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0-x-2-e-x-2-dx-




Question Number 189752 by TUN last updated on 21/Mar/23
∫^∞ _0 x^2 .e^(−x^2 ) dx=¿
$$\underset{\mathrm{0}} {\int}^{\infty} {x}^{\mathrm{2}} .{e}^{−{x}^{\mathrm{2}} } {dx}=¿ \\ $$
Answered by MJS_new last updated on 21/Mar/23
∫x^2 e^(−x^2 ) dx=       u′=xe^(−x^2 )  → u=(1/2)e^(−x^2 )        v=x → v′=1  =−(1/2)xe^(−x^2 ) −(1/2)∫e^(−x^2 ) dx=  =−(1/2)xe^(−x^2 ) −((√π)/4)erf x +C  ⇒ answer is ((√π)/4)
$$\int{x}^{\mathrm{2}} \mathrm{e}^{−{x}^{\mathrm{2}} } {dx}= \\ $$$$\:\:\:\:\:{u}'={x}\mathrm{e}^{−{x}^{\mathrm{2}} } \:\rightarrow\:{u}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{e}^{−{x}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:{v}={x}\:\rightarrow\:{v}'=\mathrm{1} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{e}^{−{x}^{\mathrm{2}} } −\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{e}^{−{x}^{\mathrm{2}} } {dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}\mathrm{e}^{−{x}^{\mathrm{2}} } −\frac{\sqrt{\pi}}{\mathrm{4}}\mathrm{erf}\:{x}\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\pi}}{\mathrm{4}} \\ $$

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