CHAPTER 71 NUMERICAL INTEGRATION
EXERCISE 280 Page 759
1. Evaluate using the trapezoidal rule, giving the answers correct to 3 decimal places:
∫
Since
∫
2
dx
0 1 + x2
1
(use 8 intervals)
2
1− 0
= 0.125
d x , width of interval =
0 1 + x2
8
1
x
2
1 + x2
0
0.125
2.0000
0.250
0.375
0.500
0.625
0.750
0.875
1.000
1.9692 1.8824 1.7534 1.6000 1.4382 1.2800 1.1327 1.0000
Hence, using the trapezoidal rule
∫
2
dx
0 1 + x2
1
1

≈ (0.125)  ( 2.0000 + 1.0000 ) + 1.9692 + 1.8824 + 1.7534 + 1.6000 + 1.4382 + 1.2800 + 1.1327 
2

= 0.125[12.5559]
= 1.569
2. Evaluate using the trapezoidal rule, giving the answers correct to 3 decimal places:
∫
3
1
Since
∫
3
1
2 ln 3 x d x , width of interval =
x
2 ln 3x
(use 8 intervals)
2 ln 3 x d x
3 −1
= 0.25
8
1
1.25
1.50
1.75
2.0
2.25
2.50
2.75
3.0
2.1972 2.6435 3.0082 3.3165 3.5835 3.8191 4.0298 4.2204 4.3944
Hence, using the trapezoidal rule
∫
3
1
2 ln 3 x d x
1

≈ (0.25)  ( 2.1972 + 4.3944 ) + 2.6435 + 3.0082 + 3.3165 + 3.5835 + 3.8191 + 4.0298 + 4.2204 
2

= 0.25[27.9168] = 6.979
1120
© 2014, John Bird
3. Evaluate using the trapezoidal rule, giving the answers correct to 3 decimal places:
∫
π /3
0
(sin θ ) d θ
(use 6 intervals)
π
∫
Since
π /3
0
−0
π
3
=
or 10°
(sin θ ) d θ , width of interval =
6
18
θ
sin θ
0
0
π/18
0.4167
2π/18 3π/18 4π/18 5π/18 6π/18
0.5848 0.7071 0.8017 0.8752 0.9306
Hence, using the trapezoidal rule
∫
π /3
0
(sin θ ) d θ
 π  1

≈    ( 0 + 0.9306 ) + 0.4167 + 0.5848 + 0.7071 + 0.8017 + 0.8752 
 18   2

π 
= ≈   [3.8508]
 18 
= 0.672
4. Evaluate using the trapezoidal rule, giving the answers correct to 3 decimal places:
∫
Since ∫
1.4
0
1.4
0
e − x2 d x
e − x2 d x , width of interval =
x
e − x2
0
1.0
(use 7 intervals)
1.4 − 0
= 0.2
7
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0.9608 0.8521 0.6977 0.5273 0.3679 0.2369 0.1409
Hence, using the trapezoidal rule,
∫
1.4
0
1

e − x2 d x ≈ (0.2)  (1.0 + 0.1409 ) + 0.9608 + 0.8521 + 0.6977 + 0.5273 + 0.3679 + 0.2369 
2

