Evaluate the double integral of x*y*e^(x^2*y) over the region D = {-2 <= x <= 0, 0 <= y <= 5pi/2}.

Step-by-step solution

1. 1

   Choose the inner variable to make a substitution available. The exponent $x^2y$ differentiates to $2xy$ with respect to $x$, and the integrand already carries a factor $xy$. So integrate in $x$ first; integrating in $y$ first would require integration by parts instead.

2. 2

   Substitute u = x^2 y in the inner integral. Then $du=2xy\,dx$, so $xy\,dx=\tfrac12\,du$. The limits map as $x=-2\Rightarrow u=4y$ and $x=0\Rightarrow u=0$ (note $x$ is negative but $x^2$ is not).

3. 3

   Evaluate the inner integral. $\int_{-2}^{0}xy\,e^{x^2y}\,dx=\frac12\int_{4y}^{0}e^u\,du=\frac12\left(e^0-e^{4y}\right)=\frac{1-e^{4y}}{2}$. The reversed $u$-limits are what make the result negative for $y>0$.

4. 4

   Integrate the result in y. $\int_0^{5\pi/2}\frac{1-e^{4y}}{2}\,dy=\frac12\left[y-\frac{e^{4y}}{4}\right]_0^{5\pi/2}=\frac12\left(\frac{5\pi}{2}-\frac{e^{10\pi}}{4}+\frac14\right)$.

5. 5

   Distribute the one half. The exact value is $\frac{5\pi}{4}+\frac18-\frac{e^{10\pi}}{8}$. The exponential term dominates completely, since $e^{10\pi}\approx 4.4031\times 10^{13}$.

6. 6

   Numerical check. The closed form evaluates to $-5.5039382325749\times 10^{12}$, and adaptive quadrature of $\int_0^{5\pi/2}\frac{1-e^{4y}}{2}dy$ returns $-5.5039382325749\times 10^{12}$ as well, agreeing to 13 significant figures.