how to solve $\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$?

Question

$$\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$$

I tried this this: $$\int_{y}^{1} e^{x^2}dx=\sum_{0}^{\infty}\frac{1}{n!}(\frac{1}{2n+1}-\frac{y^{2n+1}}{2n+1})$$ I don't know what to do after or even if this way works...

Answer 1

You can do it as follows:\begin{align}\int_0^1\int_y^1e^{x^2}\,\mathrm dx\,\mathrm dy&=\int_0^1\int_0^xe^{x^2}\,\mathrm dy\,\mathrm dx\\&=\int_0^1xe^{x^2}\,\mathrm dx\\&=\left[\frac12e^{x^2}\right]_{x=0}^{x=1}\\&=\frac{e-1}2.\end{align}

Answer 2

OP's way $$I=\int_{0}^{1}\int_{y}^{1} e^{x^2}dxdy$$ Then $$I=\int_{0}^{1}\int_{y}^{1} e^{x^2}dx=\int_{0}^{1}\sum_{0}^{\infty}\frac{1}{n!}\left(\frac{1}{2n+1}-\frac{y^{2n+1}}{2n+1}\right) dy$$ $$\implies I=\sum_{0}^{\infty}\frac{1}{n!}\left(\frac{1}{2n+1}-\frac{1}{(2n+1)(2n+2)}\right).$$ $$I=\sum_{n=0}^{\infty}\frac{1}{n!(2n+2)}=\frac{1}{2}\sum_{n=0}^{\infty}\frac{1}{(n+1)!}= \frac{1}{2}\sum_{m=1}^{\infty}\frac{1}{m!}=\frac{e-1}{2}$$