# Signed Polynomial Integral and Finite Left Sum Foundations

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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-20261008-r07-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. Signed Moment Integral of a Constant

On all R, f(x)=3. Use the scoped signed polynomial moments to find I(f,2,2), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

The constant moment c(b−a) gives 0/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0001/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 2. Signed Moment Integral of a Constant

On all R, f(x)=-1. Use the scoped signed polynomial moments to find I(f,-11,-9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-2/1

The constant moment c(b−a) gives -2/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0002/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 3. Signed Moment Integral of a Constant

On all R, f(x)=2. Use the scoped signed polynomial moments to find I(f,-10,-7), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation6/1

The constant moment c(b−a) gives 6/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0003/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 4. Signed Moment Integral of a Constant

On all R, f(x)=-3. Use the scoped signed polynomial moments to find I(f,-9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation12/1

The constant moment c(b−a) gives 12/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0004/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 5. Signed Moment Integral of a Constant

On all R, f(x)=4. Use the scoped signed polynomial moments to find I(f,-8,-3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation20/1

The constant moment c(b−a) gives 20/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0005/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 6. Signed Moment Integral of a Constant

On all R, f(x)=-5. Use the scoped signed polynomial moments to find I(f,-7,-1), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-30/1

The constant moment c(b−a) gives -30/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0006/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 7. Signed Moment Integral of a Constant

On all R, f(x)=6. Use the scoped signed polynomial moments to find I(f,-6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-42/1

The constant moment c(b−a) gives -42/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0007/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 8. Signed Moment Integral of a Constant

On all R, f(x)=-7. Use the scoped signed polynomial moments to find I(f,-5,3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-56/1

The constant moment c(b−a) gives -56/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0008/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 9. Signed Moment Integral of a Constant

On all R, f(x)=8. Use the scoped signed polynomial moments to find I(f,-4,5), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation72/1

The constant moment c(b−a) gives 72/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0009/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 10. Signed Moment Integral of a Constant

On all R, f(x)=-9. Use the scoped signed polynomial moments to find I(f,-3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation90/1

The constant moment c(b−a) gives 90/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0010/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 11. Signed Moment Integral of a Constant

On all R, f(x)=10. Use the scoped signed polynomial moments to find I(f,-2,9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation110/1

The constant moment c(b−a) gives 110/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0011/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 12. Signed Moment Integral of a Constant

On all R, f(x)=-11. Use the scoped signed polynomial moments to find I(f,-1,11), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-132/1

The constant moment c(b−a) gives -132/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0012/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 13. Signed Moment Integral of a Constant

On all R, f(x)=12. Use the scoped signed polynomial moments to find I(f,0,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-156/1

The constant moment c(b−a) gives -156/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0013/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 14. Signed Moment Integral of a Constant

On all R, f(x)=-13. Use the scoped signed polynomial moments to find I(f,1,15), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-182/1

The constant moment c(b−a) gives -182/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0014/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 15. Signed Moment Integral of a Constant

On all R, f(x)=14. Use the scoped signed polynomial moments to find I(f,2,17), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation210/1

The constant moment c(b−a) gives 210/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0015/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 16. Signed Moment Integral of a Constant

On all R, f(x)=-15. Use the scoped signed polynomial moments to find I(f,3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation240/1

The constant moment c(b−a) gives 240/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0016/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 17. Signed Moment Integral of a Constant

On all R, f(x)=16. Use the scoped signed polynomial moments to find I(f,4,21), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation272/1

The constant moment c(b−a) gives 272/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0017/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 18. Signed Moment Integral of a Constant

On all R, f(x)=-17. Use the scoped signed polynomial moments to find I(f,5,23), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-306/1

The constant moment c(b−a) gives -306/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0018/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 19. Signed Moment Integral of a Constant

On all R, f(x)=18. Use the scoped signed polynomial moments to find I(f,6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-342/1

The constant moment c(b−a) gives -342/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0019/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 20. Signed Moment Integral of a Constant

On all R, f(x)=-19. Use the scoped signed polynomial moments to find I(f,7,27), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-380/1

The constant moment c(b−a) gives -380/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0020/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 21. Signed Moment Integral of a Constant

On all R, f(x)=20. Use the scoped signed polynomial moments to find I(f,8,29), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation420/1

The constant moment c(b−a) gives 420/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0021/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 22. Signed Moment Integral of a Constant

On all R, f(x)=-21. Use the scoped signed polynomial moments to find I(f,9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation462/1

The constant moment c(b−a) gives 462/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0022/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 23. Signed Moment Integral of a Constant

On all R, f(x)=22. Use the scoped signed polynomial moments to find I(f,10,33), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation506/1

