Oriented Integral, Parity, and Finite-Sum Error Depth
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batch-20261007-advance-20261009-r06-definite-integral-foundations · 2026-10-07 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [0], constant term first. For arbitrary oriented endpoints a=0,b=1,c=2, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(0/1, 0/1, 0/1)
Exact signed moments give (0/1, 0/1, 0/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
2. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [3], constant term first. For arbitrary oriented endpoints a=3,b=-1,c=2, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-12/1, 9/1, -3/1)
Exact signed moments give (-12/1, 9/1, -3/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
3. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [1, 2, 3], constant term first. For arbitrary oriented endpoints a=0,b=2,c=0, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(14/1, -14/1, 0/1)
Exact signed moments give (14/1, -14/1, 0/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
4. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [-4, 4, 5, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-6,c=-11, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(160/3, 20015/12, 6885/4)
Exact signed moments give (160/3, 20015/12, 6885/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
5. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [-3, 5, 6, 1], constant term first. For arbitrary oriented endpoints a=-1,b=0,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-15/4, 828/1, 3297/4)
Exact signed moments give (-15/4, 828/1, 3297/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
6. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [-2, 6, 7, 1], constant term first. For arbitrary oriented endpoints a=0,b=-6,c=-4, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-60/1, 92/3, -88/3)
Exact signed moments give (-60/1, 92/3, -88/3); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
7. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [-1, 7, 8, 1], constant term first. For arbitrary oriented endpoints a=-2,b=1,c=-7, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(27/4, -424/3, -1615/12)
Exact signed moments give (27/4, -424/3, -1615/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
8. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [0, 8, 9, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-9,c=-5, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-224/1, 104/1, -120/1)
Exact signed moments give (-224/1, 104/1, -120/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
9. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [1, 9, 10, 1], constant term first. For arbitrary oriented endpoints a=0,b=1,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(109/12, 14435/12, 1212/1)
Exact signed moments give (109/12, 14435/12, 1212/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
10. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [2, 10, 11, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-12,c=-23, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-1340/3, 340835/12, 111825/4)
Exact signed moments give (-1340/3, 340835/12, 111825/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
11. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [3, 11, 12, 1], constant term first. For arbitrary oriented endpoints a=-1,b=2,c=4, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(261/4, 356/1, 1685/4)
Exact signed moments give (261/4, 356/1, 1685/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
12. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [4, 12, 13, 1], constant term first. For arbitrary oriented endpoints a=0,b=-12,c=-9, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-1488/1, 1677/4, -4275/4)
Exact signed moments give (-1488/1, 1677/4, -4275/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
13. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [5, 13, 14, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-1,c=-15, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(173/12, -5110/3, -20267/12)
Exact signed moments give (173/12, -5110/3, -20267/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
14. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [6, 14, 15, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-15,c=-10, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-2730/1, 3495/4, -7425/4)
Exact signed moments give (-2730/1, 3495/4, -7425/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
15. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [7, 15, 16, 1], constant term first. For arbitrary oriented endpoints a=0,b=3,c=9, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(1011/4, 5946/1, 24795/4)
Exact signed moments give (1011/4, 5946/1, 24795/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
16. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [8, 16, 17, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-18,c=-35, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-12992/3, 1752887/12, 566973/4)
Exact signed moments give (-12992/3, 1752887/12, 566973/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
17. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [9, 17, 18, 1], constant term first. For arbitrary oriented endpoints a=-1,b=0,c=3, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(25/4, 1143/4, 292/1)
Exact signed moments give (25/4, 1143/4, 292/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
18. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [10, 18, 19, 1], constant term first. For arbitrary oriented endpoints a=0,b=-18,c=-14, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-7956/1, 5416/3, -18452/3)
Exact signed moments give (-7956/1, 5416/3, -18452/3); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
19. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [11, 19, 20, 1], constant term first. For arbitrary oriented endpoints a=-2,b=1,c=-19, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(243/4, -29860/3, -118711/12)
Exact signed moments give (243/4, -29860/3, -118711/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
20. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [12, 20, 21, 1], constant term first. For arbitrary oriented endpoints a=-1,b=-21,c=-15, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-12040/1, 3150/1, -8890/1)
Exact signed moments give (-12040/1, 3150/1, -8890/1); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
21. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [13, 21, 22, 1], constant term first. For arbitrary oriented endpoints a=0,b=1,c=3, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(373/12, 962/3, 1407/4)
Exact signed moments give (373/12, 962/3, 1407/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
22. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [14, 22, 23, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-24,c=-47, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-50996/3, 5575499/12, 1790505/4)
Exact signed moments give (-50996/3, 5575499/12, 1790505/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
23. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [15, 23, 24, 1], constant term first. For arbitrary oriented endpoints a=-1,b=2,c=6, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(621/4, 2412/1, 10269/4)
Exact signed moments give (621/4, 2412/1, 10269/4); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
24. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [16, 24, 25, 1], constant term first. For arbitrary oriented endpoints a=0,b=-24,c=-19, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(-25728/1, 62135/12, -246601/12)
Exact signed moments give (-25728/1, 62135/12, -246601/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
25. Three Oriented Polynomial-Moment Integrals
On all R, f has ascending coefficients [17, 25, 26, 1], constant term first. For arbitrary oriented endpoints a=-2,b=-1,c=-27, give (I(f,a,b),I(f,b,c),I(f,a,c)) using the scoped polynomial moments. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(437/12, -87178/3, -348275/12)
Exact signed moments give (437/12, -87178/3, -348275/12); the first two sum to the third, retaining orientation even when the middle endpoint lies outside the other interval. This is not a new general integral proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-additivity-triple · candidate
26. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [0, 2, 0, 1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(Yes, 0/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (Yes, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
27. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-3, 0, 1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 0/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
28. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [1, 0, 2], constant term first. For r=0≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 0/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 0/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
29. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-9, 4, 4, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 296/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 296/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
30. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-8, 1, 5, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 1010/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1010/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
31. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-7, 2, 6, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, -10/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, -10/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
32. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-6, 3, 7, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 40/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 40/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
33. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-5, 4, 8, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 114/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 114/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
34. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-4, 1, 9, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 352/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 352/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
35. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-3, 2, 10, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 2410/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 2410/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
36. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-2, 3, 11, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 10/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 10/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
37. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [-1, 4, 12, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 60/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 60/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
38. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [0, 1, 13, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 234/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 234/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
39. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [1, 2, 14, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 1816/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1816/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
40. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [2, 3, 15, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 1270/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1270/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
41. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [3, 4, 16, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 50/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 50/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
42. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [4, 1, 17, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 320/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 320/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
43. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [5, 2, 18, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 354/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 354/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
44. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [6, 3, 19, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 2576/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 2576/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
45. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [7, 4, 20, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 5210/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 5210/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
46. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [8, 1, 21, -1], constant term first. For r=1≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 30/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 30/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
47. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [9, 2, 22, -1], constant term first. For r=2≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 460/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 460/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
48. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [10, 3, 23, -1], constant term first. For r=3≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 474/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 474/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
49. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [11, 4, 24, -1], constant term first. For r=4≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 1112/1)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 1112/1). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
50. Oddness Eligibility and Symmetric Integral Pair
On all R, f has ascending coefficients [12, 1, 25, -1], constant term first. For r=5≥0, give (is f odd?,I(f,−r,r)) as Yes/No followed by its exact signed polynomial moment. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
(No, 6610/3)
Oddness requires all even-power coefficients to be zero; the scoped symmetric integral is computed separately. The pair is (No, 6610/3). A zero symmetric integral does not imply oddness, including the r=0 case.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-parity-symmetric-pair · candidate
51. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(0)x+(2). On a=0≤b=2, use N=3 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
0/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=0/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
52. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-2)x+(-11). On a=-1≤b=1, use N=1 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-4/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-4/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
53. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(3)x+(-10). On a=0≤b=3, use N=3 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
9/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
54. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-4)x+(-9). On a=1≤b=5, use N=4 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-8/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
55. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(5)x+(-8). On a=-2≤b=3, use N=5 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
25/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
56. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-6)x+(-7). On a=-1≤b=5, use N=6 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-18/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
57. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(7)x+(-6). On a=0≤b=1, use N=7 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
1/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
58. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-8)x+(-5). On a=1≤b=3, use N=8 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-2/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
59. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(9)x+(-4). On a=-2≤b=1, use N=9 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
9/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
60. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-10)x+(-3). On a=-1≤b=3, use N=10 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-8/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
61. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(11)x+(-2). On a=0≤b=5, use N=11 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
25/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
62. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-12)x+(-1). On a=1≤b=7, use N=12 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-18/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
63. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(13)x+(0). On a=-2≤b=-1, use N=13 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
1/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
64. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-14)x+(1). On a=-1≤b=1, use N=14 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-2/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
65. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(15)x+(2). On a=0≤b=3, use N=15 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
9/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
66. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-16)x+(3). On a=1≤b=5, use N=16 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-8/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
67. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(17)x+(4). On a=-2≤b=3, use N=17 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
