# Continuous FTC and Oriented Endpoint 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-20261009-r01-fundamental-theorem-calculus-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. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [0], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation0/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 0/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0001/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 2. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation3/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 3/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0002/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 3. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-10, 3, 1, 3], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-10/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -10/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0003/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 4. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-9, 4, -1, 4], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-2/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -2/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0004/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 5. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-8, 5, 0, 1], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation10/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 10/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0005/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 6. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-7, 6, 1, 2], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-31/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -31/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0006/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 7. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-6, 7, -1, 3], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-17/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -17/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0007/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 8. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-5, 8, 0, 4], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-5/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -5/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0008/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 9. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-4, 9, 1, 1], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation7/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 7/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0009/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 10. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-3, 10, -1, 2], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation29/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 29/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0010/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 11. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-2, 11, 0, 3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-48/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -48/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0011/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 12. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [-1, 12, 1, 4], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-16/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -16/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0012/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 13. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [0, 13, -1, 1], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation0/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 0/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0013/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 14. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [1, 14, 0, 2], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation17/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 17/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0014/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 15. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [2, 15, 1, 3], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation60/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 60/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0015/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 16. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [3, 16, -1, 4], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-65/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -65/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0016/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 17. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [4, 17, 0, 1], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-14/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -14/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0017/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 18. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [5, 18, 1, 2], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation5/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 5/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0018/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 19. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [6, 19, -1, 3], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation27/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 27/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0019/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 20. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [7, 20, 0, 4], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation79/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 79/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0020/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 21. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [8, 21, 1, 1], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(-2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-38/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -38/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0021/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 22. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [9, 22, -1, 2], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(-1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-16/1

Under the stated full-real continuity premise, G′(x)=f(x), giving -16/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0022/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 23. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [10, 23, 0, 3], constant term first. Let G(x)=∫_0^x f(t)dt with oriented endpoints. Find G′(0). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation10/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 10/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0023/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 24. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [11, 24, 1, 4], constant term first. Let G(x)=∫_1^x f(t)dt with oriented endpoints. Find G′(1). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation40/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 40/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0024/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 25. Point Derivative of a Continuous Accumulating Integral

On all R, the polynomial f is continuous and f has ascending coefficients [12, 25, -1, 1], constant term first. Let G(x)=∫_-1^x f(t)dt with oriented endpoints. Find G′(2). Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation66/1

Under the stated full-real continuity premise, G′(x)=f(x), giving 66/1. This exact polynomial instance is not a new general FTC formal proof.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0025/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-accumulation-derivative · candidate





