Evaluate the double integral of x^2 - 8y over the region D = {-2 <= x <= -1, 0 <= y <= 1}.

Step-by-step solution

1. 1

   Recognise a rectangle with separable terms. Over a rectangle with constant limits, an integral of a sum splits into a sum of integrals, and each term here is a function of a single variable, so each becomes a product of one-dimensional integrals.

2. 2

   Split the integrand. $\iint_D\left(x^2-8y\right)dA=\iint_D x^2\,dA-8\iint_D y\,dA$. Handling the two pieces separately avoids carrying both variables through one antiderivative.

3. 3

   Evaluate the x^2 piece. $\int_{-2}^{-1}x^2\,dx=\left[\frac{x^3}{3}\right]_{-2}^{-1}=\frac{-1}{3}-\frac{-8}{3}=\frac73$, and $\int_0^1 dy=1$, so this piece contributes $\frac73$. Note $x^2$ stays positive even though $x$ is negative.

4. 4

   Evaluate the -8y piece. $\int_{-2}^{-1}dx=1$ and $\int_0^1 y\,dy=\frac12$, so this piece contributes $-8\cdot 1\cdot\frac12=-4$.

5. 5

   Combine. $\frac73-4=\frac{7-12}{3}=-\frac53\approx -1.6667$. The negative sign is expected: the $-8y$ term outweighs $x^2$, which never exceeds $4$ on this rectangle while $8y$ reaches $8$.

6. 6

   Numerical check. Adaptive quadrature over the same rectangle returns $-1.6666667$, matching $-\frac53$.