Signed Polynomial Integral and Finite Left Sum Foundations
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batch-20261007-advance-20261008-r07-definite-integral-foundations · 2026-10-07 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Signed Moment Integral of a Constant
On all R, f(x)=3. Use the scoped signed polynomial moments to find I(f,2,2), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
The constant moment c(b−a) gives 0/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
2. Signed Moment Integral of a Constant
On all R, f(x)=-1. Use the scoped signed polynomial moments to find I(f,-11,-9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-2/1
The constant moment c(b−a) gives -2/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
3. Signed Moment Integral of a Constant
On all R, f(x)=2. Use the scoped signed polynomial moments to find I(f,-10,-7), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
6/1
The constant moment c(b−a) gives 6/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
4. Signed Moment Integral of a Constant
On all R, f(x)=-3. Use the scoped signed polynomial moments to find I(f,-9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
12/1
The constant moment c(b−a) gives 12/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
5. Signed Moment Integral of a Constant
On all R, f(x)=4. Use the scoped signed polynomial moments to find I(f,-8,-3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
20/1
The constant moment c(b−a) gives 20/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
6. Signed Moment Integral of a Constant
On all R, f(x)=-5. Use the scoped signed polynomial moments to find I(f,-7,-1), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-30/1
The constant moment c(b−a) gives -30/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
7. Signed Moment Integral of a Constant
On all R, f(x)=6. Use the scoped signed polynomial moments to find I(f,-6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-42/1
The constant moment c(b−a) gives -42/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
8. Signed Moment Integral of a Constant
On all R, f(x)=-7. Use the scoped signed polynomial moments to find I(f,-5,3), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-56/1
The constant moment c(b−a) gives -56/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
9. Signed Moment Integral of a Constant
On all R, f(x)=8. Use the scoped signed polynomial moments to find I(f,-4,5), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
72/1
The constant moment c(b−a) gives 72/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
10. Signed Moment Integral of a Constant
On all R, f(x)=-9. Use the scoped signed polynomial moments to find I(f,-3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
90/1
The constant moment c(b−a) gives 90/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
11. Signed Moment Integral of a Constant
On all R, f(x)=10. Use the scoped signed polynomial moments to find I(f,-2,9), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
110/1
The constant moment c(b−a) gives 110/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
12. Signed Moment Integral of a Constant
On all R, f(x)=-11. Use the scoped signed polynomial moments to find I(f,-1,11), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-132/1
The constant moment c(b−a) gives -132/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
13. Signed Moment Integral of a Constant
On all R, f(x)=12. Use the scoped signed polynomial moments to find I(f,0,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-156/1
The constant moment c(b−a) gives -156/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
14. Signed Moment Integral of a Constant
On all R, f(x)=-13. Use the scoped signed polynomial moments to find I(f,1,15), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-182/1
The constant moment c(b−a) gives -182/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
15. Signed Moment Integral of a Constant
On all R, f(x)=14. Use the scoped signed polynomial moments to find I(f,2,17), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
210/1
The constant moment c(b−a) gives 210/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
16. Signed Moment Integral of a Constant
On all R, f(x)=-15. Use the scoped signed polynomial moments to find I(f,3,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
240/1
The constant moment c(b−a) gives 240/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
17. Signed Moment Integral of a Constant
On all R, f(x)=16. Use the scoped signed polynomial moments to find I(f,4,21), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
272/1
The constant moment c(b−a) gives 272/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
18. Signed Moment Integral of a Constant
On all R, f(x)=-17. Use the scoped signed polynomial moments to find I(f,5,23), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-306/1
The constant moment c(b−a) gives -306/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
19. Signed Moment Integral of a Constant
On all R, f(x)=18. Use the scoped signed polynomial moments to find I(f,6,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-342/1
The constant moment c(b−a) gives -342/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
20. Signed Moment Integral of a Constant
On all R, f(x)=-19. Use the scoped signed polynomial moments to find I(f,7,27), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-380/1
The constant moment c(b−a) gives -380/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
21. Signed Moment Integral of a Constant
On all R, f(x)=20. Use the scoped signed polynomial moments to find I(f,8,29), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
420/1
The constant moment c(b−a) gives 420/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
22. Signed Moment Integral of a Constant
On all R, f(x)=-21. Use the scoped signed polynomial moments to find I(f,9,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
462/1
The constant moment c(b−a) gives 462/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
23. Signed Moment Integral of a Constant
On all R, f(x)=22. Use the scoped signed polynomial moments to find I(f,10,33), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
506/1
The constant moment c(b−a) gives 506/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
24. Signed Moment Integral of a Constant
On all R, f(x)=-23. Use the scoped signed polynomial moments to find I(f,11,35), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-552/1
The constant moment c(b−a) gives -552/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
25. Signed Moment Integral of a Constant
On all R, f(x)=24. Use the scoped signed polynomial moments to find I(f,12,-13), with oriented endpoints. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-600/1
The constant moment c(b−a) gives -600/1. Equal endpoints give zero and reversal changes the sign; this is not automatically nonnegative geometric area or a general FTC proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-constant · candidate
26. Signed Moment Integral of a Monomial
On all R, f(x)=(2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
27. Signed Moment Integral of a Monomial
On all R, f(x)=(-2)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
4/1
The restricted polynomial moment gives 4/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
28. Signed Moment Integral of a Monomial
On all R, f(x)=(3)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
28/1
The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
29. Signed Moment Integral of a Monomial
On all R, f(x)=(-4)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
15/1
The restricted polynomial moment gives 15/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
30. Signed Moment Integral of a Monomial
On all R, f(x)=(5)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
31/1
The restricted polynomial moment gives 31/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
31. Signed Moment Integral of a Monomial
On all R, f(x)=(-6)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
729/1
The restricted polynomial moment gives 729/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
32. Signed Moment Integral of a Monomial
On all R, f(x)=(7)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-14/1
The restricted polynomial moment gives -14/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
33. Signed Moment Integral of a Monomial
On all R, f(x)=(-8)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
