Question

The value of ${\int }_{-1}^{2}f\left(x\right)\mathrm{dx},\mathrm{where}f\left(x\right)=\left(x+1\right|+\left(x\right|+\left(x-1\right|$ is

• A. $\frac{9}{2}$
• B. $\frac{19}{2}$
• C. $\frac{15}{2}$
• D. $\frac{21}{2}$

Answer 1

The correct option is B $\frac{19}{2}$

We can redefine f as
$f\left(x\right)=(\begin{array}{c}2-x,\mathrm{if}-1<x\le 0\\ x+2,\mathrm{if}0<x\le 1\\ 3x,\mathrm{if}1<x\le 2\end{array}\therefore {\int }_{-1}^{2}f(x)\mathrm{dx}={\int }_{-1}^{0}f(x)\mathrm{dx}+{\int }_0^{1}f(x)\mathrm{dx}+{\int }_1^{2}f(x)\mathrm{dx}={\int }_{-1}^0(2-x)\mathrm{dx}+{\int }_0^1(x+2)\mathrm{dx}+{\int }_1^{2}3x\mathrm{dx}={\left(2x-\frac{x^2}{2}\right)}_{-1}^0+{\left(\frac{x^2}{2}+2x\right)}_0^1+3\cdot {\left(\frac{x^2}{2}\right)}_1^2=(2+\frac{1}{2})+(\frac{1}{2}+2)+3\cdot (2-\frac{1}{2})=\frac{19}{2}$