Introduction to integration 329
For example, the increase in the value of the integral
x^2 asxincreases from 1 to 3 is written as
∫ 3
1 x(^2) dx.
Applying the limits gives
∫ 3
1
x^2 dx=
[
x^3
3
+c
] 3
1
(
33
3
+c
)
−
(
13
3
+c
)
=( 9 +c)−
(
1
3
+c
)
= 8
2
3
Note that thecterm always cancels out when limits are
applied and it need not be shown with definite integrals.
Problem 17. Evaluate
∫ 2
1
3 xdx
∫ 2
1
3 xdx=
[
3 x^2
2
] 2
1
{
3
2
( 2 )^2
}
−
{
3
2
( 1 )^2
}
= 6 − 1
1
2
= 4
1
2
Problem 18. Evaluate
∫ 3
− 2
( 4 −x^2 )dx
∫ 3
− 2
( 4 −x^2 )dx=
[
4 x−
x^3
3
] 3
− 2
{
4 ( 3 )−
( 3 )^3
3
}
−
{
4 (− 2 )−
(− 2 )^3
3
}
={ 12 − 9 }−
{
− 8 −
− 8
3
}
={ 3 }−
{
− 5
1
3
}
= 8
1
3
Problem 19. Evaluate
∫ 2
0
x( 3 + 2 x)dx
∫ 2
0
x( 3 + 2 x)dx=
∫ 2
0
( 3 x+ 2 x^2 )dx=
[
3 x^2
2
2 x^3
3
] 2
0
{
3 ( 2 )^2
2
2 ( 2 )^3
3
}
−{ 0 + 0 }
= 6 +
16
3
= 11
1
3
or 11. 33
Problem 20. Evaluate
∫ 1
− 1
(
x^4 − 5 x^2 +x
x
)
dx
∫ 1
− 1
(
x^4 − 5 x^2 +x
x
)
dx
∫ 1
− 1
(
x^4
x
−
5 x^2
x
x
x
)
dx
∫ 1
− 1
(
x^3 − 5 x+ 1
)
dx=
[
x^4
4
−
5 x^2
2
+x
] 1
− 1
{
1
4
−
5
2
1
}
−
{
(− 1 )^4
4
−
5 (− 1 )^2
2
+(− 1 )
}
{
1
4
−
5
2
- 1
}
−
{
1
4
−
5
2
− 1
}
= 2
Problem 21. Evaluate
∫ 2
1
(
1
x^2
2
x
)
dxcorrect
to 3 decimal places
∫ 2
1
(
1
x^2
2
x
)
dx
∫ 2
1
{
x−^2 + 2
(
1
x
)}
dx=
[
x−^2 +^1
− 2 + 1
+2lnx
] 2
1
[
x−^1
− 1
+2lnx
] 2
1
[
−
1
x
+2lnx
] 2
1
(
−
1
2
+2ln2
)
−
(
−
1
1
+2ln1
)
= 1. 886
Problem 22. Evaluate
∫π/ 2
0
3sin2xdx
∫π/ 2
0
3sin2xdx
[
( 3 )
(
−
1
2
)
cos2x
]π/ 2
0
[
−
3
2
cos2x
]π/ 2
0
{
−
3
2
cos2
(π
2
)}
−
{
−
3
2
cos2( 0 )
}
{
−
3
2
cosπ
}
−
{
−
3
2
cos0
}