# Question and Answers Forum

Integration Questions

Question Number 8672 by swapnil last updated on 20/Oct/16

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{x}.\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{dx} \\$$ $$\mathrm{evaluate}\:\mathrm{above}\:\mathrm{expression}. \\$$

Answered by 123456 last updated on 21/Oct/16

$${u}=−{x}^{\mathrm{2}} \\$$ $${du}=−\mathrm{2}{xdx}\Rightarrow{xdx}=−\frac{{du}}{\mathrm{2}} \\$$ $${x}=\mathrm{0},{u}=\mathrm{0} \\$$ $${x}=\epsilon,{u}=−\epsilon^{\mathrm{2}} \\$$ $$\underset{\mathrm{0}} {\overset{\infty} {\int}}{xe}^{−{x}^{\mathrm{2}} } {dx}=\underset{\epsilon\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{0}} {\overset{\epsilon} {\int}}{xe}^{−{x}^{\mathrm{2}} } {dx} \\$$ $$=\underset{\epsilon\rightarrow\infty} {\mathrm{lim}}−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{−\epsilon^{\mathrm{2}} } {\int}}{e}^{{u}} {du} \\$$ $$=\underset{\epsilon\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{2}}\underset{−\epsilon^{\mathrm{2}} } {\overset{\mathrm{0}} {\int}}{e}^{{u}} {du} \\$$ $$=\underset{\epsilon\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}−{e}^{−\epsilon^{\mathrm{2}} } }{\mathrm{2}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}} \\$$

Commented byswapnil last updated on 21/Oct/16

$$\mathrm{is}\:\epsilon>\mathrm{0}\:? \\$$ $$\\$$