Question

${\int }_a^b\left(\frac{\left|x\right|}{x}\right)dx,$ where $a<0<b=$

• A. $\left|b\right|-\left|a\right|$
• B. $\left|b\right|+\left|a\right|$
• C. $\left|a-b\right|$
• D. None of these

Answer 1

The correct option is A

$\left|b\right|-\left|a\right|$

Explanation for the correct option:

Finding the value for the given integral:

Let $I={\int }_a^b\left(\frac{\left|x\right|}{x}\right)dx,$ where $a<0<b$

$={\int }_a^0\frac{\left|x\right|}{x}dx+{\int }_0^b\frac{\left|x\right|}{x}dx$

$={\int }_a^0-\frac{x}{x}dx+{\int }_0^b\frac{x}{x}dx$ $\left[\sin ce\left|x\right|=\left\{\begin{array}{l}x,x\ge 0\\ -x,x<0\end{array}\right.\right]$

$={\int }_a^0-1dx+{\int }_0^{b}dx$

$={-\left[x\right]}_a^0+{\left[x\right]}_0^b$

$=a+b$

$=\left|b\right|-\left|a\right|$

Thus, option A is the correct answer.