Answer
$\frac{28}{3}$
Work Step by Step
Step 1. Given the function $y=-x^2-2x=-x(x+2)$ and the interval $[-3,2]$, find the zeros at $x=-2,0$; thus we need to divide the integration into three regions: $[-3,-2],[-2,0],[0,2]$
Step 2. We have $A_1=\int_{-3}^{-2}(-x^2-2x)dx=(-\frac{1}{3}x^3-x^2)|_{-3}^{-2}=(-\frac{1}{3}(-2)^3-(-2)^2)-(-\frac{1}{3}(-3)^3-(-3)^2)=-\frac{4}{3}$
Step 3. Similarly, $A_2=\int_{-2}^{0}(-x^2-2x)dx=(-\frac{1}{3}x^3-x^2)|_{-2}^{0}=(-\frac{1}{3}(0)^3-(0)^2)-(-\frac{1}{3}(-2)^3-(-2)^2)=\frac{4}{3}$
Step 4. For the third part, $A_3=\int_{0}^{2}(-x^2-2x)dx=(-\frac{1}{3}x^3-x^2)|_{0}^{2}=(-\frac{1}{3}(2)^3-(2)^2)-(-\frac{1}{3}(0)^3-(0)^2)=-\frac{20}{3}$
Step 5. The total area is given by $A=|A_1|+|A_2+|A_3|=\frac{28}{3}$
