Compute the double integral of x² − 8y over the rectangle where x runs from −2 to −1 and y runs from 0 to 1.

Step-by-step solution

1. 1

   Recognise the easy case. $D$ is a rectangle, so both limits are constants and the order of integration does not matter. Take $x$ first:

   $\int_0^1 \int_{-2}^{-1} (x^2-8y)\,dx\,dy$

2. 2

   Integrate with respect to $x$, holding $y$ constant.

   $\int (x^2-8y)\,dx = \frac{x^3}{3}-8yx$

3. 3

   Substitute the $x$ limits. At $x=-1$: $-\tfrac{1}{3}+8y$. At $x=-2$: $-\tfrac{8}{3}+16y$. Subtracting,

   $\left(-\tfrac{1}{3}+8y\right)-\left(-\tfrac{8}{3}+16y\right) = \tfrac{7}{3}-8y$

   The negative $x$ values are the usual place to slip a sign — note $(-2)^3 = -8$, not $8$.

4. 4

   Integrate the result over $y$ from $0$ to $1$.

   $\int_0^1\left(\tfrac{7}{3}-8y\right)dy = \left[\tfrac{7}{3}y-4y^2\right]_0^1 = \tfrac{7}{3}-4 = -\tfrac{5}{3}$