Evaluating a piecewise integral via area formulas

On $[-3,-1]$, the line $y=-x-1$ forms a right triangle above the x-axis with vertices at $(-3,2)$, $(-1,0)$, and $(-3,0)$; its base is $2$ and height is $2$, so the area is $\frac{1}{2}(2)(2)=2$. On $[1,3]$, the line $y=x-1$ forms a right triangle above the x-axis with vertices at $(1,0)$, $(3,2)$, and $(3,0)$; its base is $2$ and height is $2$, giving area $\frac{1}{2}(2)(2)=2$. Both triangles lie entirely above the x-axis, so both contribute positive area to the integral.

On $[-1,1]$, $y=\sqrt{1-x^2}$ traces the upper half of the unit circle, so this piece is a semicircle of radius $1$ above the x-axis, with area $\frac{1}{2}\pi(1)^2=\frac{\pi}{2}$. Using additivity over adjacent intervals, $\int_{-3}^{3} f(x)\,dx = 2 + \frac{\pi}{2} + 2 = 4+\frac{\pi}{2}.$

The middle piece is a semicircle. On the interval from negative one to one, the function equals the square root of one minus x squared, which traces the upper half of a unit circle sitting above the horizontal axis.