How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #g(x) = int(1 / (t^3 + 1)) dt# from [1,x]?

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Answer 1 · 2015-05-14T12:01:17

The theorem says:

Given the function:

#y=int_(h(x))^g(x)f(t)dt#

then:

#y'=f(g(x))*g'(x)-f(h(x))*h'(x)#.

So:

#g'(x)=1/(x^3+1)*1-1/(1^3+1)*0=1/(x^3+1)#.