600/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof."},"skills":["Signed Moment Integral of a Constant"],"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-20261008-r07-definite-integral-0026","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7526e9a179276897cd6320fa40b97b77937849e863e4804fec94eb7e2d3d90a","family_id":"qf-integral-monomial","seed":202610087026,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"zero_integral\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0027","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bfc592dd51482ba6e301570ff60b12992576f345668ca02ef95f195528bee8e5","family_id":"qf-integral-monomial","seed":202610087027,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-2\",1,\"-2\",\"0\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"4/1","explanation":"The restricted polynomial moment gives 4/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0028","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7152bab4c55984aeb44199b62aa75cba9858f0d8719084529324e752fc942423","family_id":"qf-integral-monomial","seed":202610087028,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"3\",2,\"-3\",\"1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(3)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"28/1","explanation":"The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0029","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f06fb2037c71e54f0d16dc9439a7ad952f668010f9477dcea10c76d137024371","family_id":"qf-integral-monomial","seed":202610087029,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-4\",3,\"2\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-4)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"15/1","explanation":"The restricted polynomial moment gives 15/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0030","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1c25d9f0cc4dc52bb89b62b876212bf3d35fc1da378870662800e0e1cd71eea","family_id":"qf-integral-monomial","seed":202610087030,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"5\",4,\"-2\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(5)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"31/1","explanation":"The restricted polynomial moment gives 31/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0031","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dc28c8ec0f6f4ddd1e20850d35c58b73746015d16b89be82110a2e66f1b0d7f","family_id":"qf-integral-monomial","seed":202610087031,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-6\",5,\"-3\",\"0\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-6)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"729/1","explanation":"The restricted polynomial moment gives 729/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0032","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12bb49553540c862f3fd9d03bd33049a391f1b07b2b82a244dc0d933a422db91","family_id":"qf-integral-monomial","seed":202610087032,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"7\",\"constant_width\",\"-2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(7)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-14/1","explanation":"The restricted polynomial moment gives -14/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0033","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3adbd308e6fd7722ef77a183f180d801332cd19cd7da533651f45060ca4ddd4b","family_id":"qf-integral-monomial","seed":202610087033,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-8\",1,\"-2\",\"2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-8)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0034","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8dac256874ec9e538132c9e83a301432881c80d87c854e51e7a9c0861a808c80","family_id":"qf-integral-monomial","seed":202610087034,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"9\",2,\"-3\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(9)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"78/1","explanation":"The restricted polynomial moment gives 78/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0035","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46c030e9cfb56c2cb29703739c35f14450c6d995d75d42c5c26dcc86ec3c2f50","family_id":"qf-integral-monomial","seed":202610087035,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-10\",3,\"0\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-10)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-5/2","explanation":"The restricted polynomial moment gives -5/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0036","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d4b6747f9cc81d0587911a7ee941463d91a09112932c77e743bfaa3e3c4ceaa","family_id":"qf-integral-monomial","seed":202610087036,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"11\",4,\"-2\",\"1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(11)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"363/5","explanation":"The restricted polynomial moment gives 363/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0037","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b87740f8c735cb8bb88f83bff50380e70f36db84f54eecf5c7d8bfc2ed4673f","family_id":"qf-integral-monomial","seed":202610087037,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-12\",5,\"-3\",\"2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-12)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"1330/1","explanation":"The restricted polynomial moment gives 1330/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0038","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4be430bb2d68f2b269b0748b436a6cbc725e6ea051f4de2314e12b97d1fe8fd","family_id":"qf-integral-monomial","seed":202610087038,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"13\",\"constant_width\",\"-2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(13)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-26/1","explanation":"The restricted polynomial moment gives -26/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0039","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d087648253b68ff05d2b4110de25834519dac890060434c937b7babfe13e935","family_id":"qf-integral-monomial","seed":202610087039,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-14\",1,\"-2\",\"0\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-14)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"28/1","explanation":"The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0040","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1ecd5737d25218518b0c0fdb7772fca1fb913efea3e8ac6d4212d608e169c73","family_id":"qf-integral-monomial","seed":202610087040,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"15\",2,\"-3\",\"1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(15)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"140/1","explanation":"The restricted polynomial moment gives 140/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0041","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fae7346ee9db78f8e955ed4e9da46dc75a7675775e7af075aa6c0659460e8bb3","family_id":"qf-integral-monomial","seed":202610087041,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-16\",3,\"2\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-16)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"60/1","explanation":"The restricted polynomial moment gives 60/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0042","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be18b58dfddf90a347070b3306591280baffd9b355beede668cd9b29c21abfb9","family_id":"qf-integral-monomial","seed":202610087042,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"17\",4,\"-2\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(17)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"527/5","explanation":"The restricted polynomial moment gives 527/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0043","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f3967bf87e2c3bc2cf81184af6152e6c3ae3c78fa845cfdf14a25bd3cea53a1","family_id":"qf-integral-monomial","seed":202610087043,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-18\",5,\"-3\",\"0\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-18)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"2187/1","explanation":"The restricted polynomial moment gives 2187/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0044","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"779b29edcea20f6498eb84fbb6828f7f34ccdd83e9b15d5e3ef0e229788560c0","family_id":"qf-integral-monomial","seed":202610087044,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"19\",\"constant_width\",\"-2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(19)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-38/1","explanation":"The restricted polynomial moment gives -38/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0045","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"830971748c7f985a204e9519bada56cf699cb50513d8903f4274577bb009b661","family_id":"qf-integral-monomial","seed":202610087045,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-20\",1,\"-2\",\"2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-20)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0046","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"555a361d82485950bf2729f2e36fc5783ce23e30669d94c902f9760385153475","family_id":"qf-integral-monomial","seed":202610087046,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"21\",2,\"-3\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(21)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"182/1","explanation":"The restricted polynomial moment gives 182/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0047","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88aba4c8b38ffbfdb825fdea44b3648d287c04ea16c8fd2c128834d63ea53ba1","family_id":"qf-integral-monomial","seed":202610087047,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-22\",3,\"0\",\"-1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-22)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-11/2","explanation":"The restricted polynomial moment gives -11/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0048","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d0cd5215f322ffbfcc55e678b0afebfc6874febce93b896f2623e04a4a9bfb7b","family_id":"qf-integral-monomial","seed":202610087048,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"23\",4,\"-2\",\"1\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(23)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"759/5","explanation":"The restricted polynomial moment gives 759/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0049","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a9d63508b0d36ee80e90022fc0a258125e3464ee0406e1a3425ad4f41f8111df","family_id":"qf-integral-monomial","seed":202610087049,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"-24\",5,\"-3\",\"2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(-24)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"2660/1","explanation":"The restricted polynomial moment gives 2660/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0050","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3323dd233b076b70db7a75c832759daaa378419902076a6f61f955b7b06bfed3","family_id":"qf-integral-monomial","seed":202610087050,"semantic_template":"qf-integral-monomial/{\"inputs\":[\"25\",\"constant_width\",\"-2\"],\"operation\":\"monomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Monomial","stem":"On all R, f(x)=(25)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-50/1","explanation":"The restricted polynomial moment gives -50/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim."