Introduction
Derivation
Examples
Summary
Test
TRAPEZIUM RULE
NUMERICAL METHODS 3
INU0114/514 (M ATHS 1)
Dr Adrian Jannetta MIMA CMath FRAS
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Background
of complex shapes. Kepler used the
equivalent of a method later called
Simpson’s Rule to calculate the volume of
wine barrels.
Mathematicians now use definite
integration to calculate quantities such as
distance, area, volume, mass, electric
charge and others.
You will have already seen that the definite
integral
Z b
f (x) dx
a
Johannes Kepler (1571 — 1630)
Several mathematicians, including
Johannes Kepler, discovered approximate
methods for calculating areas and volumes
Trapezium Rule
can found by doing indefinite integration
and then substituting the limits. But in
some cases this is not possible.
Methods of numerical (or approximate)
integration exist and don’t rely rely on
indefinite integration to find the value of
the integral.
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Numerical Integration
Definite integrals of the form
Z
b
f (x) dx
a
represent the area beneath the curve by y = f (x) on a graph.
Numerical integration is useful when:
• We can’t find the indefinite integral of f (x).
• We have data in the form of samples or measurements (e.g. of electric
current at a given time points).
Numerical methods divide the region under the curve into regular regions whose
area can be calculated and summed.
• Trapezium rule (partitions the region using trapezoids).
• Simpson’s rule (replaces sections of the curve with quadratic curves).
In this presentation we will examine the Trapezium rule. Simpson’s Rule will
come next time!
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
The Trapezium Rule
A trapezium is a four sided polygon with two parallel sides.
q
p
The area of a trapezium is equal to h2 (p + q).
h
Consider the function y = f (x). We want to find the area beneath the curve which is
bounded by the lines x = a, x = b and the x-axis.
y
y4
y2
y = f (x)
y3
y1
y0
h
x0
a
x1
x2
x3
x4
x
b
We can approximate the area beneath the curve by covering it with trapezia of width h as
shown above. The x-values are regularly spaced across the interval. The y-values (called
ordinates) can be calculated using y = f (x).
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
y
y2
y4
y = f (x)
y3
y1
y0
h
x0
a
x1
x2
x3
x4
x
b
In the picture the total area beneath the curve is being approximated by the combined area
of four trapezia.
Area =
h
h
h
h
(y0 + y1 ) + (y1 + y2 ) + (y2 + y3 ) + (y3 + y4 )
2
2
2
2
Factorise and collect the like-terms together:
Area =
Z
Area
=
h
(y0 + y1 + y1 + y2 + y2 + y3 + y3 + y4 )
2
f (x) dx
=
h
(y0 + 2y1 + 2y2 + 2y3 + y4 )
2
b
a
Using more strips should approximate the area much more closely.
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
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The general case of dividing the interval into n strips of width h leads to the
following expression for the approximate area beneath the curve:
Z
f (x) dx = 21 (y0 + y1 )h + 12 (y1 + y2 )h + . . . + 21 (yn−2 + yn−1 )h + 12 (yn−1 + yn )h
which can be factorised and simplified to give
Z b
h
f (x) dx = (y0 + 2y1 + . . . + 2yn−1 + yn )
2
a
Or more compactly
Z
b
f (x) dx =
a
h
y0 + 2(y1 + . . . + yn−1 ) + yn
2
,h=
b−a
n
This is the trapezium rule. An easier version of this formula to remember is:
Z
b
f (x) dx =
a
h
F + L + 2(everything else)
2
,h=
b−a
n
where ‘F’ and ‘L’ are first the first ordinates.
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Area under a curve
Trapezium Rule
Use the trapezium rule with 4 strips to calculate the area beneath the curve y = x2
bounded by the lines x = 1, x = 4 and the x-axis.
In this case we have a = 1, b = 4 and n = 4.
The strip width is h =
4−1
4
=
3
4
Construct a table of values to calculate the values of y:
Z
x
1
1.75
2.5
3.25
4
y
1
3.0625
6.25
10.5625
16
4
x2 dx
1
≈
≈
Z
≈
4
x2 dx
1
≈
3
4
2
3
8
3
8
F + L + 2(everything else)
[1 + 16 + 2(3.0625 + 6.25 + 10.5625)]
× 56.75
21.28125
So the area beneath the curve is 21.28 units2 (to 2 decimal places).
