Trapezoidal Rule of Integration
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Trapezoidal Rule of Integration
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What is IntegrationIntegration:
b
adx)x(fI
The process of measuring the area under a function plotted on a graph.
Where: f(x) is the integranda= lower limit of integrationb= upper limit of integration
f(x)
a b
b
a
dx)x(f
y
x
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Basis of Trapezoidal Rule
b
adx)x(fI
Trapezoidal Rule is based on the Newton-Cotes Formula that states if one can approximate the integrand as an nth order polynomial…
where )x(f)x(f n
nn
nnn xaxa...xaa)x(f
1110and
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Basis of Trapezoidal Rule
b
an
b
a)x(f)x(f
Then the integral of that function is approximated by the integral of that nth order polynomial.
Trapezoidal Rule assumes n=1, that is, the area under the linear polynomial,
2)b(f)a(f)ab(
b
adx)x(f
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Derivation of the Trapezoidal Rule
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Method Derived From Geometry
The area under the curve is a trapezoid. The integral
trapezoidofAreadxxfb
a
)(
)height)(sidesparallelofSum(21
)ab()a(f)b(f 21
2)b(f)a(f)ab(
Figure 2: Geometric Representation
f(x)
a b
b
a
dx)x(f1
y
x
f1(x)
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Example 1The vertical distance covered by a rocket from t=8 to t=30 seconds is given by:
30
8
892100140000
1400002000 dtt.t
lnx
a) Use single segment Trapezoidal rule to find the distance covered.
b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).
tEa
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Solution
2
)b(f)a(f)ab(I
a)8a 30b
t.t
ln)t(f 892100140000
1400002000
)(.)(
ln)(f 88982100140000
14000020008
)(.)(
ln)(f 3089302100140000
140000200030
s/m.27177
s/m.67901
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Solution (cont)
2
6790127177830 ..)(I
m11868
a)
b)The exact value of the above integral is
30
8
892100140000
1400002000 dtt.t
lnx m11061
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Solution (cont)b) ValueeApproximatValueTrueEt
1186811061
m807
c) The absolute relative true error, , would be t
10011061
1186811061
t %.29597
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Multiple Segment Trapezoidal Rule
In Example 1, the true error using single segment trapezoidal rule was large. We can divide the interval [8,30] into [8,19] and [19,30] intervals and apply Trapezoidal rule over each segment.
t.t
ln)t(f 892100140000
1400002000
30
19
19
8
30
8dt)t(fdt)t(fdt)t(f
2
30191930
2198
819)(f)(f
)()(f)(f
)(
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Multiple Segment Trapezoidal Rule
With
s/m.)(f 271778
s/m.)(f 7548419
s/m.)(f 6790130
267.90175.484)1930(
275.48427.177)819()(
30
8
dttf
m11266
Hence:
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Multiple Segment Trapezoidal Rule
1126611061 tE
m205
The true error is:
The true error now is reduced from -807 m to -205 m.
Extending this procedure to divide the interval into equal segments to apply the Trapezoidal rule; the sum of the results obtained for each segment is the approximate value of the integral.
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Multiple Segment Trapezoidal Rule
f(x)
a b
y
x
4aba
42 aba
4
3 aba
Figure 4: Multiple (n=4) Segment Trapezoidal Rule
Divide into equal segments as shown in Figure 4. Then the width of each segment is:
n
abh
The integral I is:
b
adx)x(fI
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Multiple Segment Trapezoidal Rule
b
h)n(a
h)n(a
h)n(a
ha
ha
ha
adx)x(fdx)x(f...dx)x(fdx)x(f
1
1
2
2
The integral I can be broken into h integrals as:
b
adx)x(f
Applying Trapezoidal rule on each segment gives:
b
adx)x(f
)b(f)iha(f)a(f
nab n
i
1
12
2
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Example 2The vertical distance covered by a rocket from to seconds is given by:
30
889
21001400001400002000 dtt.
tlnx
a) Use two-segment Trapezoidal rule to find the distance covered.
b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).
atE
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Solutiona) The solution using 2-segment Trapezoidal rule is
)b(f)iha(f)a(f
nabI
n
i
1
12
2
2n 8a 30b
2830
n
abh 11
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Solution (cont)
)(f)iha(f)(f
)(I
i3028
22830 12
1
)(f)(f)(f 3019284
22
6790175484227177422 .).(.
m11266
Then:
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Solution (cont)
30
8
892100140000
1400002000 dtt.t
lnx m11061
b) The exact value of the above integral is
so the true error is
ValueeApproximatValueTrueEt
1126611061
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Solution (cont)
c)The absolute relative true error, , would be t
100Value TrueError True
t
10011061
1126611061
%8534.1
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Solution (cont)Table 1 gives the values obtained using multiple segment Trapezoidal rule for:
n Value Et
1 11868 -807 7.296 ---2 11266 -205 1.853 5.3433 11153 -91.4 0.8265 1.0194 11113 -51.5 0.4655 0.35945 11094 -33.0 0.2981 0.16696 11084 -22.9 0.2070 0.0908
27 11078 -16.8 0.1521 0.0548
28 11074 -12.9 0.1165 0.0356
0
30
889
21001400001400002000 dtt.
tlnx
Table 1: Multiple Segment Trapezoidal Rule Values
%t %a
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Example 3Use Multiple Segment Trapezoidal Rule to find the area under the curve
xex)x(f
1300 from to 0x 10x
.
Using two segments, we get 52
010
h
01
03000
0
e)(
)(f 039101
53005
5.
e)(
)(f
13601
1030010
10.
e)(
)(f
and
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Solution
)b(f)iha(f)a(f
nabI
n
i
1
12
2
)(f)(f)(f
)( i105020
22010 12
1
)(f)(f)(f 10520
410
136003910204
10 .).(
53550.
Then:
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Solution (cont)So what is the true value of this integral?
59246130010
0.dx
exx
Making the absolute relative true error:
%.
..t 100
592465355059246
%.50679
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Solution (cont)
n Approximate Value
1 0.681 245.91 99.724%2 50.535 196.05 79.505%4 170.61 75.978 30.812%8 227.04 19.546 7.927%16 241.70 4.887 1.982%32 245.37 1.222 0.495%64 246.28 0.305 0.124%
Table 2: Values obtained using Multiple Segment Trapezoidal Rule for:
10
01300 dx
exx
tE t
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