},"skills":["Signed Moment Integral of a Monomial"],"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-20261008-r07-definite-integral-0051","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f5437c58aa478589cae058c5d3606d7f7fee9cae3e85cc08327f8c2f3f59201","family_id":"qf-integral-polynomial","seed":202610087051,"semantic_template":"qf-integral-polynomial/{\"inputs\":[\"zero_integral\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=0, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 0/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0052","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29c5757d40f1a5a9c29a9a62ec23cf654cd6f2d59f35ea5a6ace63f1d781f5ff","family_id":"qf-integral-polynomial","seed":202610087052,"semantic_template":"qf-integral-polynomial/{\"inputs\":[\"3\",\"constant_width\",\"-4\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-12/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -12/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0053","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae7140b63dd3c823dac3e06250ac89f5411b732ab2542bc28cf8b0dceb8fc067","family_id":"qf-integral-polynomial","seed":202610087053,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-10\",\"3\",\"1\",\"3\"],\"-3\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-10, 3, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-42/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -42/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0054","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5ddf5bbfe78d477a115f0b1a5cefb11a22feb904911d023b3b82be9a76d11e0","family_id":"qf-integral-polynomial","seed":202610087054,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-9\",\"4\",\"-1\",\"4\"],\"4\",\"-1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-9, 4, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-655/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -655/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0055","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"237f3031d934b2fa4719adc140fcb9255f6c086cb9e60d6340afa7b9cb290601","family_id":"qf-integral-polynomial","seed":202610087055,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-8\",\"5\",\"0\",\"1\"],\"-2\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-8, 5, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-141/4","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -141/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0056","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f0e370ba8de23f05e4180d967bc82ea32673a5d1d16ee541305dfec02576355","family_id":"qf-integral-polynomial","seed":202610087056,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-7\",\"6\",\"1\",\"2\"],\"2\",\"-3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-7, 6, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"425/6","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 425/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0057","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e64e00a5b5e170db3fe924d05961127ed5596626d88347d475193342b8b19e53","family_id":"qf-integral-polynomial","seed":202610087057,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-6\",\"7\",\"-1\",\"3\"],\"-1\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-6, 7, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"164/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 164/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0058","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1285dae124a2a0ee413895e478590759da53dd676ca0668148b1bf47d718903","family_id":"qf-integral-polynomial","seed":202610087058,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-5\",\"8\",\"0\",\"4\"],\"4\",\"-2\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-5, 8, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-258/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -258/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0059","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4de55ac195a4ccd09ced7428b49004bee3627b94f7b6a53498e91f80e8165ee6","family_id":"qf-integral-polynomial","seed":202610087059,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-4\",\"9\",\"1\",\"1\"],\"-3\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-4, 9, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-188/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0060","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"517ffc8d053c9068295f40325a82335929883638bc4f2e72b0ec14e6525bafca","family_id":"qf-integral-polynomial","seed":202610087060,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-3\",\"10\",\"-1\",\"2\"],\"2\",\"-1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-3, 10, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-21/2","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -21/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0061","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d930299db3b887533eb7b09a18c57f60d50cec8017f995f67ef0a0abcbdfaff8","family_id":"qf-integral-polynomial","seed":202610087061,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-2\",\"11\",\"0\",\"3\"],\"-2\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-2, 11, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"265/4","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 265/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0062","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26e9eb6a4a0cce5d1ba0d7ae65de4cc6867c4d02e82f2de92e0f307c1d8e2b20","family_id":"qf-integral-polynomial","seed":202610087062,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"-1\",\"12\",\"1\",\"4\"],\"4\",\"-3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [-1, 12, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-721/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -721/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0063","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cf64cadfd90d7591e82ea125c9cd7c00bc72358b8d0616419963a74c417f1bf","family_id":"qf-integral-polynomial","seed":202610087063,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"0\",\"13\",\"-1\",\"1\"],\"-1\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [0, 13, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-2/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -2/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0064","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3de51121bb40b40910d85b80d2f3e05fb635ee34fc79f06bd7361fd6c93e0d3","family_id":"qf-integral-polynomial","seed":202610087064,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"1\",\"14\",\"0\",\"2\"],\"2\",\"-2\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [1, 14, 0, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-4/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -4/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0065","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e82c11a04bfb737cd983106b106ab42da63cba4e615b028c93a16b15f1fbb9d","family_id":"qf-integral-polynomial","seed":202610087065,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"2\",\"15\",\"1\",\"3\"],\"-3\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [2, 15, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"30/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 30/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0066","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad6879d89b14f77accb257013639d40dec848fe92f6b6a1d2d04a581476b2fb8","family_id":"qf-integral-polynomial","seed":202610087066,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"3\",\"16\",\"-1\",\"4\"],\"4\",\"-1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [3, 16, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-1105/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1105/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0067","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68d3eb723640ef1a48610afe7f93586f2e77eeeaf16325c762e975c7ad812d50","family_id":"qf-integral-polynomial","seed":202610087067,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"4\",\"17\",\"0\",\"1\"],\"-2\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [4, 17, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-69/4","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -69/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0068","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cca4497d8766d3fa12ec019cd74e810de0f8b7e28daae56c3b1aa143d8eef4d","family_id":"qf-integral-polynomial","seed":202610087068,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"5\",\"18\",\"1\",\"2\"],\"2\",\"-3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [5, 18, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"245/6","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 245/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0069","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e57a62dad09f374b4729e71cd1ed153767414e05b97ab8f871dffec9e1d5f4a","family_id":"qf-integral-polynomial","seed":202610087069,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"6\",\"19\",\"-1\",\"3\"],\"-1\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [6, 19, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"452/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 452/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0070","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13bfd40a54ff45da15714497c39a7236473599013bab83889bb463851e8acc24","family_id":"qf-integral-polynomial","seed":202610087070,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"7\",\"20\",\"0\",\"4\"],\"4\",\"-2\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [7, 20, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-402/1","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -402/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0071","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"262972d4ab9aa10744e0e21c90cacbde179404d4cc73ce4120fd46608982a336","family_id":"qf-integral-polynomial","seed":202610087071,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"8\",\"21\",\"1\",\"1\"],\"-3\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [8, 21, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-188/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0072","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc88f0997e9a99d945be66dacd3f092b55d461e85ba584006e71b4563806bd53","family_id":"qf-integral-polynomial","seed":202610087072,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"9\",\"22\",\"-1\",\"2\"],\"2\",\"-1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [9, 22, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-129/2","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -129/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0073","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e34221adb858b8c1b06252a4cd9f4e0933ce60d43eab31e76a64067a5443a164","family_id":"qf-integral-polynomial","seed":202610087073,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"10\",\"23\",\"0\",\"3\"],\"-2\",\"3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [10, 23, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"625/4","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 625/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0074","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c9674d48e61597373e9f7d2e65ae44b5403ab261653d37ca5e54fdc6e3d671d","family_id":"qf-integral-polynomial","seed":202610087074,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"11\",\"24\",\"1\",\"4\"],\"4\",\"-3\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [11, 24, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-1099/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1099/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0075","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a919c50b83ccd72cd445f776718e538a2181e22777bee0d92d788096662ec506","family_id":"qf-integral-polynomial","seed":202610087075,"semantic_template":"qf-integral-polynomial/{\"inputs\":[[\"12\",\"25\",\"-1\",\"1\"],\"-1\",\"1\"],\"operation\":\"polynomial_integral\"}"},"fields":{"title":"Signed Moment Integral of a Polynomial","stem":"On all R, f has ascending coefficients [12, 25, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"70/3","explanation":"Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 70/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge."},"skills":["Signed Moment Integral of a Polynomial"],"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-20261008-r07-definite-integral-0076","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5eeb4c63bb7b86c8954e55965156554583fc76e68995c315949b0eae58f5d18a","family_id":"qf-integral-affine-left-sum","seed":202610087076,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"2\",\"1\",\"2\",\"2\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(2)x+(1). For a=0≤b=2 and N=2 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"4/1","explanation":"Exact termwise rational accumulation gives the finite sum 4/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0077","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b43ec6ad2e646602b81539407f64f5bd368f7c6bb47a660034cee52c65ebbdb1","family_id":"qf-integral-affine-left-sum","seed":202610087077,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"constant_sum\",\"2\",\"3\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(0)x+(2). For a=-1≤b=2 and N=3 