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Question Number 189752 by TUN last updated on 21/Mar/23

$${\underset{0}{\int }}^{\mathrm{\infty }}{x}^2.e^{-x^2}dx=¿$$

Answered by MJS_new last updated on 21/Mar/23

$$\int x^2{\mathrm{e}}^{-x^2}dx=$$$$u^{\prime }=x{\mathrm{e}}^{-x^2}\to u=\frac{1}{2}{\mathrm{e}}^{-x^2}$$$$v=x\to v^{\prime }=1$$$$=-\frac{1}{2}x{\mathrm{e}}^{-x^2}-\frac{1}{2}\int {\mathrm{e}}^{-x^2}dx=$$$$=-\frac{1}{2}x{\mathrm{e}}^{-x^2}-\frac{\sqrt{\pi }}{4}\mathrm{erf}x+C$$$$\Rightarrow \mathrm{answer}\mathrm{is}\frac{\sqrt{\pi }}{4}$$

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