= (0.2)[4.21315]
= 0.843
1121
© 2014, John Bird
EXERCISE 281 Page 761
1. Evaluate using the mid-ordinate rule, giving the answers correct to 3 decimal places:
∫
Since ∫
2
0
2
0
3
dt
1+ t2
(use 8 intervals)
2−0
3
= 0.25
d t , width of interval =
2
8
1+ t
Hence, ordinates occur at 0, 0.25, 0.50, 0.75, 1.0, 1.25, 1.50, 1.75 and 2.0,
and mid-ordinates occur at 0.125, 0.375, 0.625, 0.875, 1.125, 1.375, 1.625 and 1.875
t
3
1+ t2
0.125
2.9538
0.375
0.625
0.875
1.125
1.375
1.625
1.875
2.6301 2.1573 1.6991 1.3241 1.0378
0.8240
0.6644
Hence, using the mid-ordinate rule,
∫
2
0
3
d t ≈ ( 0.25 ) [ 2.9538 + 2.6301 + 2.1573 + 1.6991 + 1.3241 + 1.0378 + 0.8240 + 0.6644]
1+ t2
= ( 0.25 ) [13.2906]
= 3.323
2. Evaluate using the mid-ordinate rule, giving the answers correct to 3 decimal places:
∫
π /2
0
1
1 + sin θ
(use 6 intervals)
π
Since
∫
π /2
0
−0
π
1
= 2
=
rad or 15°
d θ , width of interval
6
12
1 + sin θ
Hence, ordinates occur at 0°, 15°, 30°, 45°, 60°, 75° and 90°,
and mid-ordinates occur at 7.5°, 22.5°, 37.5°, 52.5°, 67.5° and 82.5°.
θ
1
1 + sin θ
7.5°
22.5°
37.5°
52.5°
67.5°
82.5°
0.8845 0.7232 0.6216 0.5576 0.5198 0.5021
Hence, using the mid-ordinate rule
1122
© 2014, John Bird
∫
π /2
0
1
π 
d θ ≈   [ 0.8845 + 0.7232 + 0.6216 + 0.5576 + 0.5198 + 0.5021]
1 + sin θ
 12 
π 
=   [3.8088]
 12 
= 0.997
3. Evaluate using the mid-ordinate rule, giving the answers correct to 3 decimal places:
∫
3
1
Since ∫
3
1
ln x
dx
x
(use 10 intervals)
ln x
3 −1
= 0.2
d x , width of interval =
10
x
Hence, ordinates occur at 1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2, 2.4, 2.6, 2.8, 3.0,
and mid-ordinates occur at 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7 and 2.9.
x
1.1
ln x
x
1.3
1.5
1.7
1.9
2.1
2.3
2.5
2.7
2.9
0.0866 0.2018 0.2703 0.3121 0.3378 0.3533 0.3621 0.3665 0.3679 0.3671
Hence, using the mid-ordinate rule
∫
3
1
ln x
d x ≈ (0.2)[ 0.0866 + 0.2018 + 0.2703 + 0.3121 + 0.3378 + 0.3533 + 0.3621 + 0.3665
x
+ 0.3679 + 0.3671]
= (0.2)[3.0255]
= 0.605
4. Evaluate using the mid-ordinate rule, giving the answers correct to 3 decimal places:
∫
π /3
0
(use 6 intervals)
(cos3 x) d x
π
Since
∫
π /3
0
−0
π
3
,
width
of
interval
=
rad or 10°
=
(cos x) d x
6
18
3
Hence, ordinates occur at 0°, 10°, 20°, 30°, 40°, 50° and 60°,
and mid-ordinates occur at 5°, 15°, 25°, 35°, 45° and 55°
1123
© 2014, John Bird
θ
5°
( cos 2 x )
0.9943
15°
0.9493
25°
0.8628
35°
45°
55°
0.7414
0.5946
0.4344
Hence, using the mid-ordinate rule,
∫
π /3
0
π 
(cos3 x) d x ≈   [ 0.9943 + 0.9493 + 0.8628 + 0.7414 + 0.5946 + 0.4344]
 18 
π 
=   [ 4.5768]
 18 
= 0.799
1124
© 2014, John Bird
EXERCISE 282 Page 764
1. Evaluate using Simpson’s rule, giving the answers correct to 3 decimal places:
∫
π /2
0
(use 6 intervals)
(sin x) d x
π
Since
∫
π /2
0
x
−0
π
= 2
=
rad or 15°
(sin x) d x , width of interval
6
12
0
(sin x)
π
2π
12
12
3π
12
5π
12
4π
12
6π
12
0 0.5087 0.7071 0.8409 0.9306 0.9828 1.0000
Hence, using Simpson’s rule,
∫
1 π 
(sin x) d x ≈   ( 0 + 1.000 ) + 4 ( 0.5087 + 0.8409 + 0.9828 ) + 2 ( 0.7071 + 0.9306 ) 
3  12 
π /2
0
π 
=   [1.0000 + 9.3296 + 3.2754]
 36 
π 
=   [13.605] = 1.187
 36 
2. Evaluate using Simpson’s rule, giving the answers correct to 3 decimal places:
∫
Since
∫
1.6
0
1.6
0
1
dθ
1+θ 4
(use 8 intervals)
1.6 − 0
1
= 0.2
d θ , width of interval =
8
1+θ 4
θ
0
1
1+θ 4
1.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
0.9984 0.9750 0.8853 0.7094 0.5000 0.3254 0.2065
0.1324
Hence, using Simpson’s rule,
∫
1.6
0
1
dθ
1+θ 4
1125
© 2014, John Bird
1
≈ (0.2) (1.0 + 0.1324 ) + 4 ( 0.9984 + 0.8853 + 0.5000 + 0.2065 ) + 2 ( 0.9750 + 0.7094 + 0.3254 ) 
3
=
1
(0.2) [1.1324 + 10.3608 + 4.0196]
3
=
1
(0.2) [15.5128] = 1.034
3
3. Evaluate using Simpson’s rule, giving the answers correct to 3 decimal places:
∫
Since ∫
1.0
sin θ
0.2
θ
θ
sin θ
θ
1.0
sin θ
0.2
θ
dθ
(use 8 intervals)
d θ , width of interval =
0.2
0.3
1.0 − 0.2
= 0.1
8
0.4
0.5
(note that values of θ are in radians)
0.6
0.7
0.8
0.9933 0.9851 0.9735 0.9589 0.9411 0.9203 0.8967
0.9
1.0
0.8704 0.8415
Hence, using Simpson’s rule,
∫
1.0
( 0.9933 + 0.8415 ) + 4 ( 0.9851 + 0.9589 + 0.9203 + 0.8704 )