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"6/1","explanation":"Exact termwise rational accumulation gives the finite sum 6/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0078","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"77326bc31ef06dc8984ce4d23eeb81b13f7479cff321ea42791d6d49b907bbae","family_id":"qf-integral-affine-left-sum","seed":202610087078,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"zero_width\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(3)x+(-2). For a=1≤b=1 and N=4 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0079","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9224277f4715ab82d19d0956f75f91c031c37e3392cf6df743abe611b5f98688","family_id":"qf-integral-affine-left-sum","seed":202610087079,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-4\",\"-13\",\"4\",\"1\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-4)x+(-9). For a=1≤b=5 and N=1 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-52/1","explanation":"Exact termwise rational accumulation gives the finite sum -52/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0080","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96291a5a695a8bef6f503dce67ba904db6956af32c5da22383175e3112eadbda","family_id":"qf-integral-affine-left-sum","seed":202610087080,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"5\",\"-18\",\"5\",\"5\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(5)x+(-8). For a=-2≤b=3 and N=5 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-40/1","explanation":"Exact termwise rational accumulation gives the finite sum -40/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0081","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53e802f2dd37c75f4c69b79020268eeb92870f69867cccc23f3bb6b348940977","family_id":"qf-integral-affine-left-sum","seed":202610087081,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-6\",\"-1\",\"1\",\"6\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-6)x+(-7). For a=-1≤b=0 and N=6 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-7/2","explanation":"Exact termwise rational accumulation gives the finite sum -7/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0082","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2059638995ce48575dbf8b3479b61c3cc77bdb4ce7c8596361a2106d249d3b7","family_id":"qf-integral-affine-left-sum","seed":202610087082,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"7\",\"-6\",\"2\",\"7\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(7)x+(-6). For a=0≤b=2 and N=7 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"0/1","explanation":"Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0083","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfef484ab45b031bc1360b938528f4f71f9f93feb7bebbecfa00f382414966a2","family_id":"qf-integral-affine-left-sum","seed":202610087083,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-8\",\"-13\",\"3\",\"8\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-8)x+(-5). For a=1≤b=4 and N=8 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-141/2","explanation":"Exact termwise rational accumulation gives the finite sum -141/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0084","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb8b456ab5e0fd95c2cf363a3103a377295180db790cd221acdba797b6ae942c","family_id":"qf-integral-affine-left-sum","seed":202610087084,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"9\",\"-22\",\"4\",\"9\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(9)x+(-4). For a=-2≤b=2 and N=9 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-24/1","explanation":"Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0085","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e507e4fb996a8ff43ad261cf9e389b29d71fd4f0d5a6b3a143e16f56d0eb9aa","family_id":"qf-integral-affine-left-sum","seed":202610087085,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-10\",\"7\",\"5\",\"10\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-10)x+(-3). For a=-1≤b=4 and N=10 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-155/2","explanation":"Exact termwise rational accumulation gives the finite sum -155/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0086","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"44cf62a4747511a781330dfdc137e15ff7ce435c4a467f69cda6c261c1ec1463","family_id":"qf-integral-affine-left-sum","seed":202610087086,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"11\",\"-2\",\"1\",\"11\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(11)x+(-2). For a=0≤b=1 and N=11 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"3/1","explanation":"Exact termwise rational accumulation gives the finite sum 3/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0087","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"315565daf75f0628d68c6cfd071cdf40e3738538974de907a959f5c61e4ad82a","family_id":"qf-integral-affine-left-sum","seed":202610087087,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-12\",\"-13\",\"2\",\"12\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-12)x+(-1). For a=1≤b=3 and N=12 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-48/1","explanation":"Exact termwise rational accumulation gives the finite sum -48/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0088","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4f27fcdfa7bc9511ddab7bb646a3509be40d3747c458d8e7c8be2a938ea4670","family_id":"qf-integral-affine-left-sum","seed":202610087088,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"13\",\"-26\",\"3\",\"13\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(13)x+(0). For a=-2≤b=1 and N=13 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-24/1","explanation":"Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0089","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0501f764fdaf6c3df40cd5e4fd5baf89e13e6770d4db08bb1de8c8bbd6ec89f7","family_id":"qf-integral-affine-left-sum","seed":202610087089,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-14\",\"15\",\"4\",\"14\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-14)x+(1). For a=-1≤b=3 and N=14 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-44/1","explanation":"Exact termwise rational accumulation gives the finite sum -44/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0090","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fb249717fabbfe8d8377e882c7948c3f8162b8ebcfcd68633ed8bb93d855192","family_id":"qf-integral-affine-left-sum","seed":202610087090,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"15\",\"2\",\"5\",\"15\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(15)x+(2). For a=0≤b=5 and N=15 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"185/1","explanation":"Exact termwise rational accumulation gives the finite sum 185/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0091","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9a84f25c527ac5010e9dcd44b32efadbc4f85d284dd8ddda8478fd07c5272ec","family_id":"qf-integral-affine-left-sum","seed":202610087091,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-16\",\"-13\",\"1\",\"16\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-16)x+(3). For a=1≤b=2 and N=16 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-41/2","explanation":"Exact termwise rational accumulation gives the finite sum -41/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0092","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ccb86d43855f63a92a4455926f5ca4f79fcba77a9c714cddeffcf333dc43c26","family_id":"qf-integral-affine-left-sum","seed":202610087092,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"17\",\"-30\",\"2\",\"17\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(17)x+(4). For a=-2≤b=0 and N=17 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-28/1","explanation":"Exact termwise rational accumulation gives the finite sum -28/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0093","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a81542d6a870b1aea324c4a72c8a7f6826ee1d0d04ff9e5d2b6e5309157efcda","family_id":"qf-integral-affine-left-sum","seed":202610087093,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-18\",\"23\",\"3\",\"18\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-18)x+(5). For a=-1≤b=2 and N=18 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-15/2","explanation":"Exact termwise rational accumulation gives the finite sum -15/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0094","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7151cc4dc90f381e155bf7ac4b88ae5b81f90b79d1191b60ce2b70ba1029670","family_id":"qf-integral-affine-left-sum","seed":202610087094,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"19\",\"6\",\"4\",\"19\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(19)x+(6). For a=0≤b=4 and N=19 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"168/1","explanation":"Exact termwise rational accumulation gives the finite sum 168/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0095","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd3a0bba868ac23a4f04c6da74e3dddb32f9399f9dbb5e50c7f47c7695bdd14c","family_id":"qf-integral-affine-left-sum","seed":202610087095,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-20\",\"-13\",\"5\",\"20\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-20)x+(7). For a=1≤b=6 and N=20 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-605/2","explanation":"Exact termwise rational accumulation gives the finite sum -605/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0096","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2ccfb3dbc5a5208680e13b7d97417f460ab3b96f68da9362e5328eb7566f694","family_id":"qf-integral-affine-left-sum","seed":202610087096,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"21\",\"-34\",\"1\",\"21\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(21)x+(8). For a=-2≤b=-1 and N=21 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-24/1","explanation":"Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0097","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dbf2ec2ff8e843428932c9e9c706428a4b5a4139f4631aec3784df1021f5934","family_id":"qf-integral-affine-left-sum","seed":202610087097,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-22\",\"31\",\"2\",\"22\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-22)x+(9). For a=-1≤b=1 and N=22 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"20/1","explanation":"Exact termwise rational accumulation gives the finite sum 20/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0098","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04c281ea72c040033cd5ebf3e5e9d4b09fe1ee97eb4a2857256423fd4adedda6","family_id":"qf-integral-affine-left-sum","seed":202610087098,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"23\",\"10\",\"3\",\"23\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(23)x+(10). For a=0≤b=3 and N=23 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"129/1","explanation":"Exact termwise rational accumulation gives the finite sum 129/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0099","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"277bd8619d17823b71393f4770726c7b7b74536e60b6d6b02420b9d49c772df8","family_id":"qf-integral-affine-left-sum","seed":202610087099,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"-24\",\"-13\",\"4\",\"24\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(-24)x+(11). For a=1≤b=5 and N=24 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"-236/1","explanation":"Exact termwise rational accumulation gives the finite sum -236/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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-20261008-r07-definite-integral-0100","batch_id":"batch-20261007-advance-20261008-r07-definite-integral-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1664eca412eac2416d1eb84b1244342c91944c03ac1be08b95e49b7a3305df39","family_id":"qf-integral-affine-left-sum","seed":202610087100,"semantic_template":"qf-integral-affine-left-sum/{\"inputs\":[\"25\",\"-38\",\"5\",\"50\"],\"operation\":\"affine_left_sum\"}"},"fields":{"title":"Finite Left-Endpoint Sum of an Affine Function","stem":"On all R, f(x)=(25)x+(12). For a=-2≤b=3 and N=50 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.","answer":{"kind":"expression","value":"465/4","explanation":"Exact termwise rational accumulation gives the finite sum 465/4. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof."},"skills":["Finite Left-Endpoint Sum of an Affine Function"],"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."}}