Consider the following for the two (02) items that follow: Let the...

The curve intersects the $x$-axis when $f(x) = 0$.

$x^2 - 1 = 0 \Rightarrow x = -1, 1$

The required area is given by $A = \int_{-1}^{1} |x^2 - 1| dx$.

Since $x^2 - 1 \le 0$ for $x \in [-1,1]$, we have:

$A = \int_{-1}^{1} (1 - x^2) dx$

$A =$\left[ x - \dfrac{x^3}{3} \right]_{-1}^{1}$<br><br>$A = \left(1 - \dfrac{1}{3} \right) -\left( -1 + \dfrac{1}{3} \right)

$A =$\dfrac{2}{3} -\left( -\dfrac{2}{3} \right) = \dfrac{4}{3}$square units.<br><br>Answer:$\dfrac{4}{3} square units