A graph of the function \(y=f(x)\) is sketched on the axes below;
The value of \(\displaystyle\int_{-1}^1 f(x^2-1) \:dx\) equals (a) \(\dfrac{1}{4}\qquad\) (b) \(\dfrac{1}{3}\qquad\) (c) \(\dfrac{3}{5}\qquad\) (d) \(\dfrac{2}{3}\).
We can write the function as \[ f(x) = \begin{cases} x+1, & -1 \le x \le 0; \\ 1-x, & 0 \le x \le 1; \\ 0, & \text{otherwise.} \end{cases} \]
By substituting we can say, \[ f(x^2-1) = \begin{cases} x^2-1+1 = x^2, & -1 \le x^2-1 \le 0; \\ 1-x^2+1 = 2-x^2, & 0 \le x^2-1 \le 1; \\ \qquad 0, & \text{otherwise.} \end{cases} \]
The first of these intervals can be rewritten, \[ -1 \le x^2-1 \le 0 \quad\implies\quad 0\le x^2\le 1 \quad\implies\quad -1\le x \le 1. \] We only need to integrate over this interval, so we do not need to worry about the second interval.
So we have \[\int_{-1}^1 f(x^2-1) \:dx = \int_{-1}^1 x^2 \:dx = \left[\frac{x^3}{3}\right]_{-1}^1 = \dfrac{2}{3}\] and the answer is (d).
