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I am having trouble with the following question.

If $F(x)=\int_{0}^x xf(t) dt$, find F'(x).

Please help

Thank you in advance

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Do you know about the Fundamental Theorem of Calculus? – user17794 Oct 25 '12 at 15:38
up vote 1 down vote accepted

Well, notice first that you can pull $x$ out of the integral. Then, you can use the product rule and the fundamental theorem of calculus to obtain the answer. Remember that $$\frac{\text{d}}{\text{d}x}\int_{0}^x f(t)\text{d}t=f(x).$$

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I get $F'(x)=\int_{0}^x f(t)dt +x(f(x)-f(0))$ – user43418 Oct 25 '12 at 15:48
Why is f(0)=0 ? – user43418 Oct 25 '12 at 15:50
@user43418 let's say that we have $G(t)=\int f(t) dt$, so $\int_0^x f(t) dt=G(x)-G(0)$. When you take the derivative the constant term drops out, and you're left with $G'(x)=f(x)$. – Robert Mastragostino Oct 25 '12 at 15:55
So $F'(x)= \int_{0}^{x}f(t)dt+xf(x)$ – user43418 Oct 25 '12 at 15:58
Is that correct ? – user43418 Oct 25 '12 at 16:01

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