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Area under a curve
In the previous example we can compare the exact answer to to the Trapezium
rule estimate by evaluating the definite integral
Z
4
x2 dx =
1
x3
3
4
= 21
1
y
The trapezium rule produced a slight
overestimate of the actual area.
y = x2
8
6
This could have been anticipated.
Consider the shape of the curve and
the shape of the trapezia on a graph.
4
2
The dashed line is the straight edge at
the top of the trapezium; you can see
that area of each trapezium is slightly
greater than the area under the curve.
Trapezium Rule
x
1
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2
3
Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Area under a curve
Trapezium Rule
p
Use the trapezium rule with 4 ordinates to calculate the area beneath the curve y = sin θ
bounded by the axis and the lines θ = 0 and θ = π/2. Give your answer to four decimal
places.
The use of 4 ordinates means using 3 strips.
We have a = 0 and b = π/2 and h =
π/2−0
3
Construct a table of values using y =
Z
p
0
π
6
π
3
π
2
y
0
0.70711
0.93060
1
sin θ dθ
≈
≈
π/2
0
p
sin θ
0
≈
Z
π
6.
θ
π/2
p
=
sin θ dθ
≈
h
F + L + 2(everything else)
2
π
12
π
12
[0 + 2(0.70711 + 0.93060) + 1]
× 4.27542
1.11930
The estimated area is 1.1193 units2 .
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Area under a curve
Is the answer we obtained an underestimate or overestimate of the true area? We
can answer this by examining the graph of the function.
y
1
y=
π
6
π
3
π
2
p
sin θ
θ
The curve is convex to the trapezia; the trapezia have an area less than the area
beneath the curve, so the method gave a slight underestimate of the true answer.
As always, a more accurate answer could be obtained by using more strips.
Trapezium Rule
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Introduction
Derivation
Examples
Summary
Test
Using data points
Numerical integration methods can be used with data points or measurements — useful
when the underlying function is not known.
Trapezium rule with data points
Calculate the value of the integral
t
0
5
10
15
20
Z
50
v dt from the following data:
0
25
30
35
40
45
50
v 5.3 17.6 32.4 44.8 59.1 71.9 86.1 99.7 112.5 126.6 140.1
In this case the measurements can simply be substituted into the Trapezium rule formula.
Clearly the interval width is h = 5.
Z 50
h
F + L + 2(everything else)
v dt ≈
2
0
≈
Z
Trapezium Rule
≈
50
v dt
0
≈
5
2
5
2
[5.3 + 2(17.6 + 32.4 + . . . + 126.5) + 140.1]
× [145.4 + 2 (650.7)]
3617
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Introduction
Derivation
Examples
Summary
Test
Summary
The trapezium rule is a method for calculating the approximate value of definite
integrals. This might be essential when:
• The integration is too difficult or impossible to carry out.
• The function is only specified in the form of data points.
In either case the formula for evaluating the integral is:
Z
b
f (x) dx =
a
h
y0 + 2(y1 + . . . + yn−1 ) + yn
2
...or this easier to remember version:
Zb
h
F + L + 2(everything else)
f (x) dx =
2
a
where h =
b−a
.
n
The trapezium rule estimate can always be improved by using more strips (so
larger n or smaller h).
Trapezium Rule
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Adrian Jannetta
Introduction
Derivation
Examples
Summary
Test
Test yourself...
You should be able to solve the following problems using the trapezium rule if
you have understood these notes.
1
2
3
Use h = 0.5 to evaluate
Z
3
(x2 + 1) dx
1
Find the exact value of the integral and hence find the percentage error in
the estimate in (1).
Z − π2
cos x dx to 4 decimal places.
Use four strips to evaluate
π
2
4
Does the value you found in (3) under or over estimate the true value?
Answers:
43
1 10.75 (or 4 )
25
32
2 Exact value (using integration) is 3 . Percentage error is 32 % (≈ 0.78%)
3 1.8961 (to 4 decimal places)
4 It’s an underestimate; sketch the graph of cos x and the trapezia to verify this.
Trapezium Rule
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Adrian Jannetta