For continuous Random Variable

$ \int_{\infty}^{-\infty} f(x) \,\, dx = 1 \\ \int_{0}^{\infty} f(x) \,\, dx = 1 \hspace{0.25cm} [x \geq 0] \\ \int_{0}^{\infty} kx^2 e^{-x} \,\,dx = 1 \\ k [x^2 \frac{e^{-x}}{-1} - 2x \frac{e^{-x}}{(-1)^2} + 2 \frac{e^{-x}}{(-1)^3}]_0^{\infty} = 1 \\ k [-2 e^{-0}] = 1 \\ \therefore k = \frac{1}{2} $

$E(X) = \int_0^{\infty} = xf(x) \,\, dx $

$ = \int_0^{\infty} x[kx^2 e^{-x}] \,\,dx \\ = k \int_0^{\infty} x^3 e^{-x}] \,\,dx \\ = k [x^3 \frac{e^{-x}}{-1} - 3x^2 \frac{e^{-x}}{(-1)^2} + 6x \frac{e^{-x}}{(-1)^3} - 6 \frac{e^{-x}}{(-1)^4} ]_0^{\infty} \\ = \frac{1}{2} [6] = 3 $

$ E(X^2) = \int_{0}^{\infty} x^2 f(x) \,\, dx \\ = \int_0^{\infty} x^2 [kx^2e^{-x}] \,\, dx \\ = k \int_0^{\infty} x^4 e^{-x}] \,\, dx \\ = k[x^4 \frac{e^{-x}}{-1} - 4x^3 \frac{e^{-x}}{(-1)^2} + 12x^2 \frac{e^{-x}}{(-1)^3} - 24x \frac{e^{-x}}{(-1)^4} + 24 \frac{e^{-x}}{(-1)^5}]_0^{\infty} \\ = \frac{1}{2} [-24 (\frac{1}{(-1)^5})] = 12 \\ Var(X) = E(X^2) - E(X)^2 = 12 - 9 = 3 $