The constant moment c(b−a) gives 506/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0023/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 24. Signed Moment Integral of a Constant

On all R, f(x)=-23. Use the scoped signed polynomial moments to find I(f,11,35), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-552/1

The constant moment c(b−a) gives -552/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0024/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 25. Signed Moment Integral of a Constant

On all R, f(x)=24. Use the scoped signed polynomial moments to find I(f,12,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-600/1

The constant moment c(b−a) gives -600/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0025/)

Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate





## 26. Signed Moment Integral of a Monomial

On all R, f(x)=(2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0026/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 27. Signed Moment Integral of a Monomial

On all R, f(x)=(-2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation4/1

The restricted polynomial moment gives 4/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0027/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 28. Signed Moment Integral of a Monomial

On all R, f(x)=(3)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation28/1

The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0028/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 29. Signed Moment Integral of a Monomial

On all R, f(x)=(-4)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation15/1

The restricted polynomial moment gives 15/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0029/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 30. Signed Moment Integral of a Monomial

On all R, f(x)=(5)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation31/1

The restricted polynomial moment gives 31/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0030/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 31. Signed Moment Integral of a Monomial

On all R, f(x)=(-6)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation729/1

The restricted polynomial moment gives 729/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0031/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 32. Signed Moment Integral of a Monomial

On all R, f(x)=(7)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-14/1

The restricted polynomial moment gives -14/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0032/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 33. Signed Moment Integral of a Monomial

On all R, f(x)=(-8)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0033/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 34. Signed Moment Integral of a Monomial

On all R, f(x)=(9)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation78/1

The restricted polynomial moment gives 78/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0034/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 35. Signed Moment Integral of a Monomial

On all R, f(x)=(-10)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-5/2

The restricted polynomial moment gives -5/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0035/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 36. Signed Moment Integral of a Monomial

On all R, f(x)=(11)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation363/5

The restricted polynomial moment gives 363/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0036/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 37. Signed Moment Integral of a Monomial

On all R, f(x)=(-12)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation1330/1

The restricted polynomial moment gives 1330/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0037/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 38. Signed Moment Integral of a Monomial

On all R, f(x)=(13)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-26/1

The restricted polynomial moment gives -26/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0038/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 39. Signed Moment Integral of a Monomial

On all R, f(x)=(-14)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation28/1

The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0039/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 40. Signed Moment Integral of a Monomial

On all R, f(x)=(15)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation140/1

The restricted polynomial moment gives 140/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0040/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 41. Signed Moment Integral of a Monomial

On all R, f(x)=(-16)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation60/1

The restricted polynomial moment gives 60/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0041/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 42. Signed Moment Integral of a Monomial

On all R, f(x)=(17)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation527/5

The restricted polynomial moment gives 527/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0042/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 43. Signed Moment Integral of a Monomial

On all R, f(x)=(-18)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation2187/1

The restricted polynomial moment gives 2187/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0043/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 44. Signed Moment Integral of a Monomial

On all R, f(x)=(19)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-38/1

The restricted polynomial moment gives -38/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0044/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 45. Signed Moment Integral of a Monomial

On all R, f(x)=(-20)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0045/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 46. Signed Moment Integral of a Monomial

On all R, f(x)=(21)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation182/1

The restricted polynomial moment gives 182/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0046/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 47. Signed Moment Integral of a Monomial

On all R, f(x)=(-22)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-11/2

The restricted polynomial moment gives -11/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0047/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 48. Signed Moment Integral of a Monomial

On all R, f(x)=(23)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation759/5

The restricted polynomial moment gives 759/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0048/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 49. Signed Moment Integral of a Monomial

On all R, f(x)=(-24)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation2660/1

The restricted polynomial moment gives 2660/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0049/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 50. Signed Moment Integral of a Monomial

On all R, f(x)=(25)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-50/1

The restricted polynomial moment gives -50/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0050/)

Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate





## 51. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=0, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 0/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0051/)

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





## 52. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-12/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -12/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0052/)

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





## 53. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-10, 3, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-42/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -42/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0053/)

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





## 54. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-9, 4, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-655/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -655/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0054/)

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





## 55. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-8, 5, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-141/4

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -141/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0055/)

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





## 56. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-7, 6, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation425/6

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 425/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0056/)

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





## 57. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-6, 7, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation164/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 164/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0057/)

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





## 58. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-5, 8, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-258/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -258/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0058/)

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





## 59. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-4, 9, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-188/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0059/)

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





## 60. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-3, 10, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-21/2

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -21/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0060/)

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





## 61. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-2, 11, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation265/4

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 265/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0061/)

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





## 62. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [-1, 12, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-721/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -721/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0062/)

