How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function #y = int sin^3 t dt# from #[e^x, 0]#?

Question

12017 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-20T09:07:08

The theorem says:

Given the function:

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

than:

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

So, since #y=int_(e^x)^0sin^3tdt#:

#y'=sin^3 0*0-sin^3e^x*e^x=-e^xsin^3e^x#.