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Manipal Entrance Test 2013 Math Paper

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Question : 59 of 80

Marks: +1, -0

The value of

\int\limits_{a}^{b} \frac{|x|}{x} \, dx

is

$|b|-|a|$
$|a|-|b|$
$|b|+|a|$
$-|b|-|a|$

Solution:

Case I. If

0 \leq a < b,

then

\frac{|x|}{x}=1

∴

I=\int\limits_{a}^{b} 1 \, dx=b-a=|b|-|a|

Case

|:

If

a < b \leq 0,

then

|x|=-x

∴

I=\int\limits_{a}^{b} \frac{-x}{x} \, dx=\int\limits_{a}^{b} (-1) \, dx

=\left[ -x \right]_{a}^{b}

=-b-(-a)=|b|-|a|

Case III If

a < 0 < b

.

then

|x|=-x

when

a < x < 0

and

|x|=x

when

0 < x < b

I=\int\limits_{a}^{b} \frac{|x|}{x} \, dx

=\int\limits_{a}^{0} \frac{|x|}{x} \, dx+\int\limits_{0}^{b} \frac{|x|}{x} \, dx

=\int\limits_{a}^{0} \frac{-x}{x} \, dx+\int\limits_{0}^{b} \frac{x}{x} \, dx

=\int\limits_{a}^{0} (-1) \, dx+\int\limits_{0}^{b} (1) \, dx

=\left[ -x \right]_{a}^{0}+\left[ x \right]_{0}^{b}

=a+b=b-(-a)

=|b|-|a|

Hence, in all cases,

I=\int\limits_{a}^{b} \frac{|x|}{x} \, dx=|b|-|a|

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