Definite integral

Observe that the function $f(x) = 1 - x^2$ takes negative values everywhere except at $-1 <x < +1$ .

To visualize it, here's the graph of $y = 1- x^2$ : imgur

thus, taking its absolute value inverts the sign of the values of $y$ from $x=(-\infty,-1)$ and $x=(1,+\infty)$ $y=|1-x^2|$ imgur

So, we evaluate $\int_{-2}^{2} |1-x^2|dx$ as follows: $\int_{-2}^{-1}-(1-x^2)dx + \int_{-1}^{1}(1-x^2)dx + \int_{1}^{2}-(1-x^2) dx = \boxed{4}$