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Question Number 38651 by rahul 19 last updated on 28/Jun/18

$$\mathrm{If}{\int }_0^1{\mathrm{e}}^{-x^2}dx=a,\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}$$$$\mathrm{of}{\int }_0^1{x}^2{e}^{-x^2}dxintermsof{}^{\prime }{a}^{\prime }?$$

Answered by MrW3 last updated on 28/Jun/18

$${\int }_0^1{x}^2{e}^{-x^2}dx$$$$=-\frac{1}{2}{\int }_0^1{xde}^{-x^2}$$$$=-\frac{1}{2}{[{xe}^{-x^2}-{\int }_0^1{e}^{-x^2}dx]}_0^1$$$$=-\frac{1}{2}{\left[{xe}^{-x^2}\right]}_0^1+\frac{1}{2}{\int }_0^1{e}^{-x^2}dx$$$$=-\frac{1}{2e}+\frac{a}{2}$$$$=\frac{1}{2}(a-\frac{1}{e})$$

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

$$excellent...$$

Commented by rahul 19 last updated on 28/Jun/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

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