Question

Evaluate: Evaluate :
3
( 1)( 2)
x
dx  x x

Answer 1

Answer:

∫

−1

2

∣x

3

−x∣dx

It is clear that

x

3

−x⩾0on[−1,0]

x

3

−x⩽0on[0,1]

x

3

−x⩾0on[1,2]

Hence the interval of the integral can be subdivided as

∫

−1

2

∣x

3

−x∣dx=∫

0

−1

(x

3

−x)dx+∫

0

1

−(x

3

−x)dx+∫

1

2

(x

3

−x)dx

=∫

−1

0

(x

3

−x)dx+∫

0

1

(x−x

3

)dx+∫

1

2

(x

3

−x)dx

=[

4

x

4

−

2

x

2

]

−1

0

+[

2

x

2

−

4

x

4

]

0

1

+[

4

x

4

−

2

x

2

]

1

2

[∫x

n

dx=

n+

x

n+1

]

=−(

4

1

−

2

1

)+(

2

1

−

4

1

)+(4−2)−(

4

1

−

2

1

)

=

4

−1

+

2

1

+

2

1

−

4

1

+2−

4

1

+

2

1

=

2

3

−

4

3

+2

=

4

11

Ans