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





## 63. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [0, 13, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-2/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -2/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0063/)

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





## 64. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [1, 14, 0, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-4/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -4/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0064/)

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





## 65. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [2, 15, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation30/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 30/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0065/)

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





## 66. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [3, 16, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-1105/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1105/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0066/)

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





## 67. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [4, 17, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-69/4

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -69/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0067/)

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





## 68. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [5, 18, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation245/6

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 245/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0068/)

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





## 69. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [6, 19, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation452/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 452/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0069/)

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





## 70. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [7, 20, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-402/1

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -402/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0070/)

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





## 71. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [8, 21, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-188/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0071/)

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





## 72. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [9, 22, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-129/2

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -129/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0072/)

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





## 73. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [10, 23, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation625/4

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 625/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0073/)

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





## 74. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [11, 24, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-1099/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1099/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0074/)

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





## 75. Signed Moment Integral of a Polynomial

On all R, f has ascending coefficients [12, 25, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation70/3

Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 70/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0075/)

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





## 76. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(2)x+(1). For a=0≤b=2 and N=2 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation4/1

Exact termwise rational accumulation gives the finite sum 4/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0076/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 77. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(0)x+(2). For a=-1≤b=2 and N=3 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation6/1

Exact termwise rational accumulation gives the finite sum 6/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0077/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 78. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(3)x+(-2). For a=1≤b=1 and N=4 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0078/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 79. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-4)x+(-9). For a=1≤b=5 and N=1 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-52/1

Exact termwise rational accumulation gives the finite sum -52/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0079/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 80. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(5)x+(-8). For a=-2≤b=3 and N=5 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-40/1

Exact termwise rational accumulation gives the finite sum -40/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0080/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 81. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-6)x+(-7). For a=-1≤b=0 and N=6 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-7/2

Exact termwise rational accumulation gives the finite sum -7/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0081/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 82. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(7)x+(-6). For a=0≤b=2 and N=7 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation0/1

Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0082/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 83. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-8)x+(-5). For a=1≤b=4 and N=8 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-141/2

Exact termwise rational accumulation gives the finite sum -141/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0083/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 84. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(9)x+(-4). For a=-2≤b=2 and N=9 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-24/1

Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0084/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 85. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-10)x+(-3). For a=-1≤b=4 and N=10 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-155/2

Exact termwise rational accumulation gives the finite sum -155/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0085/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 86. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(11)x+(-2). For a=0≤b=1 and N=11 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation3/1

Exact termwise rational accumulation gives the finite sum 3/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0086/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 87. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-12)x+(-1). For a=1≤b=3 and N=12 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-48/1

Exact termwise rational accumulation gives the finite sum -48/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0087/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 88. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(13)x+(0). For a=-2≤b=1 and N=13 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-24/1

Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0088/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 89. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-14)x+(1). For a=-1≤b=3 and N=14 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-44/1

Exact termwise rational accumulation gives the finite sum -44/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0089/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 90. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(15)x+(2). For a=0≤b=5 and N=15 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation185/1

Exact termwise rational accumulation gives the finite sum 185/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0090/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 91. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-16)x+(3). For a=1≤b=2 and N=16 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-41/2

Exact termwise rational accumulation gives the finite sum -41/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0091/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 92. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(17)x+(4). For a=-2≤b=0 and N=17 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-28/1

Exact termwise rational accumulation gives the finite sum -28/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0092/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 93. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-18)x+(5). For a=-1≤b=2 and N=18 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-15/2

Exact termwise rational accumulation gives the finite sum -15/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0093/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 94. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(19)x+(6). For a=0≤b=4 and N=19 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation168/1

Exact termwise rational accumulation gives the finite sum 168/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0094/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 95. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-20)x+(7). For a=1≤b=6 and N=20 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-605/2

Exact termwise rational accumulation gives the finite sum -605/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0095/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 96. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(21)x+(8). For a=-2≤b=-1 and N=21 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-24/1

Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0096/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 97. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-22)x+(9). For a=-1≤b=1 and N=22 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation20/1

Exact termwise rational accumulation gives the finite sum 20/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0097/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 98. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(23)x+(10). For a=0≤b=3 and N=23 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation129/1

Exact termwise rational accumulation gives the finite sum 129/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0098/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 99. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(-24)x+(11). For a=1≤b=5 and N=24 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation-236/1

Exact termwise rational accumulation gives the finite sum -236/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0099/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate





## 100. Finite Left-Endpoint Sum of an Affine Function

On all R, f(x)=(25)x+(12). For a=-2≤b=3 and N=50 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.

View answer and explanation465/4

Exact termwise rational accumulation gives the finite sum 465/4. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261008-r07-definite-integral-0100/)

Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate



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