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If $f(x)$ is continuous on $[0,1]$ and $$\int^{1}_{0} f(x) \ \mathrm{d}x = \sqrt{2}$$ compute $$\int_{0}^{1} \int^{1}_{x} f(x)f(y) \ \mathrm{d}y \ \mathrm{d}x$$

First I change the order of integration: $$\int_{?}^{?} \int^{?}_{?} f(x)f(y) \ \mathrm{d}x \ \mathrm{d}y$$ but what happens to the limits? What do I change them to?

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up vote 4 down vote accepted

Draw a picture. The region $$\{(x,y):0\leqslant x\leqslant 1\;,\;x\leqslant y\leqslant 1\}$$ is the same region as $$\{(x,y):0\leqslant y\leqslant 1\;,\;0\leqslant x\leqslant y\}$$

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How do you draw a picture? I don't know what $f$ is – user2850514 Apr 30 '14 at 1:34
@user2850514 Draw the region of integration, my point is. – Pedro Tamaroff Apr 30 '14 at 1:35

Setting $$ F(x)=\int_0^xf(t)\,dt $$ we have \begin{eqnarray} \int_0^1\int_x^1f(x)f(y)\,dxdy&=&\int_0^1f(x)\left(\int_x^1f(y)\,dy\right)\,dx =\int_0^1f(x)(F(1)-F(x))\,dx\\ &=&F(1)\int_0^1f(x)\,dx-\int_0^1f(x)F(x)\,dx =(F(1))^2-\int_0^1F'(x)F(x)\,dx\\ &=&(F(1))^2-\frac12\left[(F(x))^2\right]_0^1 =(F(1))^2-\frac12(F(1))^2\\ &=&\frac12(F(1))^2 =\frac12(\sqrt{2})^2=1. \end{eqnarray}

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$$ \int_{0}^{1} \int^{y}_{0} f(x)f(y) \ \mathrm{d}x \ \mathrm{d}y $$

Added: Here is the region

enter image description here

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How does this help the OP? There's no explanation whatsoever... – Pedro Tamaroff Apr 30 '14 at 1:36
Perhaps it should be explained, rather than simply asserting that this is the result of interchanging limits. – hardmath Apr 30 '14 at 1:59

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