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I am having a problem with this exercise. Please help.

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

Thank you in advance

3 Answers 3

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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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    @user43418: yes.2012-10-25
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Hint: if $G$ is any indefinite integral of $g$, we have that $\int_{x}^{x^2} g(t)dt=G(x^2)-G(x)$

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Hint

  1. $\int\limits_x^{x^2}{g(t)dt}=\int\limits_x^{c}{g(t)dt}+\int\limits_c^{x^2}{g(t)dt} \quad (\forall c\in\mathbb{R});$
  2. For differentiating use Newton-Leibniz formula and the chain rule.