## 26. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation1/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0026/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 27. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [0], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(3)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation0/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 0/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0027/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 28. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [3, 1], constant term first. Let lower l(x)=(2)x+(1) and upper u(x)=(2)x+(1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation0/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 0/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0028/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 29. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. Let lower l(x)=(-1)x+(-3) and upper u(x)=(0)x+(-9). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-297/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -297/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0029/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 30. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. Let lower l(x)=(2)x+(0) and upper u(x)=(4)x+(-8). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-16552/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -16552/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0030/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 31. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. Let lower l(x)=(-3)x+(-1) and upper u(x)=(0)x+(-7). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation75/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 75/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0031/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 32. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. Let lower l(x)=(1)x+(-2) and upper u(x)=(2)x+(-6). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-1416/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1416/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0032/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 33. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. Let lower l(x)=(-2)x+(-3) and upper u(x)=(0)x+(-5). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-1090/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1090/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0033/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 34. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. Let lower l(x)=(3)x+(0) and upper u(x)=(6)x+(-4). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-23214/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -23214/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0034/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 35. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. Let lower l(x)=(-1)x+(-1) and upper u(x)=(0)x+(-3). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-3/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -3/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0035/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 36. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. Let lower l(x)=(2)x+(-2) and upper u(x)=(4)x+(-2). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-96/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -96/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0036/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 37. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. Let lower l(x)=(-3)x+(-3) and upper u(x)=(0)x+(-1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-2703/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -2703/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0037/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 38. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(0). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-226/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -226/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0038/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 39. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. Let lower l(x)=(-2)x+(-1) and upper u(x)=(0)x+(1). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation34/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 34/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0039/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 40. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. Let lower l(x)=(3)x+(-2) and upper u(x)=(6)x+(2). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation504/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 504/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0040/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 41. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. Let lower l(x)=(-1)x+(-3) and upper u(x)=(0)x+(3). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-333/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -333/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0041/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 42. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. Let lower l(x)=(2)x+(0) and upper u(x)=(4)x+(4). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-256/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -256/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0042/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 43. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. Let lower l(x)=(-3)x+(-1) and upper u(x)=(0)x+(5). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation183/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 183/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0043/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 44. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. Let lower l(x)=(1)x+(-2) and upper u(x)=(2)x+(6). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation1524/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1524/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0044/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 45. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. Let lower l(x)=(-2)x+(-3) and upper u(x)=(0)x+(7). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-1186/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -1186/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0045/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 46. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. Let lower l(x)=(3)x+(0) and upper u(x)=(6)x+(8). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation150/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 150/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0046/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 47. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. Let lower l(x)=(-1)x+(-1) and upper u(x)=(0)x+(9). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation9/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 9/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0047/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 48. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. Let lower l(x)=(2)x+(-2) and upper u(x)=(4)x+(10). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=0. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation13080/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 13080/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0048/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 49. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. Let lower l(x)=(-3)x+(-3) and upper u(x)=(0)x+(11). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=1. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation-2883/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining -2883/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0049/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 50. Derivative of an Oriented Integral with Moving Bounds

On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. Let lower l(x)=(1)x+(0) and upper u(x)=(2)x+(12). Find the derivative of ∫_l(x)^u(x) f(t)dt at x=-2. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation1370/1

Use f(u(x))*u′(x)−f(l(x))*l′(x) with the declared continuous integrand and affine bounds, obtaining 1370/1. Bounds coinciding at this point do not force equal velocities or a zero derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0050/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-moving-bounds-derivative · candidate





## 51. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [0], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 0/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 0/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0051/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 52. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [3], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 3/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 3/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0052/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 53. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-10, 3, 1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -3/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -3/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0053/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 54. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 27/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 27/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0054/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 55. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -14/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -14/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0055/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 56. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -7/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -7/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0056/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 57. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 3/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 3/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0057/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 58. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 43/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 43/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0058/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 59. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -13/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -13/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0059/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 60. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -3/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -3/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0060/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 61. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 12/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 12/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0061/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 62. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 59/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 59/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0062/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 63. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -15/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -15/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0063/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 64. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 1/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 1/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0064/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 65. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 21/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 21/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0065/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 66. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 63/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 63/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0066/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 67. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -14/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -14/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0067/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 68. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 5/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 5/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0068/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 69. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 27/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 27/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0069/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 70. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 79/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 79/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0070/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 71. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -13/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -13/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0071/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 72. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. For G(x)=∫_0^x f(t)dt, give (G(0),G′(0)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 9/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 9/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0072/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 73. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. For G(x)=∫_1^x f(t)dt, give (G(1),G′(1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 36/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 36/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0073/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 74. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. For G(x)=∫_2^x f(t)dt, give (G(2),G′(2)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, 95/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, 95/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0074/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 75. Accumulation Value and Derivative at the Anchor

On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. For G(x)=∫_-1^x f(t)dt, give (G(-1),G′(-1)) in that order. Use reduced positive-denominator fractions, retaining /1 for integers.

