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8TECHNIQUES OF INTEGRATIONTECHNIQUES OF INTEGRATION
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There are two situations in which it is
impossible to find the exact value of
a definite integral.
TECHNIQUES OF INTEGRATION
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TECHNIQUES OF INTEGRATION
The first situation arises from the fact that,
in order to evaluate using the
Fundamental Theorem of Calculus (FTC),
we need to know an antiderivative of f.
( )b
af x dx
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TECHNIQUES OF INTEGRATION
However, sometimes, it is difficult, or
even impossible, to find an antiderivative
(Section 7.5).
For example, it is impossible to evaluate the following integrals exactly:
21 1 3
0 11xe dx x dx
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TECHNIQUES OF INTEGRATION
The second situation arises when the function
is determined from a scientific experiment
through instrument readings or collected data.
There may be no formula for the function (as we will see in Example 5).
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TECHNIQUES OF INTEGRATION
In both cases, we need to find
approximate values of definite
integrals.
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7.7Approximate Integration
In this section, we will learn:
How to find approximate values
of definite integrals.
TECHNIQUES OF INTEGRATION
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APPROXIMATE INTEGRATION
We already know one method for
approximate integration.
Recall that the definite integral is defined as a limit of Riemann sums.
So, any Riemann sum could be used as an approximation to the integral.
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APPROXIMATE INTEGRATION
If we divide [a, b] into n subintervals
of equal length ∆x = (b – a)/n, we have:
where xi* is any point in the i th subinterval
[xi -1, xi].
1
( ) ( *)nb
iai
f x dx f x x
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Ln APPROXIMATION
If xi* is chosen to be the left endpoint of
the interval, then xi* = xi -1 and we have:
The approximation Ln is called the left endpoint approximation.
11
( ) ( )nb
n iai
f x dx L f x x
Equation 1
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If f(x) ≥ 0, the integral represents an area
and Equation 1 represents an approximation
of this area by the rectangles shown here.
Ln APPROXIMATION
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If we choose xi* to be the right endpoint,
xi* = xi and we have:
The approximation Rn is called right endpoint approximation.
Equation 2
1
( ) ( )nb
n iai
f x dx R f x x
Rn APPROXIMATION
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APPROXIMATE INTEGRATION
In Section 5.2, we also considered the case
where xi* is chosen to be the midpoint
of the subinterval [xi -1, xi].
ix
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Mn APPROXIMATION
The figure shows
the midpoint
approximation Mn.
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Mn APPROXIMATION
Mn appears to be better
than either Ln or Rn.
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THE MIDPOINT RULE
where
and
1 2
( )
[ ( ) ( ) ... ( )]
b
na
n
f x dx M
x f x f x f x
b ax
n
11 12 ( ) midpoint of [ , ]i i i i ix x x x x
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TRAPEZOIDAL RULE
Another approximation—called the
Trapezoidal Rule—results from averaging
the approximations in Equations 1 and 2,
as follows.
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TRAPEZOIDAL RULE
11 1
11
0 1 1 2
1
0 1 2
1
1( ) ( ) ( )
2
( ( ) ( ))2
( ( ) ( )) ( ( ) ( ))2
... ( ( ) ( ))
( ) 2 ( ) 2 ( )2
... 2 ( ) ( )
n nb
i iai i
n
i ii
n n
n n
f x dx f x x f x x
xf x f x
xf x f x f x f x
f x f x
xf x f x f x
f x f x
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THE TRAPEZOIDAL RULE
where ∆x = (b – a)/n and xi = a + i ∆x
0 1 2
1
( )
( ) 2 ( ) 2 ( )2
... 2 ( ) ( )
b
na
n n
f x dx T
xf x f x f x
f x f x
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TRAPEZOIDAL RULE
The reason for the name can be seen
from the figure, which illustrates the case
f(x) ≥ 0.
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TRAPEZOIDAL RULE
The area of the trapezoid that lies above
the i th subinterval is:
If we add the areas of all these trapezoids,we get the right side of the Trapezoidal Rule.
