For the integral of (x − 4y) over the rectangle −1 ≤ x ≤ 1, 0 ≤ y ≤ 2: describe the region, evaluate the double integral, then reverse the order of integration and evaluate again.

Step-by-step solution

1. 1

   (a) Identify the region. All four limits are constants, so the region is the rectangle

   $D=\{(x,y):-1\le x\le 1,\ 0\le y\le 2\}$

   with corners $(-1,0)$, $(1,0)$, $(1,2)$ and $(-1,2)$.

2. 2

   (b) Use the symmetry of the x-interval. The interval $[-1,1]$ is symmetric about $0$ and $x$ is an odd function, so

   $\int_{-1}^{1}x\,dx=\left[\frac{x^{2}}{2}\right]_{-1}^{1}=\frac12-\frac12=0$

   This kills the entire $x$ term before any $y$ work is done.

3. 3

   Integrate the remaining term over x. With $y$ held constant,

   $\int_{-1}^{1}(-4y)\,dx=-4y\big[x\big]_{-1}^{1}=-4y(2)=-8y$

   so the inner integral is $0-8y=-8y$.

4. 4

   Integrate over y.

   $\int_{0}^{2}(-8y)\,dy=\left[-4y^{2}\right]_{0}^{2}=-16$

5. 5

   (c) Reverse the order and confirm.

   $\int_{-1}^{1}\!\!\int_{0}^{2}(x-4y)\,dy\,dx=\int_{-1}^{1}\left[xy-2y^{2}\right]_{0}^{2}dx=\int_{-1}^{1}(2x-8)\,dx=\big[x^{2}-8x\big]_{-1}^{1}=(1-8)-(1+8)=-16$

   $\boxed{-16}$

6. 6

   Sanity-check the sign and size. Over this rectangle $y$ ranges up to $2$, so $-4y$ reaches $-8$ while $x$ contributes nothing on average; the mean value of the integrand is therefore $-4\bar y=-4$, and the area of the rectangle is $2\times 2=4$, giving $-4\times 4=-16$ ✓.