Uppgift 6

Eftersom $x\ge0$, $x^2\le2-x^2\iff x\ge0$, $x^2\le1\iff0\le x\le1$, är $D=\{(x,y)\mid x^2\le y\le2-x^2,\ 0\le x\le1\}$ (rita en figur). Så
$$\begin{aligned}\iint_Dxe^y\,dx\,dy&=\int_0^1\left(\int_{x^2}^{2-x^2}xe^y\,dy\right)dx=\int_0^1[xe^y]_{y=x^2}^{y=2-x^2}\,dx=\int_0^1(xe^{2-x^2}-xe^{x^2})\,dx\\&=\int_0^1(e^2xe^{-x^2}-xe^{x^2})\,dx=[x^2=t,\ 2x\,dx=dt]=\frac12\int_0^1(e^2e^{-t}-e^t)\,dt\\&=\frac12[-e^2e^{-t}-e^t]_0^1=\frac12(-e^2e^{-1}+e^2-e+1)=\frac12(e^2-2e+1)=\frac12(e-1)^2.\end{aligned}$$