Manipal Entrance Test 2013 Math Paper

Case I. If 0≤a<b, then

|x|

x

=1
∴ I=

b

∫

a

1.dx=b−a=|b|−|a|
Case |: If a<b≤0, then |x|=−x
∴ I=

b

∫

a

−x

x

dx=

b

∫

a

(−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<b
I=

b

∫

a

|x|

x

dx =

0

∫

a

|x|

x

dx+

b

∫

0

|x|

x

dx
=

0

∫

a

−x

x

dx+

b

∫

0

x

x

dx
=

0

∫

a

(−1)dx+

b

∫

0

(1)dx
=[−x]a0+[x]0b
=a+b=b−(−a)
=|b|−|a|
Hence, in all cases,
I=

b

∫

a

|x|

x

dx=|b|−|a|