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 …