Review question

# If we know $f(x)$, can we find $\int_{-1}^1 f(x^2-1) \:dx$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5549

## Solution

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).