How do you calculate the derivative of #int3(sin(t))^4 dt# from #[e^x,1]#?

1 Answer
Jun 12, 2015

Answer:

Use the Fundamental Theorem of Calculus, Part 1 (after rewriting) and use the chain rule.

Explanation:

g(x) = #int_(e^x)^1 3sin^4t dt#

To use the fundamental theorem in one form, we must have the constant as the lower limit of integration, so we rewrite:

g(x) = #-int_1^(e^x) 3sin^4t dt#

With #u=e^x#, we have: #g(x) = h(u) = -int_1^(u) 3sin^4t dt#

Use the chain rule to find #g'(x) = h'(u) (du)/dx#.

By FTC 1, #h'(u) = 3sin^4 u#

We also have #(du)/dx = d/dx(e^x) = x^e#.

So,

#g'(x) = 3sin^4u (du)/dx = 3sin^4(e^x)* e^x#