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How would one interpret: $$\frac{\mathrm d}{\mathrm dx}\int_0^x (F(y)-F(x))\,\mathrm dy$$ I don't think I can use the fundamental theorem of calculus here, can I?

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For this problem you need that $F$ is a differentiable function. If it is, then indeed the fundamental theorem of calculus comes up here, but it's not immediate. Have you tried anything yet? – Ragib Zaman Oct 8 '11 at 14:40
@Ragib: F is differentiable. But then I just get F(x)-F(x) = 0 which doesn't seem to make sense. – Angada Oct 8 '11 at 14:41
@Ragib: Ok, Mathematica is giving me -x *F'(x) as the answer. but I can't figure out why exactly! – Angada Oct 8 '11 at 14:47
That's not exactly what J.M. did. The fundamental theorem of calculus states that if $f$ is differentiable, then $ \frac{d}{dx}\int^x_a f(t) dt = f(x) $. Note that the integrand $f(t)$ does not depend on the upper limit of integration, $x$. So to be able to use FTC, J.M. first had to remove the $F(x)$ term out of the integral. Note also that $F(x)$ is held as a constant whilst integrating with respect to $y$. – Ragib Zaman Oct 8 '11 at 14:56
No problem. I'm just telling you that the check mark is an important piece of feedback, IMHO. If you aren't sure/satisfied with an answer you've gotten, it is perfectly acceptable to hold off accepting an answer... – J. M. Oct 8 '11 at 16:14
up vote 7 down vote accepted

Certainly doable:

$$\begin{split}\frac{\mathrm d}{\mathrm dx}\int_0^x (F(y)-F(x))\,\mathrm dy&=\frac{\mathrm d}{\mathrm dx}\left(\int_0^x F(y)\mathrm dy-F(x)\int_0^x \,\mathrm dy\right)\\&=\frac{\mathrm d}{\mathrm dx}\left(\int_0^x F(y)\mathrm dy-x\,F(x)\right)\\&=F(x)-x\,F^\prime(x)-F(x)\\&=-x\,F^\prime (x)\end{split}$$

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In the general case, it intuitively ought to hold (under appropriate assumptions) that $$\frac{d}{dx}\int_0^x f(x,y)dy=\int_0^x \frac{\partial f}{\partial x} dy + f(x,x)$$ which agrees with this solution for the case where $f$ can be separated into terms depending only on $x$ and $y$. – Henning Makholm Oct 8 '11 at 15:09
@HenningMakholm under what conditions it is going to hold? I am solving a similar problem, but the bounds of integration are $(-\infty, \infty)$ – Fazzolini May 5 '14 at 7:31

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