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
![=−(1/2)∫_0 ^2 e^(4−x^2 ) d(4−x^2 ) =−(1/2)[e^(4−x^2 ) ]_0 ^2 =−(1/2)[1−e^4 ] =((e^4 −1)/2)](Q153037.png)
$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {e}^{\mathrm{4}−{x}^{\mathrm{2}} } {d}\left(\mathrm{4}−{x}^{\mathrm{2}} \right) \\ $$ $$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{e}^{\mathrm{4}−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$ $$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{e}^{\mathrm{4}} \right] \\ $$ $$=\frac{{e}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}} \\ $$
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
$$\mathrm{u}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \\ $$ $$\mathrm{du}=−\mathrm{2xdx} \\ $$ $$\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{2x}} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{xe}^{\mathrm{u}} −\frac{\mathrm{du}}{\mathrm{2x}} \\ $$ $$\frac{\mathrm{1}}{\mathrm{2}}\int−\mathrm{e}^{\mathrm{u}} \mathrm{du} \\ $$