Question

Urgent help needed will give brainiest

Answer 1

We can write the integration domain as

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

so that the integral is

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

Compute the integral with respect to $x$.

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

Compute the remaining integral.

$\displaystyle \int_{-1}^1 (\cos(3y-2) - \cos(1)) \, dy = \left(\frac13 \sin(3y-2) - \cos(1) y\right) \bigg|_{y=-1}^{y=1} \\\\ ~~~~~~~~ = \left(\frac13 \sin(3-2) - \cos(1)\right) - \left(\frac13 \sin(-3-2) + \cos(1)\right) \\\\ ~~~~~~~~ = \boxed{\frac13 \sin(1) - 2 \cos(1) + \frac13 \sin(5)}$