Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 71

$$9$$

Given $$\int_{0}^{6}|3-x| d x $$ Since $$ 3-x \geq 0\ \ \text{ for } \ 0\leq x\leq 3\\ 3-x \leq 0\ \ \text{ for } \ 3\leq x\leq 6$$ Then \begin{align*} \int_{0}^{6}|3-x| d x&=\int_{0}^{3}(3-x) d x+\int_{3}^{6}(x-3) d x\\ &=(3x-\frac{1}{2}x^2)\bigg|_{0}^{3}+ (\frac{1}{2}x^2-3x)\bigg|_{3}^{6}\\ &= \frac{9}{2}+\frac{9}{2}=9 \end{align*}