For the integral of (x - 4y) with x from -1 to 2 and y from 0 to 1: 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. The inner variable $x$ runs from $-1$ to $2$ and the outer variable $y$ runs from $0$ to $1$, both with constant limits. The region is therefore the axis-aligned rectangle $R=[-1,2]\times[0,1]$, three units wide and one unit tall, sitting just above the $x$-axis.

2. 2

   Do the inner integral in $x$, treating $y$ as a constant. $\displaystyle\int_{-1}^{2}(x-4y)\,dx=\left[\frac{x^2}{2}-4yx\right]_{-1}^{2}=(2-8y)-\left(\tfrac12+4y\right)=\tfrac32-12y$.

3. 3

   Do the outer integral in $y$. $\displaystyle\int_{0}^{1}\left(\tfrac32-12y\right)dy=\left[\tfrac32y-6y^2\right]_0^1=\tfrac32-6=-\tfrac92$.

4. 4

   Reverse the order. Because both limits are constants, the region does not have to be re-described: the reversed integral is simply $\displaystyle\int_{-1}^{2}\!\!\int_{0}^{1}(x-4y)\,dy\,dx$. This is the easy case of changing order — only regions with variable limits need re-cutting.

5. 5

   Evaluate in the new order. Inner: $\displaystyle\int_0^1(x-4y)\,dy=\left[xy-2y^2\right]_0^1=x-2$. Outer: $\displaystyle\int_{-1}^{2}(x-2)\,dx=\left[\tfrac{x^2}{2}-2x\right]_{-1}^{2}=(2-4)-\left(\tfrac12+2\right)=-2-\tfrac52=-\tfrac92$.

6. 6

   Confirm that both orders agree. Both routes give $-\tfrac92$, which is what Fubini's theorem predicts for a continuous integrand on a rectangle. A composite Simpson evaluation of the nested integral also returns $-4.5$.

7. 7

   Sanity-check the sign. Over this rectangle the average of $x$ is $\tfrac12$ and the average of $y$ is $\tfrac12$, so the average of $x-4y$ is about $\tfrac12-2=-\tfrac32$; multiplied by the area $3$ that predicts $-\tfrac92$ exactly, matching the computed value.