Answer
$\int_{-3}^{1}f(x)~dx = \frac{6-\pi}{4}$
Work Step by Step
On the interval $-3 \leq x \leq -1$, the graph of $f(x) = -x-1$ forms a triangle above the x-axis. We can find the area of this triangle:
$A = \frac{1}{2}(2)(2) = 2$
On the interval $-1 \leq x \leq 0$, the graph of $f(x) = -x-1$ forms a triangle below the x-axis. We can find the area of this triangle:
$A = \frac{1}{2}(1)(1) = \frac{1}{2}$
On the interval $0 \leq x \leq 1$, the graph of $-\sqrt{1-x^2}$ forms a quarter of a circle below the x-axis. We can find the area:
$A = \frac{\pi~r^2}{4} = \frac{\pi(1)^2}{4} = \frac{\pi}{4}$
Therefore:
$\int_{-3}^{1}f(x)~dx = 2-\frac{1}{2}- \frac{\pi}{4} = \frac{6-\pi}{4}$