Standard integration 373
=
⎡
⎣
θ
3
2
3
2
+
2θ
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
3 sin2x dx.
π
2
0
3 sin2x dx
=
(3)
−
1
2
cos 2x
π
2
0
=
−
3
2
cos 2x
π
2
0
=
−
3
2
cos 2
π
2
−
−
3
2
cos 2(0)
=
−
3
2
cos π
−
−
3
2
cos 0
=
−
3
2
(−1)
−
−
3
2
(1)
=
3
2
+
3
2
= 3
Problem 15. Evaluate
2
1
4 cos 3t dt.
2
1
4 cos3t dt =
(4)
1
3
sin 3t
2
1
=
4
3
sin 3t
2
1
=
4
3
sin 6
−
4
3
sin 3
Note that limits of trigonometric functions are always
expressed in radians—thus, for example, sin 6 means
the sine of 6 radians=−0.279415 ...
Hence
2
1
4 cos 3t dt
=
4
3
(−0.279415 ...)
−
4
3
(0.141120 ...)
= (−0.37255) − (0.18816) = −0.5607
Problem 16. Evaluate
(a)
2
1
4 e
2x dx (b)
4
1
3
4u
du,
each correct to 4 significant figures.
(a)
2
1
4 e
2x dx =
4
2
e
2x
2
1
= 2[ e
2x ]
2
1 = 2[ e
4
− e
2 ]
= 2[54.5982 −7.3891] =94.42
(b)
4
1
3
4u
du =
3
4
ln u
4
1
=
3
4
[ln 4 − ln 1]
=
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).
1. (a)
4
1
5x
2 dx (b)
1
−1
−
3
4
t
2 dt
(a) 105 (b) −
1
2
2. (a)
2
−1
(3 − x
2
) dx (b)
3
1
(x
2
− 4x + 3) dx
(a) 6 (b) −1
1
3
3. (a)
π
0
3
2
cos θ dθ (b)
π
2
0
4 cos θ dθ
[(a) 0 (b) 4]
4. (a)
π
3
π
6
2 sin 2θ dθ (b)
2
0
3 sin t dt
[(a) 1 (b) 4.248]
5. (a)
1
0
5 cos3x dx (b)
π
6
0
3 sec
2 2x dx
[(a) 0.2352 (b) 2.598]