Chapter 4 - Integrals - 4.3 The Fundamental Theorem of Calculus - 4.3 Exercises - Page 329: 70

$\frac{d}{dx}\left(\int_{g(x)}^{h(x)}f(t)dt \right)=h'(x)f(h(x))-g'(x)f(g(x))$

Suppose that $F$ is an antiderivative of $f$, $F'(x)=f(x)$ so: $\displaystyle\int_{g(x)}^{h(x)}f(t)dt=[F(t)]_{g(x)}^{h(x)}=F(h(x))-F(g(x))$ $\displaystyle\int_{g(x)}^{h(x)}f(t)dt=F(h(x))-F(g(x))$ $\displaystyle\frac{d}{dx}\left(\int_{g(x)}^{h(x)}f(t)dt \right)=h'(x)F'(h(x))-g'(x)F'(g(x))$ $\displaystyle\frac{d}{dx}\left(\int_{g(x)}^{h(x)}f(t)dt \right)=h'(x)f(h(x))-g'(x)f(g(x))$