# Oriented Integral, Parity, and Finite-Sum Error Depth

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0001/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0002/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0003/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0004/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0005/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0006/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0007/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0008/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0009/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0010/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0011/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0012/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0013/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0014/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0015/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0016/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0017/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0018/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0019/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0020/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0021/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0022/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0023/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0024/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0025/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0026/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0027/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0028/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0029/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0030/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0031/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0032/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0033/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0034/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0035/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0036/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0037/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0038/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0039/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0040/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0041/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0042/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0043/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0044/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0045/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0046/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0047/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0048/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0049/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0050/)

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 explanation0/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0051/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0052/)

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 explanation9/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0053/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0054/)

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 explanation25/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0055/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0056/)

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 explanation1/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0057/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0058/)

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 explanation9/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0059/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0060/)

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 explanation25/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0061/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0062/)

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 explanation1/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0063/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0064/)

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 explanation9/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0065/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0066/)

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 explanation25/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0067/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0068/)

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 explanation1/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0069/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0070/)

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 explanation9/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0071/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0072/)

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 explanation25/2

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0073/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0074/)

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 explanation1/4

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0075/)

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 explanation3/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0076/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0077/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0078/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0079/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0080/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0081/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0082/)

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 explanation3/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0083/)

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 explanation5/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0084/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0085/)

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.

View answer and explanation9/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0086/)

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.

View answer and explanation11/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0087/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0088/)

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.

View answer and explanation15/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0089/)

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.

View answer and explanation17/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0090/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0091/)

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.

View answer and explanation21/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0092/)

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.

View answer and explanation23/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0093/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0094/)

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 explanation27/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0095/)

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.

View answer and explanation29/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0096/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0097/)

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.

View answer and explanation33/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0098/)

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 explanation35/1

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0099/)

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](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r06-definite-integral-0100/)

Signed polynomial definite integrals on finite intervals · qf-density-integral-endpoint-recovery · candidate



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