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Given the following function: $k(x,t) = (1-x)t$ for $x>t$ and $k(x,t) = (1-t)x$ for $x<t$ I would like to evaluate the following integral: $$ I = \int_0^1\int_0^1|k(x,t)|^2dxdt $$ I should definitely be able to do this (I even know the answer: 1/90) but I am failing at it. Could someone give me a tip about how to evaluate this integral?

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up vote 2 down vote accepted

Split the square of integration as two triangles along $x=t$ line, one region $x>t$ and other region $t>x$ $$ \int_0^{1}\int_0^t ((1-x)t)^2 dx dt + \int_0^{1}\int_0^x ((1-t)x)^2 dt dx $$ enter image description here

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If I evaluate the integral this way I get 19/90, which is not correct. I am fairly certain I have evaluated it correctly. What is wrong? (It should be 1/90) – Funzies Jan 17 '13 at 8:42
@Erik how did you evaluate it? I am getting 1/90 from mathematica – Santosh Linkha Jan 17 '13 at 13:39

Function $k(x,\,t)$ takes the same values over and under line $t=x.$

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Break the integral up into 2:

$$ \int_0^1 dx \: \int_x^1 |(1-t)x |^2 + \int_0^1 dx \: \int_0^x |(1-x)t |^2 $$

Evaluating these double integrals produces the desired result. A hint: this is equal to

$$ \frac{1}{3} \int_0^1 dx [x^2 (1-x)^3 + x^3 (1-x)^2] $$

Expand the binomials and evaluate.

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I agree with the second term in the second equation, the first term I am not sure is correct. Furthermore, I get 8/45 after evaluating the integrals. (It should be 1/90) – Funzies Jan 17 '13 at 8:43
I got $1/90$; so did Wolfram Alpha:… – Ron Gordon Jan 17 '13 at 13:38

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