Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 74

$$2$$

Given $$ \int_{0}^{2}\left|x^{2}-1\right| d x$$ Since $$x^2-1>0 \ \text{ for }\ \ x>1\\ x^2-1<0 \ \text{ for }\ \ x<1$$ Then \begin{align*} \int_{0}^{2}\left|x^{2}-1\right| d x&=\int_{0}^{1} 1-x^{2} d x+\int_{1}^{2} x^{2}-1 d x\\ &= (x-\frac{1}{3}x^3)\bigg|_{0}^{1}+ (\frac{1}{3}x^3-x)\bigg|_{1}^{2}\\ &=\frac{2}{3}+ \frac{4}{3}= 2 \end{align*}