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My question has to do with differentiating an integral. We are given:

[tex]f(x)=1/2\int_{0}^{x} (x-t)^2 g(t)dt[/tex]

And we are asked to prove that:

[tex]f'(x)=x\int_{0}^{x}g(t)dt - \int_{0}^{x}tg(t)dt[/tex]

My Solution:

I expanded [tex](x-t)^2[/tex] into [tex]x^2-2xt+t[/tex] and then expanded

[tex]f(x)=x^2/2\int_{0}^{x}g(t)dt - x\int_{0}^{x}tg(t)dt + \int_{0}^{x}t/2g(t)dt[/tex]

Then I differentiated using the product rule:

[tex]f'(x)=x\int_{0}^{x}g(t)dt + g(t)x^2/2 - \int_{0}^{x}tg(t)dt - xtg(t) + t/2g(t)[/tex]

This is close to what they want me to prove, but I have these extra terms:

[tex]g(t)x^2/2-xtg(t)+t/2g(t)[/tex]

Which is basically [tex](t-x)^2/2g(t)[/tex].

How do I eliminate that term? Does it evaluate to 0 somehow?

Thanks for the help!

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# Differentiating Integrals

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