Question

The value of the integral ${\int }_a^b\frac{|x|}{x}=dx,$ where $a<b$ is

• A. $b-a$
• B. $|b|-|a|$
• C. $a+b$
• D. $|a|+|b|$

Answer 1

Answer: (B)

Three cases may arise
Case (i) : let $0\le a<b,$ then $\frac{|x|}{x}=\frac{x}{x}=1$
$\therefore {\int }_a^b\frac{|x|}{x}dx={\int }_a^{b}dx=b-a=\left|b\right|-\left|a\right|$
$\left(Since|a|=aand|b|=b\right)$
Case (ii) : Let $a<b\le 0$, then $\frac{|x|}{x}=\frac{-x}{x}=-1$
$\therefore {\int }_a^b\frac{|x|}{x}dx={\int }_a^{b}dx={\int }_a^b\left(-1\right)dx=a-b=-\left|a\right|+\left|b\right|$
$\left(Since|a|=-aand|b|=-b\right)$
Case (iii) : Let a < b < 0, then
${\int }_a^b\frac{|x|}{x}=dx={\int }_a^b\frac{|x|}{x}dx={\int }_a^b\frac{|x|}{x}dx$
$={\int }_a^b\left(-1\right)dx={\int }_a^b\left(1\right)dx=a+b=-\left|a\right|+\left|b\right|<br/>\left(Since|a|=-aand|b|=b\right)$
Thus, in every case ${\int }_b^a\frac{|x|}{x}dx=\left|b\right|-\left|a\right|$