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Question Number 105167 by mohammad17 last updated on 26/Jul/20

Commented by bemath last updated on 26/Jul/20

Answered by Aziztisffola last updated on 26/Jul/20

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{xe}}^{-{\mathrm{x}}^2}\mathrm{dx}={\left[\frac{{\mathrm{e}}^{-{\mathrm{x}}^2}}{-2}\right]}_{-\mathrm{\infty }}^{\mathrm{\infty }}=0+0=0$$$${\int }_{-2}^3\frac{\mathrm{dx}}{{\mathrm{x}}^3}={\int }_{-2}^0\frac{\mathrm{dx}}{{\mathrm{x}}^3}+{\int }_0^3\frac{\mathrm{dx}}{{\mathrm{x}}^3}={\left[\frac{-1}{{2\mathrm{x}}^2}\right]}_{-2}^0+{\left[\frac{-1}{{2\mathrm{x}}^2}\right]}_0^3$$$$=\underset{x\to 0}{\lim }\frac{-1}{{2\mathrm{x}}^2}+\frac{1}{8}-\frac{1}{18}+\underset{x\to 0}{\lim }\frac{1}{{2\mathrm{x}}^2}$$

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