11
( ) ( )[ ( ) ( )]
2 2i i
i i
f x f x xx f x f x
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APPROXIMATE INTEGRATION
Approximate the integral
with n = 5, using:
a. Trapezoidal Rule
b. Midpoint Rule
Example 12
1(1/ )x dx
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APPROXIMATE INTEGRATION
With n = 5, a = 1 and b = 2,
we have: ∆x = (2 – 1)/5 = 0.2
So, the Trapezoidal Rule gives:
2
51
1 0.2[ (1) 2 (1.2) 2 (1.4)
22 (1.6) 2 (1.8) (2)]
1 2 2 2 2 10.1
1 1.2 1.4 1.6 1.8 2
0.695635
dx T f f fx
f f f
Example 1 a
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APPROXIMATE INTEGRATION
The approximation is illustrated
here.
Example 1 a
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APPROXIMATE INTEGRATION
The midpoints of the five subintervals
are: 1.1, 1.3, 1.5, 1.7, 1.9
Example 1 b
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APPROXIMATE INTEGRATION
So, the Midpoint Rule gives:
2
1
1[ (1.1) (1.3) (1.5)
(1.7) (1.9)]
1 1 1 1 1 1
5 1.1 1.3 1.5 1.7 1.9
0.691908
dx x f f fx
f f
Example 1 b
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APPROXIMATE INTEGRATION
In Example 1, we deliberately chose
an integral whose value can be computed
explicitly so that we can see how accurate
the Trapezoidal and Midpoint Rules are.
By the FTC,
2 211
1ln ] ln 2 0.693147...dx x
x
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APPROXIMATION ERROR
The error in using an approximation is
defined as the amount that needs to be
added to the approximation to make it
exact.
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APPROXIMATE INTEGRATION
From the values in Example 1, we see that
the errors in the Trapezoidal and Midpoint
Rule approximations for n = 5 are:
ET ≈ – 0.002488
EM ≈ 0.001239
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APPROXIMATE INTEGRATION
In general, we have:
( )
( )
b
T na
b
M na
E f x dx T
E f x dx M
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APPROXIMATE INTEGRATION
The tables show the results of calculations
similar to those in Example 1. However, these are for n = 5, 10, and 20 and for
the left and right endpoint approximations and also the Trapezoidal and Midpoint Rules.
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APPROXIMATE INTEGRATION
We can make several observations
from these tables.
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In all the methods. we get more accurate
approximations when we increase n. However, very large values of n result in so many
arithmetic operations that we have to beware of accumulated round-off error.
OBSERVATION 1
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OBSERVATION 2
The errors in the left and right endpoint
approximations are: Opposite in sign Appear to decrease by a factor of about 2
when we double the value of n
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OBSERVATION 3
The Trapezoidal and Midpoint Rules
are much more accurate than the endpoint
approximations.
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OBSERVATION 4
The errors in the Trapezoidal and Midpoint
Rules are: Opposite in sign Appear to decrease by a factor of about 4
when we double the value of n
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OBSERVATION 5
The size of the error in the Midpoint Rule
is about half that in the Trapezoidal Rule.
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MIDPOINT RULE VS. TRAPEZOIDAL RULE
The figure shows why we can usually expect
the Midpoint Rule to be more accurate than
the Trapezoidal Rule.
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MIDPOINT RULE VS. TRAPEZOIDAL RULE
The area of a typical rectangle in
the Midpoint Rule is the same as the area
of the trapezoid ABCD whose upper side is
tangent to the graph at P.
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MIDPOINT RULE VS. TRAPEZOIDAL RULE
The area of this trapezoid is closer to
the area under the graph than is the area
of that used in the Trapezoidal Rule.
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MIDPOINT RULE VS. TRAPEZOIDAL RULE
The midpoint error (shaded red) is smaller
than the trapezoidal error (shaded blue).
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OBSERVATIONS
These observations are corroborated
in the following error estimates—which
are proved in books on numerical
analysis.
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OBSERVATIONS
Notice that Observation 4 corresponds
to the n2 in each denominator because:
(2n)2 = 4n2
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APPROXIMATE INTEGRATION
That the estimates depend on the size of
the second derivative is not surprising if you
look at the figure.
f’’(x) measures how much the graph is curved.
Recall that f’’(x) measures how fast the slope of y = f(x) changes.
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ERROR BOUNDS
Suppose | f’’(x) | ≤ K for a ≤ x ≤ b.
If ET and EM are the errors in
the Trapezoidal and Midpoint Rules,
then3 3
2 2
( ) ( )and
12 24T M
K b a K b aE E
n n
Estimate 3
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ERROR BOUNDS
Let’s apply this error estimate to the
Trapezoidal Rule approximation in Example 1.
