How do you use the Fundamental Theorem of Calculus to find the derivative of #int 6*(sin(t))^2 dt# from 1 to e^x?

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Answer 1 · 2016-07-15T16:55:29

#= 6 e^x (sin(e^x))^2 #

we have #d/dx \ int_1^(e^x) \ 6*(sin(t))^2 \ dt#

STEP 1: FTC pt2 states that

#d/(du) int_a^u \ f(t) \ dt = f(u)#

STEP 2: from the chain rule #d/dx = d/(du) * (du)/dx#

So #d/(dx) int_a^(u(x)) \ f(t) \ dt = f(u) * (du)/dx#

here that means that

#d/dx \ int_1^(e^x) \ 6*(sin(t))^2 \ dt = 6*(sin(e^x))^2 d/dx e^x#

#= 6 e^x (sin(e^x))^2 #