330 Basic Engineering Mathematics
=
−
3
2
(−1)
−
−
3
2
(1)
=
3
2
+
3
2
= 3
Problem 23. Evaluate
2
1
4 cos3t 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 cos3t dt=
4
3
(−0.279415 ...)
−
4
3
(0.141120 ...)
= (−0.37255) − (0.18816)
= −0.5607
Problem 24. Evaluate
2
1
4e
2x dx correct to
4 significant figures
2
1
4e
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
Problem 25. Evaluate
4
1
3
4u
du correct to 4
significant figures
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 Practice Exercise
Practice Exercise 140 Definite integrals
(answers on page 355)
In problems 1 to 10, evaluate the definite integrals
(where necessary, correct to 4 significant figures).
1. (a)
2
1
x dx
(b)
2
1
(x − 1) dx
2. (a)
4
1
5x
2 dx
(b)
1
−1
−
3
4
t
2 dt
3. (a)
2
−1
(3 − x
2
)dx (b)
3
1
(x
2
− 4x + 3) dx
4. (a)
2
1
(x
3
− 3x) dx (b)
2
1
(x
2
− 3x + 3) dx
5. (a)
4
0
2
√
x dx
(b)
3
2
1
x 2 dx
6. (a)
π
0
3
2
cos θ dθ
(b)
π/2
0
4 cosθ dθ
7. (a)
π/3
π/6
2 sin2θ dθ (b)
2
0
3 sint dt
8. (a)
1
0
5 cos3x dx
(b)
π/2
π/4
(3 sin2x − 2 cos 3x) dx
9. (a)
1
0
3e
3t dt
(b)
2
−1
2
3e 2x dx
10. (a)
3
2
2
3x
dx
(b)
3
1
2x 2 + 1
x
dx
35.5 The area under a curve
The area shown shaded in Figure 35.1 may be determined using approximate methods such as the trapezoidal rule, the mid-ordinate rule or Simpson’s rule (see
Chapter 28) or, more precisely, by using integration.
0
x 5 a
x 5 b
y 5 f (x)
y
x
Figure 35.1