Probability Density Function to Cumulative Distribution Function (Integration Problem) 
Determine the CDF for the random variable having the PDF $$f(x)=2\beta xe^{(-\beta x^2)}\space ,x>0$$ where $\beta$ is a positive constant. 

My attempt. 
$$F_x(x)=2\beta\int_{0}^{x}xe^{-\beta x^2}dx$$
I tried integration by parts. But it seems not working as I couldn't integrate $e^{-\beta x^2}$ in elementary function. But the given answer is very simple, so I doubt that I would have some errors in my working. Hope someone can guide me for it. Thanks in advance.
 A: Let $-a x^2 \equiv z$ and $dz = -2 a x\ dx$.  Then use $\int x e^{-a x^2}\ dx = {-e^{-a x^2} \over 2 a}$ and note that your limit of integration should be a different variable from $x$.  
Hence get:
$$\int\limits_{x=0}^y 2 \beta x e^{- \beta x^2}\ dx = 1 - e^{-\beta y^2}.$$
A: This thing is begging loudly for a particular substitution:
$$
F_X(x)=2\beta\int_0^x ue^{-\beta u^2} \, du = \int_0^x e^{-\beta u^2} \Big( 2\beta u \, du\Big) = \int_0^{\beta x^2} e^{-w} \, dw = \cdots\cdots
$$
You wrote $F_x(x)$ where you needed $F_X(x).$ This isn't just a silly convention; its neglect can lead you into confusion. Note that $F_X(x) = \Pr(X\le x)$ and the expression $X\le x$ is incomprehensible if you don't know the difference between $X$ and $x$.
Also, you used the letter $x$ in the expression $\displaystyle \int_0^x$ but also used it to refer to the thing that goes from $0$ to $x.$
You ought to review this sort of substitution, because this is the simplest sort of example of an occasion when its use in indicated.
A: The derivative of $e^{-\beta x^2}$ is $-2\beta xe^{-\beta x^2}=-f(x)$. 
That means exactly that $F(x)=c-e^{-\beta x^2}$ for a suitable constant $c$.
Substituting $x=0$ we find $c=1$, so we end up with:$$F(x)=1-e^{-\beta x^2}\text{ for }x>0$$
To make things complete: $F(x)=0$ for $x\leq0$.
