Expodratic

$\begin{aligned} \int_0^\infty x^2 e^{-x^2} dx & = \frac{1}{2} \int_0^\infty t^\frac{1}{2} e^{-t} dt \quad \quad \small \color{#3D99F6}{\text{Let }t = x^2 \quad \Rightarrow dt = 2 x \space dx} \\ & = \frac{1}{2} \int_0^\infty t^{\frac{3}{2}-1} e^{-t} dt \\ & = \frac{1}{2} \Gamma \left(\frac{3}{2} \right) \quad \quad \quad \quad \quad \small \color{#3D99F6}{\Gamma(z) \text{ is Gamma function}} \\ & = \frac{1}{2} \times \frac{1}{2} \Gamma \left(\frac{1}{2} \right) = \frac{1}{2} \times \frac{1}{2} \sqrt{\pi} = \frac{\pi^{\frac{1}{2}}}{4} \end{aligned}$

$\Rightarrow A \times B = \frac{1}{2} \times 4 = \boxed{2}$

Consider the integral: $I \equiv \int_0^\infty e^{-px^2}dx = \sqrt{\frac{\pi}{2p}}\;.$ Therefore: $\int_0^\infty x^2e^{-x^2}dx = -\left.\frac{dI}{dp}\right|_{p=1} = \frac{\sqrt{\pi}}{4}\;.$

$\dfrac{\sqrt{\pi}}{2} = \int_0^\infty e^{-x^2} dx = \left [x \cdot e^{-x^2}\right]^{\infty}_{0} + \int_0^\infty 2x^2 \cdot e^{-x^2}\Rightarrow$ $\int_0^\infty x^2 \cdot e^{-x^2} dx = \frac{\sqrt{\pi}}{4}...$

substitute x^2=t so GE=0.5*L.T of sqrt(t) GE=sqrt(PI)/4 GE- Given Expression LT- Laplace Transform