Continuous FTC and Oriented Endpoint Foundations
This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
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 explanation
0/1
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
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 explanation
3/1
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
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
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
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 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
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
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
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
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 explanation
7/1
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
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 explanation
29/1
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
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
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
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 explanation
0/1
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
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 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
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 explanation
60/1
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
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
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
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 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
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 explanation
27/1
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
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 explanation
79/1
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
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
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
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 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
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 explanation
40/1
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
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 explanation
66/1
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
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 explanation
1/1
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
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 explanation
0/1
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
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 explanation
0/1
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
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
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
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 explanation
75/1
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
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
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
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
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
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
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
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
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 explanation
34/1
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
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 explanation
504/1
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
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
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
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 explanation
183/1
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
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 explanation
1524/1
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
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
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 explanation
150/1
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
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 explanation
9/1
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
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 explanation
13080/1
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
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
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 explanation
1370/1
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
Fundamental theorem of calculus: continuity and oriented endpoints · qf-ftc-primitive-hypothesis · candidate
Canonical question JSONL · English JSONL · Validation report