Answer
$3+ \frac{9\pi}{4}$
Work Step by Step
$\int_{-3}^{0} (1+\sqrt{9-x^2})~dx = \int_{-3}^{0} 1~dx+\int_{-3}^{0} \sqrt{9-x^2}~dx$
On the interval $-3 \leq x \leq 0$, the graph of $f(x) = 1$ forms a rectangle above the x-axis. We can find the area of this rectangle:
$A = (0-(-3))(1) = 3$
On the interval $-3 \leq x \leq 0$, the graph of $\sqrt{9-x^2}$ forms a quarter of a circle with radius=3 above the x-axis. We can find the area:
$A = \frac{\pi~r^2}{4} = \frac{9\pi}{4}$
Therefore:
$\int_{-3}^{0} (1+\sqrt{9-x^2})~dx = 3+ \frac{9\pi}{4}$
