Expected value of the area of square

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Homework Statement


Given a square with side lengths X, where X is a random variable with some probability density function (the actual pdf is not important for my question). Why isn't the expected value of the area = E[X]^2 = E[X^2]?


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The Attempt at a Solution


Intuitively I would think, if I can find the expected value of one of the sides, I can get the expected value of the area by squaring it.
On the other hand, I am well aware that E[X]^2 != E[X^2] in the general case, and this is indeed a general case (with some geometric interpretation).
 
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Wait nevermind, I got it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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