If f(x) = 1/x, then f’(x) = -1/x2 and f’’(x) = 2/x3.
As 1 ≤ x ≤ 2, we have 1/x ≤ 1; so,
3 3
2 2''( ) 2
1f x
x
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ERROR BOUNDS
So, taking K = 2, a = 1, b = 2, and n = 5
in the error estimate (3), we see:
3
2
2(2 1) 1
12(5) 150
0.006667
TE
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ERROR BOUNDS
Comparing this estimate with the actual error
of about 0.002488, we see that it can happen
that the actual error is substantially less than
the upper bound for the error given by (3).
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ERROR ESTIMATES
How large should we take n in order to
guarantee that the Trapezoidal and Midpoint
Rule approximations for are
accurate to within 0.0001?
Example 2
2
1(1/ )x dx
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ERROR ESTIMATES
We saw in the preceding calculation
that | f’’(x) | ≤ 2 for 1 ≤ x ≤ 2
So, we can take K = 2, a = 1, and b = 2 in (3).
Example 2
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ERROR ESTIMATES
Accuracy to within 0.0001 means that
the size of the error should be less than
0.0001
Therefore, we choose n so that:3
2
2(1)0.0001
12n
Example 2
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ERROR ESTIMATES
Solving the inequality for n,
we get
or
Thus, n = 41 will ensure the desired accuracy.
2 2
12(0.0001)n
140.8
0.0006n
Example 2
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ERROR ESTIMATES
It’s quite possible that a lower value
for n would suffice.
However, 41 is the smallest value for which the error-bound formula can guarantee us accuracy to within 0.0001
Example 2
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ERROR ESTIMATES
For the same accuracy with the Midpoint Rule,
we choose n so that:
This gives:
3
2
2(1)0.0001
24n
129
0.0012n
Example 2
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ERROR ESTIMATES
a. Use the Midpoint Rule with n = 10 to
approximate the integral
b. Give an upper bound for the error involved
in this approximation.
21
0.xe dx
Example 3
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ERROR ESTIMATES
As a = 0, b = 1, and n = 10, the Midpoint Rule
gives:21
0
0.0025 0.0225 0.0625 0.1225 0.2025
0.3025 0.4225 0.5625 0.7225 0.9025
[ (0.05) (0.15) ... (0.85) (0.95)]
0.1
1.460393
xe dx
x f f f f
e e e e e
e e e e e
Example 3 a
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ERROR ESTIMATES
The approximation is illustrated.
Example 3 a
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ERROR ESTIMATES
As f(x) = ex2, we have:
f’(x) = 2xex2 and f’’(x) = (2 + 4x2)ex2
Also, since 0 ≤ x ≤ 1, we have x2 ≤ 1.
Hence, 0 ≤ f’’(x) = (2 + 4x2) ex2 ≤ 6e
Example 3 b
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ERROR ESTIMATES
Taking K = 6e, a = 0, b = 1, and n = 10 in
the error estimate (3), we see that an upper
bound for the error is:3
2
6 (1)
24(10) 400
0.007
e e
Example 3 b
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ERROR ESTIMATES
Error estimates give upper bounds
for the error.
They are theoretical, worst-case scenarios.
The actual error in this case turns out to be about 0.0023
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APPROXIMATE INTEGRATION
Another rule for approximate integration
results from using parabolas instead
of straight line segments to approximate
a curve.
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APPROXIMATE INTEGRATION
As before, we divide [a, b] into n subintervals
of equal length h = ∆x = (b – a)/n.
However, this time, we assume n is an even
number.
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APPROXIMATE INTEGRATION
Then, on each consecutive pair of intervals,
we approximate the curve y = f(x) ≥ 0 by
a parabola, as shown.
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APPROXIMATE INTEGRATION
If yi = f(xi), then Pi(xi, yi) is the point on
the curve lying above xi.