25/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
68. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-18)x+(5). On a=-1≤b=5, use N=18 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-18/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
69. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(19)x+(6). On a=0≤b=1, use N=19 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
1/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=1/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
70. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-20)x+(7). On a=1≤b=3, use N=20 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-2/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-2/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
71. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(21)x+(8). On a=-2≤b=1, use N=21 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
9/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=9/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
72. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-22)x+(9). On a=-1≤b=3, use N=22 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-8/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-8/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
73. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(23)x+(10). On a=0≤b=5, use N=23 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
25/2
Compute the polynomial moment and finite left sum separately, giving I−S_N=25/2. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
74. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(-24)x+(11). On a=1≤b=7, use N=24 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-18/1
Compute the polynomial moment and finite left sum separately, giving I−S_N=-18/1. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
75. Exact Moment Integral Minus a Finite Left Sum
On all R, f(x)=(25)x+(12). On a=-2≤b=-1, use N=50 uniform left cells. Find the signed error I−S_N, exact moment integral minus the finite left sum. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
1/4
Compute the polynomial moment and finite left sum separately, giving I−S_N=1/4. The error may be negative; equality in a constant or zero-width case does not make every finite sum an integral.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-left-error · candidate
76. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=1≠0. Starting at a=3, its oriented integral has target 0/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
3/1
Solve c(b−a)=target to obtain b=3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
77. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-2≠0. Starting at a=-11, its oriented integral has target -8/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-9/1
Solve c(b−a)=target to obtain b=-9/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
78. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=3≠0. Starting at a=-10, its oriented integral has target 27/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-7/1
Solve c(b−a)=target to obtain b=-7/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
79. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-4≠0. Starting at a=-9, its oriented integral has target 16/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
80. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=5≠0. Starting at a=-8, its oriented integral has target 50/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-3/1
Solve c(b−a)=target to obtain b=-3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
81. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-6≠0. Starting at a=-7, its oriented integral has target -108/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-1/1
Solve c(b−a)=target to obtain b=-1/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
82. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=7≠0. Starting at a=-6, its oriented integral has target -49/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
83. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-8≠0. Starting at a=-5, its oriented integral has target -128/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
3/1
Solve c(b−a)=target to obtain b=3/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
84. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=9≠0. Starting at a=-4, its oriented integral has target 243/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
5/1
Solve c(b−a)=target to obtain b=5/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
85. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-10≠0. Starting at a=-3, its oriented integral has target 100/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
86. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=11≠0. Starting at a=-2, its oriented integral has target 242/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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9/1
Solve c(b−a)=target to obtain b=9/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
87. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-12≠0. Starting at a=-1, its oriented integral has target -432/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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11/1
Solve c(b−a)=target to obtain b=11/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
88. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=13≠0. Starting at a=0, its oriented integral has target -169/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
89. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-14≠0. Starting at a=1, its oriented integral has target -392/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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15/1
Solve c(b−a)=target to obtain b=15/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
90. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=15≠0. Starting at a=2, its oriented integral has target 675/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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17/1
Solve c(b−a)=target to obtain b=17/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
91. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-16≠0. Starting at a=3, its oriented integral has target 256/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
92. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=17≠0. Starting at a=4, its oriented integral has target 578/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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21/1
Solve c(b−a)=target to obtain b=21/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
93. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-18≠0. Starting at a=5, its oriented integral has target -972/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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23/1
Solve c(b−a)=target to obtain b=23/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
94. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=19≠0. Starting at a=6, its oriented integral has target -361/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
95. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-20≠0. Starting at a=7, its oriented integral has target -800/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
27/1
Solve c(b−a)=target to obtain b=27/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
96. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=21≠0. Starting at a=8, its oriented integral has target 1323/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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29/1
Solve c(b−a)=target to obtain b=29/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
97. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-22≠0. Starting at a=9, its oriented integral has target 484/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
98. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=23≠0. Starting at a=10, its oriented integral has target 1058/2. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
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33/1
Solve c(b−a)=target to obtain b=33/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
99. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=-24≠0. Starting at a=11, its oriented integral has target -1728/3. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
35/1
Solve c(b−a)=target to obtain b=35/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
100. Integer Endpoint Recovered from a Nonzero Constant Integral
On all R, the constant integrand is c=25≠0. Starting at a=12, its oriented integral has target -625/1. Recover b, which is promised to remain an integer endpoint in this slice. Use reduced positive-denominator fractions, retaining /1 for integers.
View answer and explanation
-13/1
Solve c(b−a)=target to obtain b=-13/1. The nonzero constant permits division; do not take an absolute value or discard endpoint direction.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate
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