Answer
\[
\int_{-3}^{0}(\sqrt{9-x^{2}}+2) d x=\frac{9 \pi}{4}+6
\]
Work Step by Step
By using Theorem ( 5.5.4 ), we have:
\[
\int_{-3}^{0}(\sqrt{-x^{2}+9}+2) d x=2 \int_{-3}^{0} d x+\int_{-3}^{0} \sqrt{-x^{2}+9} d x
\]
By using (area of rectangle), we're going to find $\int_{-3}^{0} d x$
\[
\int_{-3}^{0} d x=1 \cdot 3=3
\]
By using (area of the quarter-circle), we're going to find $\int_{-3}^{0} \sqrt{9-x^{2}} d x$
\[
\int_{-3}^{0} \sqrt{9-x^{2}} d x=3^{2}\cdot \pi \cdot\frac{1}{4} =\frac{9 \pi}{4}
\]
$\int_{-3}^{0}(\sqrt{-x^{2}+9}+2) d x=2 \int_{-3}^{0} d x+\int_{-3}^{0} \sqrt{-x^{2}+9} d x=3 \cdot 2+\frac{9 \pi}{4}=\frac{9 \pi}{4}+6$