A typical parabola passes through three consecutive points: Pi, Pi+1, Pi+2
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APPROXIMATE INTEGRATION
To simplify our calculations, we first
consider the case where:
x0 = -h, x1 = 0, x2 = h
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APPROXIMATE INTEGRATION
We know that the equation of
the parabola through P0, P1, and P2
is of the form
y = Ax2 + Bx + C
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APPROXIMATE INTEGRATION
Therefore, the area under the parabola
from x = - h to x = h is: 2 2
0
3
0
3
2
( ) 2 ( )
23
23
(2 6 )3
h h
h
h
Ax Bx C dx Ax C dx
xA Cx
hA Ch
hAh C
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APPROXIMATE INTEGRATION
However, as the parabola passes through
P0(- h, y0), P1(0, y1), and P2(h, y2), we have:
y0 = A(– h)2 + B(- h) + C = Ah2 – Bh + C
y1 = C
y2 = Ah2 + Bh + C
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APPROXIMATE INTEGRATION
Therefore,
y0 + 4y1 + y2 = 2Ah2 + 6C
So, we can rewrite the area under
the parabola as:0 1 2( 4 )
3
hy y y
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APPROXIMATE INTEGRATION
Now, by shifting this parabola
horizontally, we do not change
the area under it.
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APPROXIMATE INTEGRATION
This means that the area under the parabola
through P0, P1, and P2 from x = x0 to x = x2
is still:0 1 2( 4 )
3
hy y y
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APPROXIMATE INTEGRATION
Similarly, the area under the parabola
through P2, P3, and P4 from x = x2 to x = x4
is:2 3 4( 4 )
3
hy y y
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APPROXIMATE INTEGRATION
Thus, if we compute the areas under all
the parabolas and add the results, we get:
0 1 2 2 3 4
2 1
0 1 2 3 4
2 1
( ) ( 4 ) ( 4 )3 3
... ( 4 )3
( 4 2 4 23
... 2 4 )
b
a
n n n
n n n
h hf x dx y y y y y y
hy y y
hy y y y y
y y y
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APPROXIMATE INTEGRATION
Though we have derived this approximation
for the case in which f(x) ≥ 0, it is a reasonable
approximation for any continuous function f .
Note the pattern of coefficients: 1, 4, 2, 4, 2, 4, 2, . . . , 4, 2, 4, 1
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SIMPSON’S RULE
This is called Simpson’s Rule—after
the English the English mathematician
Thomas Simpson (1710–1761).
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SIMPSON’S RULE
where n is even and ∆x = (b – a)/n.
Rule
0 1 2 3
2 1
( )
[ ( ) 4 ( ) 2 ( ) 4 ( )3
... 2 ( ) 4 ( ) ( )]
b
na
n n n
f x dx S
xf x f x f x f x
f x f x f x
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SIMPSON’S RULE
Use Simpson’s Rule
with n = 10 to approximate
2
1(1/ )x dx
Example 4
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SIMPSON’S RULE
Putting f(x) = 1/x, n = 10, and ∆x = 0.1 in
Simpson’s Rule, we obtain:
2
101
1[ (1) 4 (1.1) 2 (1.2) 4 (1.3)
3... 2 (1.8) 4 (1.9) (2)]
1 4 2 4 2 4 2 40.1 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7
2 4 13
1.8 1.9 20.693150
xdx S f f f fx
f f f
Example 4
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SIMPSON’S RULE
In Example 4, notice that Simpson’s Rule
gives a much better approximation
(S10 ≈ 0.693150) to the true value of the
integral (ln 2 ≈ 0.693147) than does either:
Trapezoidal Rule (T10 ≈ 0.693771)
Midpoint Rule (M10 ≈ 0.692835)
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SIMPSON’S RULE
It turns out that the approximations in
Simpson’s Rule are weighted averages of
those in the Trapezoidal and Midpoint Rules:
Recall that ET and EM usually have opposite signs and | EM | is about half the size of | ET |.
1 22 3 3n n nS T M
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SIMPSON’S RULE
In many applications of calculus, we need to
evaluate an integral even if no explicit formula
is known for y as a function of x.
A function may be given graphically or as a table of values of collected data.
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SIMPSON’S RULE
If there is evidence that the values are not
changing rapidly, then the Trapezoidal Rule
or Simpson’s Rule can still be used to find
an approximate value for .b
ay dx
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SIMPSON’S RULE
The figure shows data traffic on the link from
the U.S. to SWITCH, the Swiss academic and
research network, on February 10, 1998.
D(t) is the data throughput, measured in megabits per second (Mb/s).
Example 5
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SIMPSON’S RULE
Use Simpson’s Rule to estimate the total
amount of data transmitted on the link up to
noon on that day.
Example 5
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SIMPSON’S RULE
Since we want the units to be consistent
and D(t) is measured in Mb/s, we convert
the units for t from hours to seconds.
Example 5
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SIMPSON’S RULE
If we let A(t) be the amount of data (in Mb)
transmitted by time t, where t is measured in
seconds, then A’(t) = D(t).
