Answer
$ \frac{32}{3} $
Work Step by Step
Step 1 of 3:
First, find the points of intersection of the curves $ x=-y $
and $ x=y^{2}+3y $
$ y^{2}+3y=-y $
$ y^{2}+4y=0 $
$ y \left( y+4 \right) =0 $
$ y=0 $
or $ y=-4 $
The y-coordinates of the points of intersection are 0 and -4.
Step 2 of 3:
Sketch the curves.
Find the area of the shaded region.
Step 3 of 3:
The area $ A= \int _{-4}^{0} \left[ \left( -y \right) - \left( y^{2}+3y \right) \right] dy $
$ A= \int _{-4}^{0} \left( -y-y^{2}-3y \right) dy $
$ A= \int _{-4}^{0} \left( -4y-y^{2} \right) dy $
$ A= \left[ -2y^{2}-\frac{y^{3}}{3} \right] _{-4}^{0} $
$ A= \left[ 0-0- \left( -2 \left( -4 \right) ^{2}-\frac{ \left( -4 \right) ^{3}}{3} \right) \right] $
$ A= \left[ - \left( -32+\frac{64}{3} \right) \right] $
$ A=32-\frac{64}{3} $
$ A=\frac{32}{3} $
