For the integral of (x - 4y) with x from -1 to 1 and y from 0 to 2: sketch the region of integration, evaluate the double integral over the rectangle, then reverse the order of integration and evaluate again.

Step-by-step solution

1. 1

   Identify the region. Both pairs of limits are constants, so the region is the rectangle $R=[-1,1]\times[0,2]$: two units wide, two units tall, symmetric about the $y$-axis.

2. 2

   Spot the symmetry before computing. The rectangle is symmetric in $x$ about $x=0$, and $x$ is an odd function of $x$, so $\int_{-1}^{1}x\,dx=0$. That kills the first half of the integrand and leaves only the $-4y$ part to integrate.

3. 3

   Do the inner integral in $x$. $\displaystyle\int_{-1}^{1}(x-4y)\,dx=\left[\frac{x^2}{2}-4yx\right]_{-1}^{1}=\left(\tfrac12-4y\right)-\left(\tfrac12+4y\right)=-8y$, confirming that the $x$ contribution cancels exactly.

4. 4

   Do the outer integral in $y$. $\displaystyle\int_{0}^{2}(-8y)\,dy=\left[-4y^2\right]_0^2=-16$.

5. 5

   Reverse the order and repeat. With constant limits the reversed integral is $\displaystyle\int_{-1}^{1}\!\!\int_{0}^{2}(x-4y)\,dy\,dx$. Inner: $\left[xy-2y^2\right]_0^2=2x-8$. Outer: $\displaystyle\int_{-1}^{1}(2x-8)\,dx=\left[x^2-8x\right]_{-1}^{1}=(1-8)-(1+8)=-16$.

6. 6

   Check the result. Both orders give $-16$. Alternatively, the average value of $x-4y$ over the rectangle is $0-4\cdot 1=-4$, and multiplying by the area $4$ gives $-16$; a numerical Simpson evaluation of the nested integral also returns $-16.0$.