1
d θ ≈ (0.1) 

θ
3
+ 2 ( 0.9735 + 0.9411 + 0.8967 ) 

sin θ
0.2
=
1
(0.1) [1.8348 + 14.9388 + 5.6226]
3
=
1
(0.1) [ 22.3962] = 0.747
3
4. Evaluate using Simpson’s rule, giving the answers correct to 3 decimal places:
∫
π /2
0
(use 6 intervals)
x cos x d x
π
Since ∫
π /2
0
x
x cos x
−0
π
= 2
=
rad or 15°
x cos x d x , width of interval
6
12
0
0
π
12
0.2529
2π
12
0.4534
3π
12
0.5554
4π
12
0.5236
1126
5π
12
0.3388
6π
12
0
© 2014, John Bird
Hence, using Simpson’s rule,
∫
π /2
0
1 π 
x cos x d x ≈   ( 0 + 0 ) + 4 ( 0.2529 + 0.5554 + 0.3388 ) + 2 ( 0.4534 + 0.5236 ) 
3  12 
π 
=   [ 0 + 4.5884 + 1.9540]
 36 
π 
=   [ 6.5424] = 0.571
 36 
5. Evaluate using Simpson’s rule, giving the answers correct to 3 decimal places:
∫
π /3
0
e x2 sin 2 x d x
(use 10 intervals)
π
Since
∫
π /3
0
x
−0
π
3
= rad
e sin 2 x d x , width of interval =
10
30
x2
0
π
30
3π
30
2π
30
5π
30
4π
30
6π
30
7π
30
8π
30
9π
30
10π
30
e sin 2 x
x2
0 0.2102 0.4250 0.6488 0.8857 1.1392 1.4114 1.7021 2.0064 2.3119 2.5929
Hence, using Simpson’s rule,
∫
π /3
0

1  π  ( 0 + 2.5929 ) + 4 ( 0.2102 + 0.6488 + 1.1392 + 1.7021 + 2.3119 )
e x2 sin 2 x d x ≈   

3  30  
+ 2 ( 0.4250 + 0.8857 + 1.4114 + 2.0064 ) 
π 
=   [ 2.5929 + 24.0488 + 9.4570]
 90 
π 
=   [36.0987 ] = 1.260
 90 
6. Evaluate using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate rule, (d) Simpson’s rule.
Give answers correct to 3 decimal places.
∫
4
1
4
dx
x3
(use 6 intervals)
1127
© 2014, John Bird
(a)
4
∫
1
4
=
dx
x3
∫
4
1
 −2   −2  
 4 x −2 
 −2 
=
4 x=
dx  =
  −    = 1.875



 −2  1  x 2  1  16   1  
4
(b) Width of interval =
x
4
x3
4
−3
4 −1
= 0.5
6
1.0
1.5
2.0
2.5
4.0000
1.1852
0.5000
0.2560
3.0
3.5
4.0
0.1481 0.0933 0.0625
Hence, using the trapezoidal rule,
∫
4
1
4
1

d x ≈ (0.5)  ( 4.0000 + 0.0625 ) + 1.1852 + 0.5000 + 0.2560 + 0.1481 + 0.0933
3
x
2

