436 Higher Engineering Mathematics
Problem 1. (a) Use integration to evaluate,
correct to 3 decimal places,
3
1
2
√
x
dx (b) Use the
trapezoidal rule with 4 intervals to evaluate the
integral in part (a), correct to 3 decimal places.
(a)
3
1
2
√
x
dx =
3
1
2x
−
1
2 dx
=
⎡
⎢
⎣
2x
−1
2
+1
−
1
2
+ 1
⎤
⎥
⎦
3
1
=
4x
1
2
3
1
= 4
√
x
3
1
= 4
√
3 −
√
1
= 2.928, correct to 3 decimal places
(b) The range of integration is the difference between
the upper and lower limits, i.e. 3 − 1 = 2. Using
the trapezoidal rule with 4 intervals gives an interval width d =
3 − 1
4
= 0.5 and ordinates situated
at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding values
of
2
√
x
are shown in the table below, each correct
to 4 decimal places (which is one more decimal
place than required in the problem).
x
2
√
x
1.0
2.0000
1.5
1.6330
2.0
1.4142
2.5
1.2649
3.0
1.1547
From equation (1):
3
1
2
√
x
dx ≈ (0.5)
1
2
(2.0000 + 1.1547)
+ 1.6330 + 1.4142 + 1.2649
= 2.945, correct to 3 decimal places
This problem demonstrates that even with just 4 intervals a close approximation to the true value of 2.928
(correct to 3 decimal places) is obtained using the
trapezoidal rule.
Problem 2. Use the trapezoidal rule with 8
intervals to evaluate,
3
1
2
√
x
dx correct to 3
decimal places.
With 8 intervals, the width of each is
3 − 1
8
i.e. 0.25
giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25,
2.50, 2.75 and 3.00. Corresponding values of
2
√
x
are
shown in the table below.
x
2
√
x
1.00
2.0000
1.25
1.7889
1.50
1.6330
1.75
1.5119
2.00
1.4142
2.25
1.3333
2.50
1.2649
2.75
1.2060
3.00
1.1547
From equation (1):
3
1
2
√
x
dx ≈ (0.25)
1
2
(2.000 + 1.1547) + 1.7889
+ 1.6330 + 1.5119 + 1.4142
+ 1.3333 + 1.2649 + 1.2060
= 2.932, correct to 3 decimal places.
This problem demonstrates that the greater the number
of intervals chosen (i.e. the smaller the interval width)
the more accurate will be the value of the definite integral. The exact value is found when the number of
intervals is infinite, which is, of course, what the process
of integration is based upon.
Problem 3. Use the trapezoidal rule to evaluate
π
2
0
1
1 + sin x
dx using 6 intervals. Give the answer
correct to 4 significant figures.
Problem 1. (a) Use integration to evaluate,
correct to 3 decimal places,
3
1
2
√
x
dx (b) Use the
trapezoidal rule with 4 intervals to evaluate the
integral in part (a), correct to 3 decimal places.
(a)
3
1
2
√
x
dx =
3
1
2x
−
1
2 dx
=
⎡
⎢
⎣
2x
−1
2
+1
−
1
2
+ 1
⎤
⎥
⎦
3
1
=
4x
1
2
3
1
= 4
√
x
3
1
= 4
√
3 −
√
1
= 2.928, correct to 3 decimal places
(b) The range of integration is the difference between
the upper and lower limits, i.e. 3 − 1 = 2. Using
the trapezoidal rule with 4 intervals gives an interval width d =
3 − 1
4
= 0.5 and ordinates situated
at 1.0, 1.5, 2.0, 2.5 and 3.0. Corresponding values
of
2
√
x
are shown in the table below, each correct
to 4 decimal places (which is one more decimal
place than required in the problem).
x
2
√
x
1.0
2.0000
1.5
1.6330
2.0
1.4142
2.5
1.2649
3.0
1.1547
From equation (1):
3
1
2
√
x
dx ≈ (0.5)
1
2
(2.0000 + 1.1547)
+ 1.6330 + 1.4142 + 1.2649
= 2.945, correct to 3 decimal places
This problem demonstrates that even with just 4 intervals a close approximation to the true value of 2.928
(correct to 3 decimal places) is obtained using the
trapezoidal rule.
Problem 2. Use the trapezoidal rule with 8
intervals to evaluate,
3
1
2
√
x
dx correct to 3
decimal places.
With 8 intervals, the width of each is
3 − 1
8
i.e. 0.25
giving ordinates at 1.00, 1.25, 1.50, 1.75, 2.00, 2.25,
2.50, 2.75 and 3.00. Corresponding values of
2
√
x
are
shown in the table below.
x
2
√
x
1.00
2.0000
1.25
1.7889
1.50
1.6330
1.75
1.5119
2.00
1.4142
2.25
1.3333
2.50
1.2649
2.75
1.2060
3.00
1.1547
From equation (1):
3
1
2
√
x
dx ≈ (0.25)
1
2
(2.000 + 1.1547) + 1.7889
+ 1.6330 + 1.5119 + 1.4142
+ 1.3333 + 1.2649 + 1.2060
= 2.932, correct to 3 decimal places.
This problem demonstrates that the greater the number
of intervals chosen (i.e. the smaller the interval width)
the more accurate will be the value of the definite integral. The exact value is found when the number of
intervals is infinite, which is, of course, what the process
of integration is based upon.
Problem 3. Use the trapezoidal rule to evaluate
π
2
0
1
1 + sin x
dx using 6 intervals. Give the answer
correct to 4 significant figures.