View answer and explanation(0/1, -15/1)

The anchor value is zero, while the derivative is f(a), so the pair is (0/1, -15/1). A zero integral value does not imply a zero instantaneous derivative.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0075/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-anchor-value-derivative · candidate





## 76. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [0], constant term first. Candidate F has ascending rational coefficients [5/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0076/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 77. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [1], constant term first. Candidate F has ascending rational coefficients [0/1, 1/1, 0/1, 0/1, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0077/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 78. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-10, 3, 1, 3], constant term first. Candidate F has ascending rational coefficients [-10/1, -10/1, 3/2, 1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0078/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 79. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-9, 4, -1, 4], constant term first. Candidate F has ascending rational coefficients [-9/1, -9/1, 2/1, -1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0079/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 80. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-8, 5, 0, 1], constant term first. Candidate F has ascending rational coefficients [-8/1, -8/1, 5/2, 0/1, 5/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0080/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 81. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-7, 6, 1, 2], constant term first. Candidate F has ascending rational coefficients [-7/1, -7/1, 3/1, 1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0081/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 82. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-6, 7, -1, 3], constant term first. Candidate F has ascending rational coefficients [-6/1, -6/1, 7/2, -1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0082/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 83. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-5, 8, 0, 4], constant term first. Candidate F has ascending rational coefficients [-5/1, -5/1, 4/1, 0/1, 2/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0083/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 84. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-4, 9, 1, 1], constant term first. Candidate F has ascending rational coefficients [-4/1, -4/1, 9/2, 1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0084/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 85. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-3, 10, -1, 2], constant term first. Candidate F has ascending rational coefficients [-3/1, -3/1, 5/1, -1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0085/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 86. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-2, 11, 0, 3], constant term first. Candidate F has ascending rational coefficients [-2/1, -2/1, 11/2, 0/1, 7/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0086/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 87. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [-1, 12, 1, 4], constant term first. Candidate F has ascending rational coefficients [-1/1, -1/1, 6/1, 1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0087/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 88. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [0, 13, -1, 1], constant term first. Candidate F has ascending rational coefficients [0/1, 0/1, 13/2, -1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0088/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 89. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [1, 14, 0, 2], constant term first. Candidate F has ascending rational coefficients [1/1, 1/1, 7/1, 0/1, 3/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0089/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 90. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [2, 15, 1, 3], constant term first. Candidate F has ascending rational coefficients [2/1, 2/1, 15/2, 1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0090/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 91. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [3, 16, -1, 4], constant term first. Candidate F has ascending rational coefficients [3/1, 3/1, 8/1, -1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0091/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 92. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [4, 17, 0, 1], constant term first. Candidate F has ascending rational coefficients [4/1, 4/1, 17/2, 0/1, 5/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0092/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 93. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [5, 18, 1, 2], constant term first. Candidate F has ascending rational coefficients [5/1, 5/1, 9/1, 1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0093/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 94. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [6, 19, -1, 3], constant term first. Candidate F has ascending rational coefficients [6/1, 6/1, 19/2, -1/3, 3/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0094/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 95. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [7, 20, 0, 4], constant term first. Candidate F has ascending rational coefficients [7/1, 7/1, 10/1, 0/1, 2/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0095/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 96. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [8, 21, 1, 1], constant term first. Candidate F has ascending rational coefficients [8/1, 8/1, 21/2, 1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0096/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 97. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [9, 22, -1, 2], constant term first. Candidate F has ascending rational coefficients [9/1, 9/1, 11/1, -1/3, 1/2]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0097/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 98. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [10, 23, 0, 3], constant term first. Candidate F has ascending rational coefficients [10/1, 10/1, 23/2, 0/1, 7/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationNo

Compare complete derivative coefficients with f, giving No, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0098/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 99. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [11, 24, 1, 4], constant term first. Candidate F has ascending rational coefficients [11/1, 11/1, 12/1, 1/3, 1/1]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0099/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate





## 100. Candidate Antiderivative Premise for the Endpoint Formula

On all R, continuous f has ascending coefficients [12, 25, -1, 1], constant term first. Candidate F has ascending rational coefficients [12/1, 12/1, 25/2, -1/3, 1/4]. Does F′=f satisfy the antiderivative premise for the endpoint formula?

View answer and explanationYes

Compare complete derivative coefficients with f, giving Yes, not a finite-sample equality claim. Rejecting this candidate does not reject other possible primitives; its constant term does not affect eligibility.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261007-advance-20261009-r01-fundamental-theorem-calculus-0100/)

Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261007-advance-20261009-r01-fundamental-theorem-calculus-foundations/validation-report.json)