34. Signed Moment Integral of a Monomial
On all R, f(x)=(9)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
78/1
The restricted polynomial moment gives 78/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
35. Signed Moment Integral of a Monomial
On all R, f(x)=(-10)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-5/2
The restricted polynomial moment gives -5/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
36. Signed Moment Integral of a Monomial
On all R, f(x)=(11)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
363/5
The restricted polynomial moment gives 363/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
37. Signed Moment Integral of a Monomial
On all R, f(x)=(-12)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
1330/1
The restricted polynomial moment gives 1330/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
38. Signed Moment Integral of a Monomial
On all R, f(x)=(13)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-26/1
The restricted polynomial moment gives -26/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
39. Signed Moment Integral of a Monomial
On all R, f(x)=(-14)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
28/1
The restricted polynomial moment gives 28/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
40. Signed Moment Integral of a Monomial
On all R, f(x)=(15)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
140/1
The restricted polynomial moment gives 140/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
41. Signed Moment Integral of a Monomial
On all R, f(x)=(-16)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
60/1
The restricted polynomial moment gives 60/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
42. Signed Moment Integral of a Monomial
On all R, f(x)=(17)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
527/5
The restricted polynomial moment gives 527/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
43. Signed Moment Integral of a Monomial
On all R, f(x)=(-18)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
2187/1
The restricted polynomial moment gives 2187/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
44. Signed Moment Integral of a Monomial
On all R, f(x)=(19)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-38/1
The restricted polynomial moment gives -38/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
45. Signed Moment Integral of a Monomial
On all R, f(x)=(-20)x^1, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
The restricted polynomial moment gives 0/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
46. Signed Moment Integral of a Monomial
On all R, f(x)=(21)x^2, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
182/1
The restricted polynomial moment gives 182/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
47. Signed Moment Integral of a Monomial
On all R, f(x)=(-22)x^3, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=0, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-11/2
The restricted polynomial moment gives -11/2, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
48. Signed Moment Integral of a Monomial
On all R, f(x)=(23)x^4, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
759/5
The restricted polynomial moment gives 759/5, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
49. Signed Moment Integral of a Monomial
On all R, f(x)=(-24)x^5, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=-3, b=2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
2660/1
The restricted polynomial moment gives 2660/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
50. Signed Moment Integral of a Monomial
On all R, f(x)=(25)x^0, 0≤n≤5. Use the exact signed moment c(b^(n+1)−a^(n+1))/(n+1) on oriented endpoints a=1, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-50/1
The restricted polynomial moment gives -50/1, including negative coefficients, reversed endpoints, and zero length. It is not a new general integration or FTC Evidence claim.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-monomial · candidate
51. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=0, b=0. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 0/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
52. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-12/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -12/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
53. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-10, 3, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-42/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -42/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
54. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-9, 4, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-655/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -655/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
55. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-8, 5, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-141/4
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -141/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
56. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-7, 6, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
425/6
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 425/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
57. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-6, 7, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
164/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 164/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
58. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-5, 8, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-258/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -258/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
59. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-4, 9, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-188/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
60. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-3, 10, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-21/2
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -21/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
61. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-2, 11, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
265/4
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 265/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
62. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [-1, 12, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-721/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -721/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
63. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [0, 13, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-2/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -2/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
64. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [1, 14, 0, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-4/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -4/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
65. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [2, 15, 1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
30/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 30/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
66. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [3, 16, -1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-1105/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1105/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
67. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [4, 17, 0, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-69/4
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -69/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
68. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [5, 18, 1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
245/6
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 245/6. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
69. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [6, 19, -1, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
452/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 452/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
70. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [7, 20, 0, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-2. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-402/1
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -402/1. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
71. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [8, 21, 1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-3, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-188/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -188/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
72. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [9, 22, -1, 2], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=2, b=-1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-129/2
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -129/2. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
73. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [10, 23, 0, 3], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-2, b=3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
625/4
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 625/4. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
74. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [11, 24, 1, 4], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=4, b=-3. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-1099/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain -1099/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
75. Signed Moment Integral of a Polynomial
On all R, f has ascending coefficients [12, 25, -1, 1], starting with its constant term. Use coefficientwise exact signed moments on oriented endpoints a=-1, b=1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
70/3
Sum a_k*(b^(k+1)−a^(k+1))/(k+1) exactly to obtain 70/3. Keep orientation and sign; restricted polynomial moment checks do not prove a general FTC bridge.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-polynomial · candidate
76. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(2)x+(1). For a=0≤b=2 and N=2 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
4/1
Exact termwise rational accumulation gives the finite sum 4/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
77. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(0)x+(2). For a=-1≤b=2 and N=3 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
6/1
Exact termwise rational accumulation gives the finite sum 6/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
78. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(3)x+(-2). For a=1≤b=1 and N=4 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
79. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-4)x+(-9). For a=1≤b=5 and N=1 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-52/1
Exact termwise rational accumulation gives the finite sum -52/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
80. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(5)x+(-8). For a=-2≤b=3 and N=5 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-40/1
Exact termwise rational accumulation gives the finite sum -40/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
81. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-6)x+(-7). For a=-1≤b=0 and N=6 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-7/2
Exact termwise rational accumulation gives the finite sum -7/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
82. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(7)x+(-6). For a=0≤b=2 and N=7 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
0/1
Exact termwise rational accumulation gives the finite sum 0/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
83. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-8)x+(-5). For a=1≤b=4 and N=8 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-141/2
Exact termwise rational accumulation gives the finite sum -141/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
84. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(9)x+(-4). For a=-2≤b=2 and N=9 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-24/1
Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
85. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-10)x+(-3). For a=-1≤b=4 and N=10 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-155/2
Exact termwise rational accumulation gives the finite sum -155/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
86. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(11)x+(-2). For a=0≤b=1 and N=11 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
3/1
Exact termwise rational accumulation gives the finite sum 3/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
87. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-12)x+(-1). For a=1≤b=3 and N=12 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-48/1
Exact termwise rational accumulation gives the finite sum -48/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
88. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(13)x+(0). For a=-2≤b=1 and N=13 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-24/1
Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
89. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-14)x+(1). For a=-1≤b=3 and N=14 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-44/1
Exact termwise rational accumulation gives the finite sum -44/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
90. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(15)x+(2). For a=0≤b=5 and N=15 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
185/1
Exact termwise rational accumulation gives the finite sum 185/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
91. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-16)x+(3). For a=1≤b=2 and N=16 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-41/2
Exact termwise rational accumulation gives the finite sum -41/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
92. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(17)x+(4). For a=-2≤b=0 and N=17 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-28/1
Exact termwise rational accumulation gives the finite sum -28/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
93. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-18)x+(5). For a=-1≤b=2 and N=18 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-15/2
Exact termwise rational accumulation gives the finite sum -15/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
94. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(19)x+(6). For a=0≤b=4 and N=19 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
168/1
Exact termwise rational accumulation gives the finite sum 168/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
95. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-20)x+(7). For a=1≤b=6 and N=20 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-605/2
Exact termwise rational accumulation gives the finite sum -605/2. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
96. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(21)x+(8). For a=-2≤b=-1 and N=21 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-24/1
Exact termwise rational accumulation gives the finite sum -24/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
97. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-22)x+(9). For a=-1≤b=1 and N=22 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
20/1
Exact termwise rational accumulation gives the finite sum 20/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
98. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(23)x+(10). For a=0≤b=3 and N=23 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
129/1
Exact termwise rational accumulation gives the finite sum 129/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
99. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(-24)x+(11). For a=1≤b=5 and N=24 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
View answer and explanation
-236/1
Exact termwise rational accumulation gives the finite sum -236/1. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
100. Finite Left-Endpoint Sum of an Affine Function
On all R, f(x)=(25)x+(12). For a=-2≤b=3 and N=50 uniform cells, compute the finite left-endpoint sum with Δx=(b−a)/N and x_i=a+iΔx for i=0,…,N−1. Give a reduced fraction with positive denominator, retaining /1 for integers.
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465/4
Exact termwise rational accumulation gives the finite sum 465/4. A finite left sum generally differs from the integral, even though constant or zero-length special cases can agree; this is not a general Riemann-integrability proof.
Reviewed step-by-step derivation
Signed polynomial definite integrals on finite intervals · qf-integral-affine-left-sum · candidate
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