The value of ∫ab |x|xdx is

Case I: If 0≤a<b, then |x|x=1∴ I=∫ab 1.dx=b−a=|b|−|a| Case II: If a<b≤0, then |x|=−xI=∫ab −xxdx=∫ab (−1)dx =[−x]ab=−b−(−a)=|b|−|a| Case III: If a<0<b then |x|=−x when a<x<0 and |x|=x when 0<x<bI=∫ab|x| |x|xdx=∫a0 |x|xdx+∫0b |x|xdx =∫a0 −xxdx+∫0b xxdx =∫a0 (−1)dx+∫0b 1dx =[−x]a0+[x]0b=a+b=b−(−a)=|b|−|a| Hence, in all the cases I=∫ab |x|xdx=|b|−|a|

The value of ∫ab |x|xdx is

Case I: If 0≤a<b, then |x|x=1∴ I=∫ab 1.dx=b−a=|b|−|a| Case II: If a<b≤0, then |x|=−xI=∫ab −xxdx=∫ab (−1)dx =[−x]ab=−b−(−a)=|b|−|a| Case III: If a<0<b then |x|=−x when a<x<0 and |x|=x when 0<x<bI=∫ab|x| |x|xdx=∫a0 |x|xdx+∫0b |x|xdx =∫a0 −xxdx+∫0b xxdx =∫a0 (−1)dx+∫0b 1dx =[−x]a0+[x]0b=a+b=b−(−a)=|b|−|a| Hence, in all the cases I=∫ab |x|xdx=|b|−|a|

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

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$| b | - | a |$

$| a | - | b |$

$| b | + | a |$

$- | b | - | a |$

answer is A.

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Detailed Solution

$Case I: If 0 \leq a < b, then \frac{| x |}{x} = 1$

$∴ I = \int_{a}^{b} 1 . d x = b - a = | b | - | a |$

$\begin{aligned} Case II: If a < b \leq 0, then | x | = - x \\ \begin{aligned} I & = \int_{a}^{b} \frac{- x}{x} d x = \int_{a}^{b} (- 1) d x \\ = [- x]_{a}^{b} = - b - (- a) = | b | - | a | \end{aligned} \end{aligned}$

$Case III: If a < 0 < b$

$then | x | = - x when a < x < 0$

$and | x | = x when 0 < x < b$

$\begin{array}{l} \begin{aligned} I & = \int_{a}^{b | x |} \frac{| x |}{x} d x = \int_{a}^{0} \frac{| x |}{x} d x + \int_{0}^{b} \frac{| x |}{x} d x \\ = \int_{a}^{0} \frac{- x}{x} d x + \int_{0}^{b} \frac{x}{x} d x \\ = \int_{a}^{0} (- 1) d x + \int_{0}^{b} 1 d x \\ = [- x]_{a}^{0} + [x]_{0}^{b} = a + b = b - (- a) = | b | - | a | \end{aligned} \\ Hence, in all the cases I = \int_{a}^{b} \frac{| x |}{x} d x = | b | - | a | \end{array}$