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Fundamental Theorem of Calculus Help?
Hello there, I've recently started learning about the fundamental theorem and have a few problems I'm not sure I understand how to do. Any help would be appreciated.
1. Find F(x), if f(x) = ∫ (3t-t^2) dt (Range; top: x, bottom: 3)
2. Find F(x), if f(x) = ∫ sqrt(1-t) dt (Range; top: sin(x), bottom: -1)
2 Answers
- az_lenderLv 71 year agoFavorite Answer
When you say "range," I guess you mean the limits on the integral.
#1. F(x) = (3/2)(x - 3) - (1/3)(x^3 - 27).
#2. F(x) = -(2/3)(1 - sin(x))^(3/2) + (2/3)(1 - (-1))^(3/2)
= -(2/3)(1 - sin(x))^(3/2) + (1/3)*2^(5/2).
- Geeganage WLv 51 year ago
1. f(x) = ∫ (3t-t^2) dt (Range; top: x, bottom: 3)
F(x) = t from 3 to x, ∫ (3t-t^2) dt = t from 3 to x,[(3t^2)/2 -(t^3)/3]
F(x) = (3x^2)/2 - (x^3)/3 -27/2+27/3 =(3x^2)/2 - (x^3)/3 - 27/6.
2. f(x) = t from -1 to sinx,∫ √(1-t) dt = t from -1 to sinx,(-2/3)[(1-t)^(3/2)].
F(x) = (-2/3)[(1-sinx)^(3/2) - 2^(3/2)].