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I know that $\frac{d}{dx} \int_{0}^{x} f(t) dt = f(x)$. What about $\frac{d}{dx} \int_{0}^{x} f(t-x) dt$? Is that just $f(x-x)$? I think I have to use the chain rule but I'm not sure how.

Thank you!

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Also, I tried to change my username back to "badatmath" but it didn't work. – badatmath Nov 16 '11 at 7:40
See here. – Ragib Zaman Nov 16 '11 at 7:48
@okay-at-math: I can make that change, if you'd like. – Zev Chonoles Nov 16 '11 at 7:49
@ZevChonoles: it's alright, apparently I'm less bad at math than I thought, if the answer is that complicated. – badatmath Nov 16 '11 at 7:52
@RagibZaman: Thank you for the link! – badatmath Nov 16 '11 at 7:52
up vote 3 down vote accepted

The following is an informal way of figuring out what the answer is. For someone like me who has a limited number of brain cells, it beats trying to remember a formula.

Let $F(u)$ be any antiderivative (indefinite integral) of $f(u)$. Then $$\int_0^x f(t-x)\,dt=\left.F(t-x)\right|_0^x=F(0)-F(-x).$$ Now differentiate with respect to $x$. By the definition of antiderivative, we have $F'(u)=f(u)$. Thus, by the Chain Rule, our derivative is $-(-f(-x))$, or more simply $f(-x)$.

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Wow that is clever :D Thank you very much! – badatmath Nov 16 '11 at 7:54

Another approach would be the following:

$$\int\limits_0^x {f\left( {t - x} \right)dt} \overbrace = ^{t - x = u}\int\limits_{ - x}^0 {f\left( u \right)du} = - \int\limits_0^{ - x} {f\left( u \right)du} $$

From this it is immediate that

$$\frac d{dx}\int\limits_0^x {f\left( {x - t} \right)dt} =-\frac d{dx} \int\limits_0^{-x} {f\left( u \right)du}=f(-x) $$

where the $-$ signs cancelled.

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eh? How do you get $t-x = 0$? – badatmath May 4 '12 at 6:41
@JackPenny It's a typo, it should be $t-x=u$ (a change of variables). – Pedro Tamaroff May 4 '12 at 14:57

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