`g(s) = int_5^s(t - t^2)^8dt` Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

Textbook Question

Chapter 5, 5.3 - Problem 9 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Hello!

Part 1 of the Fundamental Theorem of Calculus states that for a continuous function `f`
`F'_a(x)=f(x),` where `F_a(x)=int_a^xf(t)dt.`

Here `f(t)=(t-t^2)^8` and `g(x)=F_5(x).`

Therefore

`g'(x)=F'_5(x)=f(x)=(x-x^2)^8.`

Or `g'(s)=(s-s^2)^8`  (a trivial variable substitution).

 

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