Finding the derivative of the integral using the Fundamental Theorem of Calculus.
(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 $
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.
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$\begingroup$$G(x)=\int_0^{x}(2t+1)^3dt$ so $G'(x)=(2x+1)^3$ and $F(x)=G(x^7)-G(x^4)$ taking derivatives $$F'(x)=G'(x^7)7x^6-G'(x^4)4x^3=7(2x^7+1)^3x^6-4(2x^4+1)^3x^3$$
$\endgroup$ $\begingroup$If $f(t)=(2t-1)^3,a(x)=x^4,b(x)=x^7, F'(x)=f(b(x))b'(x)-f(a(x))a'(x).$
$\endgroup$ $\begingroup$Let $G'(x)=(2x-1)^3$. Then $F(x)=\int_{x^4}^{x^7}(2t-1)^3\, \mathrm{d}t=G(x^7)-G(x^4)$. Hence $$F'(x)=7x^6G'(x^7)-4x^3G'(x^4)=7x^6(2x^7-1)^3-4x^3(2x^4-1)^3$$
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