Question

Evaluate the following integrals:
$\underset{0}{\overset{2}{\int }}\left|x^2-3x+2\right|dx$

Answer 1

$$\begin{gathered} {\int }_0^2\left|x^2-3x+2\right|\mathit{d}x \\ \mathrm{We}\mathrm{know}\mathrm{that},\left|x^2-3x+2\right|=\left\{\begin{array}{l}-\left(x^2-3x+2\right),\left(x-1\right)\left(x-2\right)\le 0\mathrm{or},1\le x\le 2\\ \left(x^2-3x+2\right),x^2-3x+2\ge 0\mathrm{or},x\in \left(-\infty ,1\right)\cup \left(2,\infty \right)\end{array}\right. \\ \therefore I={\int }_0^2\left(x^2-3x+2\right)\mathit{d}x \\ \Rightarrow I={\int }_0^1\left({\mathit{x}}^{\mathit{2}}\mathit{-}\mathit{3}\mathit{x}\mathit{+}\mathit{2}\right)\mathit{}\mathit{d}x\mathit{-}{\int }_1^2\left({\mathrm{x}}^{\mathit{2}}\mathit{-}\mathit{3}\mathrm{x}\mathit{+}\mathit{2}\right)\mathit{}\mathit{d}\mathrm{x} \\ \Rightarrow I={\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]}_0^1-{\left[\frac{x^3}{3}-\frac{3x^2}{2}+2x\right]}_1^2 \\ \Rightarrow I=\frac{1}{3}-\frac{3}{2}+2-\left[\frac{8}{3}-6+4-\frac{1}{3}+\frac{3}{2}-2\right] \\ \Rightarrow I=\frac{1}{3}-\frac{3}{2}+2-\frac{8}{3}+6-2+\frac{1}{3}-\frac{3}{2} \\ \Rightarrow I=1 \end{gathered}$$