Question

$\text{Evaluate}:{\int }_{-1}^2\frac{\left|x\right|}{x}dx.$

Answer 1

$\text{Let}I={\int }_{-1}^2\frac{\left|x\right|}{x}dx$
$\Rightarrow I={\int }_{-1}^1\frac{\left|x\right|}{x}dx+{\int }_1^2\frac{\left|x\right|}{x}dx(\because {\int }_a^{b}f\left(x\right)dx={\int }_a^{c}f\left(x\right)dx+{\int }_c^{b}f\left(x\right)dx,\text{where}a<c<b)$

$\text{Consider}{\int }_{-1}^1\frac{\left|x\right|}{x}dx\text{Let}g\left(x\right)=\frac{\left|x\right|}{x}\text{Then,}g(-x)=\frac{|-x|}{-x}=-\frac{\left|x\right|}{x}=-g\left(x\right)$
$\Rightarrow g\left(x\right)$ is an odd function.
$\therefore {\int }_{-1}^1\frac{\left|x\right|}{x}dx=0$
$\text{Now,}{\int }_1^2\frac{\left|x\right|}{x}dx={\int }_1^{2}1dx(\because |x|=x\text{for}x\in [1,2]$
$=[x{]}_1^2=2-1=1$

$\therefore I=0+1=1$