Calculate the following integral: double integral over D

Calculate the following integral: $$ \iint_D x^2 y e^{xy}\,dx\,dy, $$ where $0<x<1$ and $0<y<2$.

Step 1 · Identify the region

$D=[0,1]\times[0,2]$.

Step 2 · Set up the integral

$$ I = \int_0^1\int_0^2 x^2ye^{xy}\,dy\,dx. $$

Step 3 · Integrate with respect to $y$

By parts with $u=y$, $dv=e^{xy}\,dy$ ($du=dy$, $v=\frac1xe^{xy}$):

$$ x^2\int_0^2 ye^{xy}\,dy = x^2\left[\frac{y}{x}e^{xy}\right]_0^2 - x^2\int_0^2\frac1xe^{xy}\,dy = 2xe^{2x} - x\left[\frac{e^{xy}}{x}\right]_0^2 = 2xe^{2x} - (e^{2x}-1). $$

$$ = e^{2x}(2x-1) + 1. $$

Step 4 · Integrate with respect to $x$

$$ I = \int_0^1 \Big[e^{2x}(2x-1)+1\Big]\,dx = \int_0^1(2x-1)e^{2x}\,dx + \int_0^1 1\,dx. $$

The second piece is just $1$.

For the first, by parts with $u=2x-1$, $dv=e^{2x}\,dx$ ($du=2\,dx$, $v=\frac{e^{2x}}{2}$):

$$ \int_0^1(2x-1)e^{2x}\,dx = \left[\frac{(2x-1)e^{2x}}{2}\right]_0^1 - \int_0^1 e^{2x}\,dx = \left[\frac{e^2}{2}+\frac12\right] - \left[\frac{e^2}{2}-\frac12\right] = 1. $$

Step 5 · Combine the results

$$ I = 1 + 1 = 2. $$

Additional Notes

• The exponential terms $e^2$ cancel exactly between the two boundary evaluations, leaving a clean integer answer — a good consistency check on the algebra.