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I have a continuous function $f:[0,1] \rightarrow \mathbb R$ and I would like to find the second derivative of $$G(x) = \int_x^1(t-x)f(t)dt$$

I started off by using partial integration and got:

$$G(x) = [(t-x)\int_0^tf(s)ds]|_x^1 - \int_x^1\Big ( \int_0^tf(s)ds\Big )dt=$$ $$(1-x)\int_0^1f(s)ds-\int_x^1\Big ( \int_0^tf(s)ds\Big )dt$$

This is where I get stuck. If, what I have so far is correct, I would move on trying to take the derivative of $G$ now with the Fundamental theorem of calculus. However, since I am using it for the first time, I do not quite know what to do, especially with the limits of my integral being switched up ($x$ is the lower limit instead of the upper limit). How do I proceed here? Any help is greatly appreciated!

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  • $\begingroup$ You can just swap the limits by putting in a $-$ sign in front. Also you could use $\frac{d}{dx}\int_a^x f(t)dt=f(x)$. $\endgroup$ – John Doe Apr 21 '17 at 16:19
  • $\begingroup$ @JohnDoe Then is the answer $G''(x) = f'(x) $ correct? $\endgroup$ – AxiomaticApproach Apr 21 '17 at 16:23
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    $\begingroup$ Not quite - see my answer $\endgroup$ – John Doe Apr 21 '17 at 16:26
  • $\begingroup$ @JohnDoe got it, will approve the answer once I can. Thanks! $\endgroup$ – AxiomaticApproach Apr 21 '17 at 16:26
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$$G(x)=(1-x)\int_0^1f(s)ds-\int_x^1\Big ( \int_0^tf(s)ds\Big )dt$$

$$G'(x)=-\int_0^1f(s)ds+\frac{d}{dx}\int_1^x\Big ( \int_0^tf(s)ds\Big )dt$$

$$=-\int_0^1f(s)ds+\int_0^xf(s)ds$$

$$\Rightarrow G''(x)=f(x)$$

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You can use the Leibniz integral rule: $$\frac{d}{dx}\int_{a(x)}^{b(x)} c(x,t)\,dt = \int_{a(x)}^{b(x)}\frac{dc}{dx}(x,s)\,ds + c[x,b(x)]\frac{db}{dx} - c[(x,a(x)]\frac{da}{dx}.$$

This will give $$G'(x) = \int_{x}^1 -f(t)\,dt + 0 - 0$$ $$G''(x) = 0+0+f(x).$$

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