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Question Number 153016 by joki last updated on 04/Sep/21

$$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{xe}^{\mathrm{4}−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\$$

Commented bymr W last updated on 04/Sep/21

$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {e}^{\mathrm{4}−{x}^{\mathrm{2}} } {d}\left(\mathrm{4}−{x}^{\mathrm{2}} \right) \\$$ $$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{e}^{\mathrm{4}−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} \\$$ $$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{e}^{\mathrm{4}} \right] \\$$ $$=\frac{{e}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}} \\$$

Commented bypeter frank last updated on 04/Sep/21

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{confirmation} \\$$

Answered by peter frank last updated on 04/Sep/21

$$\mathrm{u}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \\$$ $$\mathrm{du}=−\mathrm{2xdx} \\$$ $$\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{2x}} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{xe}^{\mathrm{u}} −\frac{\mathrm{du}}{\mathrm{2x}} \\$$ $$\frac{\mathrm{1}}{\mathrm{2}}\int−\mathrm{e}^{\mathrm{u}} \mathrm{du} \\$$