Standard integration 373
=
⎡
⎣θ
3
2
3
2
+
2 θ
1
2
1
2
⎤
⎦
4
1
=
[
2
3
√
θ^3 + 4
√
θ
] 4
1
=
{
2
3
√
( 4 )^3 + 4
√
4
}
−
{
2
3
√
( 1 )^3 + 4
√
( 1 )
}
=
{
16
3
+ 8
}
−
{
2
3
+ 4
}
= 5
1
3
+ 8 −
2
3
− 4 = 8
2
3
Problem 14. 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
}
=
{
−
3
2
(− 1 )
}
−
{
−
3
2
( 1 )
}
=
3
2
+
3
2
= 3
Problem 15. Evaluate
∫ 2
1
4cos3tdt.
∫ 2
1
4cos3tdt=
[
( 4 )
(
1
3
)
sin3t
] 2
1
=
[
4
3
sin3t
] 2
1
=
{
4
3
sin6
}
−
{
4
3
sin3
}
Note that limits of trigonometric functions are always
expressed in radians—thus, for example, sin6 means
the sine of 6radians=− 0. 279415 ...
Hence
∫ 2
1
4cos3tdt
=
{
4
3
(− 0. 279415 ...)
}
−
{
4
3
( 0. 141120 ...)
}
=(− 0. 37255 )−( 0. 18816 )=− 0. 5607
Problem 16. Evaluate
(a)
∫ 2
1
4e^2 xdx (b)
∫ 4
1
3
4 u
du,
each correct to 4 significant figures.
(a)
∫ 2
1
4e^2 xdx=
[
4
2
e^2 x
] 2
1
=2[e^2 x]^21 =2[e^4 −e^2 ]
=2[54. 5982 − 7 .3891]= 94. 42
(b)
∫ 4
1
3
4 u
du=
[
3
4
lnu
] 4
1
=
3
4
[ln4−ln1]
=
3
4
[1. 3863 −0]= 1. 040
Now try the following exercise
Exercise 146 Further problems on definite
integrals
In problems 1 to 8, evaluate the definite integrals
(where necessary, correct to 4 significant figures).
- (a)
∫ 4
1
5 x^2 dx (b)
∫ 1
− 1
−
3
4
t^2 dt
[
(a) 105 (b)−
1
2
]
- (a)
∫ 2
− 1
( 3 −x^2 )dx (b)
∫ 3
1
(x^2 − 4 x+ 3 )dx
[
(a) 6 (b)− 1
1
3
]
- (a)
∫π
0
3
2
cosθdθ (b)
∫ π
2
0
4cosθdθ
[(a) 0 (b) 4]
- (a)
∫ π
3
π
6
2sin2θdθ (b)
∫ 2
0
3sintdt
[(a) 1 (b) 4.248]
- (a)
∫ 1
0
5cos3xdx (b)
∫ π 6
0
3sec^22 xdx
[(a) 0.2352 (b) 2.598]