Evaluate the double integral of -2x - 2y + 2 over the region 1 <= x <= 2, x <= y <= 3.

Step-by-step solution

1. 1

   Sketch the region before integrating. The conditions $1\le x\le 2$ and $x\le y\le 3$ describe a trapezoidal strip: a vertical slice at abscissa $x$ runs from the line $y=x$ up to the line $y=3$, so its length is $3-x$, shrinking from $2$ to $1$ as $x$ goes from $1$ to $2$. Because the lower limit depends on $x$, the $y$-integration must be done first in this order.

2. 2

   Integrate in $y$ with $x$ held fixed.

   $\int_x^3\left(-2x-2y+2\right)dy=\left[-2xy-y^2+2y\right]_{y=x}^{y=3}$

   At the top, $y=3$ gives $-6x-9+6=-6x-3$. At the bottom, $y=x$ gives $-2x^2-x^2+2x=-3x^2+2x$.

3. 3

   Subtract carefully. The lower value is subtracted as a whole, so every sign in it flips:

   $(-6x-3)-\left(-3x^2+2x\right)=3x^2-8x-3$

   This is where the usual error occurs — dropping the minus sign in front of $-3x^2$ turns the final answer positive.

4. 4

   Integrate the resulting single-variable function.

   $\int_1^2\left(3x^2-8x-3\right)dx=\left[x^3-4x^2-3x\right]_1^2=(8-16-6)-(1-4-3)=-14-(-6)=-8$

5. 5

   Explain why a negative answer is correct. The integrand $2-2x-2y$ is negative on almost the whole region: on the strip, $x\ge 1$ and $y\ge 1$, so $2-2x-2y\le 2-2-2=-2<0$. An integral of a function that is negative everywhere on the region must be negative, so $-8$ is the right sign. Its magnitude is also reasonable: the region has area $\int_1^2(3-x)dx=\tfrac32$, and the integrand averages $-8/1.5\approx -5.3$, which sits inside its actual range of roughly $-10$ to $-2$.