Question

Calculus help please, will give brainiest

Answer 1

The most convenient way to capture $D$ is with the parameterization

$D = \left\{(x,y) \mid -1 \le y \le 3 \text{ and } -\dfrac{y+1}2 \le x \le y+1\right\}$

so we only need one iterated integral.

$\displaystyle \iint_D e^{x-y} \, dA = \int_{-1}^3 \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx \, dy$

Compute the integral with respect to $x$.

$\displaystyle \int_{-(y+1)/2}^{y+1} e^{x-y} \, dx = e^{x-y} \bigg|_{x=-(y+1)/2}^{x=y+1} = e^{(y+1)-y} - e^{-(y+1)/2-y} = e - e^{-(3y+1)/2}$

Compute the remaining integral.

$\displaystyle \int_{-1}^3 \left(e - e^{-(3y+1)/2}\right) \, dy = \left(ey + \frac23 e^{-(3y+1)/2}\right)\bigg|_{y=-1}^{y=3} \\\\ ~~~~~~~~ = \left(3e + \frac23 e^{-(9+1)/2}\right) - \left(-e + \frac23 e^{-(-3+1)/2}\right) \\\\ ~~~~~~~~ = \boxed{\frac{10e}3 + \frac2{3e^5}}$

If we had chosen the opposite order of variables, we would have used

$D = \left\{(x,y) \mid -2 \le x \le 4 \text{ and } \max\left(-2x-1,x-1\right)\le y\le3\right\}$

where

$\max(-2x-1, x-1) = \begin{cases} -2x-1 & \text{when } x$

so we would have needed two iterated integrals,

$\displaystyle \iint_D e^{x-y} \, dA = \int_{-2}^0 \int_{-2x-1}^3 e^{x-y} \, dy \, dx + \int_0^4 \int_{x-1}^3 e^{x-y} \, dy \, dx$