`g(x) = int_1^x(1/(t^3 + 1))dt` Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

Textbook Question

Chapter 5, 5.3 - Problem 7 - 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)=1/(t^3+1)` and `g(x)=F_1(x).`

 

Therefore

`g'(x)=F'_1(x)=f(x)=1/(x^3+1).`

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