Question

The value of $\displaystyle\int_{-1}^2 \frac {|x|}{x}dx $ is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

Let $l=\displaystyle\int_{-1}^{2} \frac{|x|}{x} d x=\displaystyle\int_{-1}^{0}\left(-\frac{x}{x}\right) d x+\displaystyle\int_{0}^{2}\left(\frac{x}{x}\right) d x$
$\left(\because|x|=\begin{cases}x, & \text { if } & x \geq 0 \\ -x_{1} & \text { if } & x<0\end{cases}\right)$
$=-\displaystyle\int_{-1}^{0} d x+\int_{0}^{2} d x$
$=-[x]_{-1}^{0}+[x]_{0}^{2}$
$=-[0+1]+[2-0]$
$=-1+2=1$