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Question Number 98672 by  M±th+et+s last updated on 15/Jun/20

$${\int }_0^4{\int }_0^{\frac{x}{4}}{e}^{x^2}dxdy$$

Answered by Ar Brandon last updated on 16/Jun/20

$$\mathrm{Let}\mathcal{I}={\int }_0^{\frac{\mathrm{x}}{4}}{\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}........\left(1\right)$$$$\Rightarrow \mathcal{I}={\int }_0^{\frac{\mathrm{x}}{4}}{\mathrm{e}}^{{\mathrm{y}}^2}\mathrm{dy}........\left(2\right)$$$$\Rightarrow {\mathcal{I}}^2={\int }_0^{\frac{\mathrm{x}}{4}}{\int }_0^{\frac{\mathrm{x}}{4}}{\mathrm{e}}^{{\mathrm{x}}^2+{\mathrm{y}}^2}\mathrm{dxdy}={\int }_0^{\frac{\pi }{2}}{\int }_0^{\frac{\mathrm{x}}{4}}{\mathrm{re}}^{{\mathrm{r}}^2}\mathrm{drd}\theta$$$$\Rightarrow {\mathcal{I}}^2={\left[\frac{\theta }{2}\right]}_0^{\frac{\pi }{2}}{\left[{\mathrm{e}}_{}^{{\mathrm{r}}^{2^{}}}\right]}_0^{\frac{\mathrm{x}}{4}}=\frac{\pi }{4}[{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{16}}-1_{}]$$$$\Rightarrow \mathcal{I}=\pm \frac{\sqrt{\pi }}{2}{[{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{16}}-1_{}]}^{\frac{1}{2}}$$$$\Rightarrow {\int }_0^4{\int }_0^{\frac{x}{4}}{e}^{x^2}dxdy=\pm 2\sqrt{\pi }{[{\mathrm{e}}^{\frac{{\mathrm{x}}^2}{16}}-1_{}]}^{\frac{1}{2}}$$

Commented by  M±th+et+s last updated on 16/Jun/20

$$niceworkandcorrectthankyou$$

Commented by Ar Brandon last updated on 16/Jun/20

I don't know if I have done it in the required manner. Can someone kindly check ?

Commented by Ar Brandon last updated on 16/Jun/20

OK, thanks for your feedback.

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