If we know $f(x)$, can we find $\int_{-1}^1 f(x^2-1) \:dx$?

If we know $f(x)$, can we find $\int_{-1}^1 f(x^2-1) \:dx$?

Question

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) $\dfrac{1}{3}\quad \quad$ (c) $\dfrac{3}{5}\quad \quad$ (d) $\dfrac{2}{3}$.

Suggestion

Can we write down an equation for $f(x)$ for each distinct interval?

Can we then write down what $f(x^2-1)$ is?

If the values of $x$ run from $-1$ to $1$, what values does $x^2 -1$ run through?

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).

There are two intervals that we are interested in, $-1 \leq x \leq 0$, and $0 \leq x \leq 1$ (everywhere else the function $f(x)$ is zero). The inequality $-1 \leq x \leq 0$ becomes on squaring $1 \geq x^2 \geq 0$, and so $0 \geq x^2 - 1 \geq -1$. The inequality $0 \leq x \leq 1$ becomes on squaring $0 \leq x^2 \leq 1$, and so $-1 \leq x^2 -1 \leq 0$. These intervals are the same, so we can simply focus on what happens to $f(x^2-1)$ for $x$ on the interval $[-1, 0]$. On this domain, $f(x) = 1+x$, and so $f(x^2-1) = 1+x^2-1 = x^2$. Thus $\int_{-1}^1 f(x^2-1) \:dx = \int_{-1}^1 x^2 \:dx = \left[\frac{x^3}{3}\right]_{-1}^1 = \dfrac{2}{3}$ and so the answer is (d). An alternative approach could be to draw up a table. $x$-2-1.5-1-0.500.511.52 $x^2-1$31.250-0.75-1-0.7501.253 $f(x^2-1)$0010.2500.25100 Plotting these points strongly suggests we are interested in the area under $y = x^2$ between $-1$ and $1$.