[Math] Finding the derivative of the integral using the Fundamental Theorem of Calculus.

Question

(1 pt) Find the derivative of the following function
$$F(x) = \int_{x^4}^{x^7} (2t – 1)^3 dt$$
using the Fundamental Theorem of Calculus.

$F'(x) = \ldots $

(original image)

I'm still not entirely solid on the concept of the Fundamental Theorem of Calculus, but I believe that the first step of the theorem will give us $$2x-1$$ which is the derivative of F(x). Usually, you would then take F(b) – F(a) to solve for the second step, but since its replaced with variables, I'm not sure how to proceed.

Answer 1

$G(x)=\int_0^{x}(2t+1)^3dt$ so $G'(x)=(2x+1)^3$ and $F(x)=G(x^7)-G(x^4)$ taking derivatives $$F'(x)=G'(x^7)7x^6-G'(x^4)4x^3=7(2x^7+1)^3x^6-4(2x^4+1)^3x^3$$