= (0.5)[4.21385] = 2.107
(c) Mid-ordinates occur at 1.25, 1.75, 2.25, 2.75, 3.25 and 3.75
x
4
x3
1.25
1.75
2.25
2.75
3.25
3.75
2.0480 0.7464 0.3512 0.1923 0.1165 0.0759
Using the mid-ordinate rule,
∫
4
1
4
d x ≈ (0.5)[ 2.0480 + 0.7464 + 0.3512 + 0.1923 + 0.1165 + 0.0759 ]
x3
= (0.5)[3.5303] = 1.765
(d) Using the table of values from part (b), using Simpson’s rule,
∫
4
1
( 4.0000 + 0.0625 ) + 4 (1.1852 + 0.2560 + 0.0933)

4
1
d x ≈ (0.5) 

3
x3
+ 2 ( 0.5000 + 0.1481) 

=
1
(0.5) [ 4.0625 + 6.1380 + 1.2962]
3
=
1
(0.5) [11.4967 ] = 1.916
3
7. Evaluate using (a) integration, (b) the trapezoidal rule, (c) the mid-ordinate rule, (d) Simpson’s rule.
Give answers correct to 3 decimal places.
∫
6
2
1
dx
(2 x − 1)
(use 8 intervals)
1128
© 2014, John Bird
(a)
∫
6
2
1
dx
(2 x − 1)
Let u = 2x – 1, then
du
=2
dx
and
dx =
du
2
1
Thus,
Hence,
1
d=
x
(2 x − 1)
∫
∫
6
2
1
dx= 

(2 x − 1)
(b) Width of interval =
x
1
( 2 x − 1)
∫
1
1 du 1
1 u2
−
u 2 d=
u
=
=
∫
2 1
u 2 2
2
6
( 2 x − 1)  2 =
u
=
( 2 x − 1)
 11 − 3  = 1.585


6−2
= 0.5
8
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
0.5774 0.5000 0.4472 0.4082 0.3780 0.3536 0.3333 0.3162 0.3015
Hence, using the trapezoidal rule,
∫
6
2
1
dx
(2 x − 1)
1

≈ (0.5)  ( 0.5774 + 0.3015 ) + 0.5000 + 0.4472 + 0.4082 + 0.3780 + 0.3536 + 0.3333 + 0.3162 
2

= (0.5)[3.17595] = 1.588
(c) Mid-ordinates occur at 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25 and 5.75
x
1
( 2 x − 1)
2.25
2.75
3.25
3.75
4.25
4.75
5.25
5.75
0.5345 0.4714 0.4264 0.3922 0.3651 0.3430 0.3244 0.3086
Using the mid-ordinate rule,
∫
6
2
1
d x ≈ (0.5)[ 0.5345 + 0.4714 + 0.4264 + 0.3922 + 0.3651 + 0.3430
(2 x − 1)
+ 0.3244 + 0.3086]
= (0.5)[3.1656] = 1.583
(d) Using the table of values from part (b), using Simpson’s rule,
∫
6
2
( 0.5774 + 0.3015 ) + 4 ( 0.5000 + 0.4082 + 0.3536 + 0.3162 )

1
1
d x ≈ (0.5) 

3
+ 2 ( 0.4472 + 0.3780 + 0.3333) 
(2 x − 1)

=
1
(0.5) [ 0.8789 + 6.312 + 2.317 ]
3
1129
© 2014, John Bird
1
(0.5) [9.5079] = 1.585
3
=
8. Evaluate
∫
3
0
(1 + x 4 ) d x using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson’s rule.
Use 6 intervals in each case and give answers correct to 3 decimal places.
(a) Width of interval =
x
(1 + x 4 )
3−0
= 0.5
6
0
0.5
1.0
1.5
2.0
2.5
3.0
1.0000
1.0308
1.4142
2.4622
4.1231
6.3295
9.0554
Hence, using the trapezoidal rule,
∫
3
0
(1 + x 4 ) d x
1

