Q119 / 200 · Manipal (MET) 2013 · Mathematics

The value of $\int_a^b \frac{|x|}{x} d x$ is

  • Correct answer
    A
    $|b|-|a|$
  • B
    $|a|-|b|$
  • C
    $|b|+|a|$
  • D
    $-|b|-|a|$

Solution

Case I. If $0 \leq a \lt b$, then $\frac{|x|}{x}=1$ $\therefore \quad I=\int_a^b 1 \cdot d x=b-a=|b|-|a|$ Case II: If $a \lt b \leq 0$, then $|x|=-x$ $\therefore \quad I=\int_a^b \frac{-x}{x} d x=\int_a^b(-1) d x=[-x]_a^b$ $=-b-(-a)=|b|-|a|$ Case III If $a \lt 0 \lt b$. then $|x|=-x \quad$ when $a \lt x \lt 0$ and $|x|=x \quad$ when $0 \lt x \lt b$ $I=\int_a^b \frac{|x|}{x} d x \quad=\int_a^0 \frac{|x|}{x} d x+\int_0^b \frac{|x|}{x} d x$ $\begin{aligned} & =\int_a^0 \frac{-x}{x} d x+\int_0^b \frac{x}{x} d x \\ & =\int_a^0(-1) d x+\int_0^b(1) d x \\ & =[-x]_a^0+[x]_0^b \\ & =a+b=b-(-a) \\ & =|b|-|a|\end{aligned}$ Hence, in all cases, $I=\int_a^b \frac{|x|}{x} d x=|b|-|a|$

Hence, the correct answer is option: 'A'.

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