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0-2-xe-4-x-2-dx-




Question Number 153016 by joki last updated on 04/Sep/21
∫_0 ^2 xe^(4−x^2 ) dx
$$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{xe}^{\mathrm{4}−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\ $$
Commented by mr 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)
$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} {e}^{\mathrm{4}−{x}^{\mathrm{2}} } {d}\left(\mathrm{4}−{x}^{\mathrm{2}} \right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[{e}^{\mathrm{4}−{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−{e}^{\mathrm{4}} \right] \\ $$$$=\frac{{e}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}} \\ $$
Commented by peter frank last updated on 04/Sep/21
thank you sir for confirmation
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{confirmation} \\ $$
Answered by peter frank last updated on 04/Sep/21
u=4−x^2   du=−2xdx  dx=−(du/(2x))  ∫_0 ^2 xe^u −(du/(2x))  (1/2)∫−e^u du
$$\mathrm{u}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{du}=−\mathrm{2xdx} \\ $$$$\mathrm{dx}=−\frac{\mathrm{du}}{\mathrm{2x}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \mathrm{xe}^{\mathrm{u}} −\frac{\mathrm{du}}{\mathrm{2x}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int−\mathrm{e}^{\mathrm{u}} \mathrm{du} \\ $$

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