# Integration+ Differentiation

I need to calculate F'(x) such that $F(x)=\int_{x}^{x^2} g(t)dt$ such that g(x) is a continuous function

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You can use this formula

$${d\over dx}\, \int_{f_1(x)}^{f_2(x)} g(t) \,dt = g[f_2(x)] {f_2'(x)} - g[f_1(x)] {f_1'(x)} \,,$$

which is known as Leibniz rule.

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So $F'(x)=2xg(x^2)-g(x)$ ? –  user43418 Oct 25 '12 at 16:32
@user43418: yes. –  Mhenni Benghorbal Oct 25 '12 at 16:33
Hint: if $G$ is any indefinite integral of $g$, we have that $\int_{x}^{x^2} g(t)dt=G(x^2)-G(x)$
1. $\int\limits_x^{x^2}{g(t)dt}=\int\limits_x^{c}{g(t)dt}+\int\limits_c^{x^2}{g(t)dt} \quad (\forall c\in\mathbb{R});$