Solving Probability Density Function for continuous random variable The probability density of a random variable $x$ is $$f(x)=a\ \cdotp x^2\ \cdotp \mathrm{e}^{−kx}\ (k&gt;0,\ 0\leq x\leq \infty)$$ Then, the coefficient $a$ equals
$$(i)\frac{k^3}{2}\ \ \ \ (ii)\ k^3 \ \ \ \ (iii)\ k^2 \ \ \ \ (iv)\ k \ \ \ \ (v) \frac{2\ \cdotp k}{\pi}$$


  Here is what I've tried:
  
  $$\int_0^\infty f(x)\mathrm{d}x = 1$$ $$\int_0^\infty\left(a\ \cdotp x^2\ \cdotp \mathrm{e}^{−kx}\right)\mathrm{d}x = 1$$
  
  solving by parts gives result as
  
  $$\left[-\frac{a\cdotp\mathrm{e}^{-kx}}{k}\cdotp\left(x^2+\frac{2x}{k}+\frac{2}{k^2}\right)\right]^\infty_0 = 1$$
  
  Now, here I'm stuck. Can't find a way to solve $a$.
What you need instead is 
$$
a=\frac{1}{\int_{0}^{\infty}x^2 e^{-kx}dx}=\frac{k^3}{2}
$$
What you need instead is 
$$
a=\frac{1}{\int_{0}^{\infty}x^2 e^{-kx}dx}=\frac{k^3}{2}
$$