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I need to find the integral for this pdf but I don't know if I need to, or can, take the integral of two variables at the same time.

$$ f(x;\theta)=\frac{x}{\theta^2} e^{-x^2/(2\theta^2)} ,\quad x>0; $$ $$ f(x;\theta)= 0,\text{ otherwise}. $$

I'm guessing you just do piecewise integration for $x$ and forget $\theta$ but I want to be sure.

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$\theta$ is a parameter in this density function (the random variable is called a Rayleigh random variable). If you want to find the CDF, it is $$F_X(t;\theta) = \int_0^t \frac{x}{\theta^2} e^{-x^2/(2\theta^2)}\,\mathrm dx$$ where the antiderivative of the integrand is known. Figure it out for yourself by computing the derivative of $\exp(-x^2/2\theta^2)$. The end result is $F_X(t;\theta) = (1-\exp(-t^2/2\theta^2))\mathbf 1_{t \geq 0}$

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$$ \int_0^x \frac u \theta e^{-(u/\theta)^2/2} \, \frac{du}\theta = \int _0^{x/\theta} v e^{-v^2/2}\, dv = \int _0^{x/\theta} e^{-v^2/2}\Big( v\, dv\Big) = \int_0^{x^2/(2\theta^2)} e^{-w} \, dw = \cdots\cdots $$

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