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It has something to do with the second part of the Fundamental Theorem of Calculus right? I've always had trouble with this theorem ever since I learned it several years ago :\ Would somebody please show step-by-step how the two are equal? I know that $f(x) = \frac{\mathrm{d}}{\mathrm{d}x} \int_0^x g(t)\,\mathrm{d}t$, but I do not know how to use that to simplify the problem.

Thanks

A

(Just for reference, this Wolfram link is what I'm referring to: http://bit.ly/15SoByj)

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Please check your question. The question you asked is different from the one shown in your reference.

I am answering the question in your reference.

$\frac{\mathrm{d}}{\mathrm{d}x} \int_0^x (x-t)g(t)\,\mathrm{d}t=\frac{\mathrm{d}}{\mathrm{d}x} x\int_0^x g(t)-\int_0^x tg(t)\,\mathrm{d}t$

Then if you set $F(x)=x\int_0^x g(t)$, by chain rule, $\frac{d}{dx}F(x)=\int_0^x g(t)+xg(x)$ by fundamental theorem of calculus,

and the second part $\frac{d}{dx}\int_0^x tg(t)\,\mathrm{d}t=xg(x)$ by fundamental theorem of calculus, then you get the result.

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  • $\begingroup$ Yes, you're right. Sorry I didn't catch that. And thanks, now I get it! I hadn't considered the chain rule :) $\endgroup$ – A is for Ambition Jan 28 '15 at 10:28
  • $\begingroup$ @AisforAmbition I have edited your question (the title), please check whether it's what you want. $\endgroup$ – John Jan 28 '15 at 10:31

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