So, by the Net Change Theorem (Section 5.4), the total amount of data transmitted by noon (when t = 12 x 602 = 43,200) is:
43,200
0(43,200) ( )A D t dt
Example 5
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SIMPSON’S RULE
We estimate the values of D(t) at hourly
intervals from the graph and compile them
here.
Example 5
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SIMPSON’S RULE
Then, we use Simpson’s Rule
with n = 12 and ∆t = 3600 to estimate
the integral, as follows.
Example 5
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SIMPSON’S RULE
43,200
0( )
[ (0) 4 (3600) 2 (7200)3... 4 (39,600) (43,200)]
3600[3.2 4(2.7) 2(1.9) 4(1.7)
32(1.3) 4(1.0) 2(1.1) 4(1.3)
2(2.8) 4(5.7) 2(7.1) 4(7.7) 7.9] 143,880
A t dt
tD D D
D D
Example 5
The total amount of data transmitted up to noon is 144,000 Mbs, or 144 gigabits.
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SIMPSON’S RULE VS. MIDPOINT RULE
The table shows how Simpson’s Rule
compares with the Midpoint Rule for
the integral , whose true value
is about 0.69314718
2
1(1/ )x dx
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SIMPSON’S RULE
This table shows how the error Es
in Simpson’s Rule decreases by
a factor of about 16 when n is doubled.
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SIMPSON’S RULE
That is consistent with the appearance of n4
in the denominator of the following error
estimate for Simpson’s Rule.
It is similar to the estimates given in (3) for the Trapezoidal and Midpoint Rules.
However, it uses the fourth derivative of f.
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ERROR BOUND (SIMPSON’S RULE)
Suppose that | f (4)(x) | ≤ K for a ≤ x ≤ b.
If Es is the error involved in using
Simpson’s Rule, then 5
4
( )
180s
K b aE
n
Estimate 4
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ERROR BOUND (SIMPSON’S RULE)
How large should we take n to guarantee
that the Simpson’s Rule approximation
for is accurate to within 0.0001?2
1(1/ )x dx
Example 6
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ERROR BOUND (SIMPSON’S RULE)
If f(x) = 1/x, then f (4)(x) = 24/x5.
Since x ≥ 1, we have 1/x ≤ 1, and so
Thus, we can take K = 24 in (4).
(4)5
24( ) 24f x
x
Example 6
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ERROR BOUND (SIMPSON’S RULE)
So, for an error less than 0.0001, we should
choose n so that:
This gives
or
5
4
24(1)0.0001
180n
4 24
180(0.0001)n
4
16.04
0.00075n
Example 6
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ERROR BOUND (SIMPSON’S RULE)
Therefore, n = 8 (n must be even)
gives the desired accuracy.
Compare this with Example 2, where we obtained n = 41 for the Trapezoidal Rule and n = 29 for the Midpoint Rule.
Example 6
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ERROR BOUND (SIMPSON’S RULE)
a. Use Simpson’s Rule with n = 10
to approximate the integral .
b. Estimate the error involved in this
approximation.
21
0
xe dx
Example 7
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ERROR BOUND (SIMPSON’S RULE)
If n =10, then ∆x = 0.1 and the rule gives:
21
0
0 0.01 0.04 0.09 0.16
0.25 0.36 0.49 0.64 0.81 1
[ (0) 4 (0.1) 2 (0.2) ...3
2 (0.8) 4 (0.9) (1)]
0.1[ 4 2 4 2
3
4 2 4 2 4 ]
1.462681
x xe f f f
f f f
e e e e e
e e e e e e
Example 7 a
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ERROR BOUND (SIMPSON’S RULE)
The fourth derivative of f(x) = ex2 is:
f(4)(x) = (12 + 48x2 + 16x4)ex2
So, since 0 ≤ x ≤ 1, we have:
0 ≤ f(4)(x) ≤ (12 + 48 +16)e1 = 76e
Example 7 b
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ERROR BOUND (SIMPSON’S RULE)
Putting K = 76e, a = 0, b = 1, and n = 10
in (4), we see that the error is at most:
Compare this with Example 3.
5
4
76 (1)0.000115
180(10)
e
Example 7 b
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ERROR BOUND (SIMPSON’S RULE)
Thus, correct to three decimal places,
we have:
21
01.463xe dx
Example 7 b