Question

$\underset{1}{\overset{3}{\int }}\left|x^2-2x\right|dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left|x^2-2x\right|=\left\{\begin{array}{l}-\left(x^2-2x\right),1\le x\le 2\\ x^2-2x,2\le x\le 3\end{array}\right. \\ \therefore {\int }_1^3\left|x^2-2x\right|dx \\ ={\int }_1^2-\left(x^2-2x\right)dx+{\int }_2^3\left(x^2-2x\right)dx \\ ={\left[-\frac{x^3}{3}+x^2\right]}_1^2+{\left[\frac{x^3}{3}-x^2\right]}_2^3 \\ =\frac{-8}{3}+4+\frac{1}{3}-1+9-9-\frac{8}{3}+4 \\ =2 \end{gathered}$$