Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 72

$$2$$

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