≈ (0.5)  (1.0000 + 9.0554 ) + 1.0308 + 1.4142 + 2.4622 + 4.1231 + 6.3295
2

= (0.5)[20.3875] = 10.194
(b) Mid-ordinates occur at 2.25, 2.75, 3.25, 3.75, 4.25, 4.75, 5.25 and 5.75
x
(1 + x 4 )
0.25
0.75
1.25
1.75
2.25
2.75
1.0020 1.1473 1.8551 3.2216 5.1603 7.6283
Using the mid-ordinate rule,
∫
3
0
(1 + x 4 ) d x ≈ (0.5)[ 1.0020 + 1.1473 + 1.8551 + 3.2216 + 5.1603 + 7.6283 ]
= (0.5)[20.0146] = 10.007
(c) Using the table of values from part (b), using Simpson’s rule,
∫
3
0
(1.0000 + 9.0554 ) + 4 (1.0308 + 2.4622 + 6.3295 )

1
(1 + x 4 ) d x ≈ (0.5) 

3
+ 2 (1.4142 + 4.1231) 

=
1
(0.5) [10.0554 + 39.2900 + 11.0746]
3
=
1
(0.5) [ 60.42] = 10.070
3
1130
© 2014, John Bird
∫
9. Evaluate
0.7
0.1
1
d y using (a) the trapezoidal rule, (b) the mid-ordinate rule, (c) Simpson’s
(1 − y 2 )
rule. Use 6 intervals in each case and give answers correct to 3 decimal places.
∫
(a) Since
1
0.7
(1 − y 2 )
0.1
d y then width of interval =
y
1
0.1
(1 − y 2 )
0.2
0.3
0.7 − 0.1
= 0.1
6
0.4
0.5
0.6
0.7
1.0050 1.0206 1.0483 1.0911 1.1547 1.2500 1.4003
Hence, using the trapezoidal rule,
∫
1

d y ≈ (0.1)  (1.0050 + 1.4003) + 1.0206 + 1.0483 + 1.0911 + 1.1547 + 1.2500 
2

(1 − y 2 )
1
0.7
0.1
= (0.1)[6.76735] = 0.677
(b) Mid-ordinates occur at 0.15, 0.25, 0.35, 0.45, 0.55 and 0.65
y
0.15
0.25
0.35
0.45
0.55
0.65
1
(1 − y 2 )
1.0114 1.0328 1.0675 1.1198 1.1974 1.3159
Using the mid-ordinate rule,
∫
1
0.7
(1 − y 2 )
0.1
d y ≈ (0.1)[1.0114 + 1.0328 + 1.0675 + 1.1198 + 1.1974 + 1.3159]
= (0.1)[6.7448] = 0.674
(c) Using the table of values from part (a), using Simpson’s rule,
∫
0.7
0.1
1
1
d y ≈ (0.1) (1.0050 + 1.4003) + 4 (1.0206 + 1.0911 + 1.2500 ) + 2 (1.0483 + 1.1547 ) 
3
(1 − y 2 )
=
1
(0.1) [ 2.4053 + 13.4468 + 4.406]
3
=
1
(0.1) [ 20.258] = 0.675
3
1131
© 2014, John Bird
10. A vehicle starts from rest and its velocity is measured every second for 8 seconds, with values as
follows:
time t (s)
0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
velocity v (ms–1)
0
0.4
1.0
1.7
2.9
4.1
6.2
8.0
9.4
The distance travelled in 8.0 seconds is given by
∫
8.0
0
vdt
Estimate this distance using Simpson’s rule, giving the answer correct to 3 significant figures.
∫
8.0
0
vdt ≈
=
1
(1.0 ) [(0 + 9.4) + 4(0.4 + 1.7 + 4.1 + 8.0) + 2(1.0 + 2.9 + 6.2)]
3
1
1
(86.4) = 28.8 m
[9.4 + 56.8 + 20.2] =
3
3
11. A pin moves along a straight guide so that its velocity v (m/s) when it is a distance x (m) from the
beginning of the guide at time t (s) is given in the table below.
t(s)
0
0.5
v(m/s) 0
0.052
1.0
1.5
2.0
2.5
3.0
3.5
4.0
0.082 0.125 0.162 0.175 0.186 0.160 0
Use Simpson’s rule with 8 intervals to determine the approximate total distance travelled by the pin
in the 4.0 second period.
Distance travelled by pin
1
≈ (0.5) [ (0 + 0) + 4(0.052 + 0.125 + 0.175 + 0.160) + 2(0.082 + 0.162 + 0.186) ]
3
=
1
(0.5) [ 0 + 2.048 + 0.86]
3
1
= (0.5) [ 2.908]
3
= 0.485 m
1132
© 2014, John Bird