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4√2

h _ 16




h = 24

Fig. 8.42

According to our analysis, the total force on the upper half of the window is thus
20 44880
h · 50 · (2h ’ 32) dh = lbs.
3
16
For the lower half of the window, we examine the isosceles right triangle in
Fig. 8.43. It has base 24 ’ h. Therefore, for h ranging from 20 to 24, we have
w(h) = 2(24 ’ h) = 48 ’ 2h.
According to our analysis, the total force on the lower half of the window is
24 51200
h · 50 · (48 ’ 2h) dh = lbs.
3
20
The total force on the entire window is thus
44880 51200 96080
+ = lbs.
3 3 3
You Try It: A tank of water has ¬‚at sides. In one side, with center 4 feet below the
surface of the water, is a circular window of radius 1 foot. What is the total force
on the window?
252 CHAPTER 8 Applications of the Integral

h = 16

4√2




24 _ h


h = 24

Fig. 8.43




8.7 Numerical Methods of Integration
While there are many integrals that we can calculate explicitly, there are many
others that we cannot. For example, it is impossible to evaluate

e’x dx.
2
(—)

That is to say, it can be proved mathematically that no closed-form antiderivative can
be written down for the function e’x . Nevertheless, (—) is one of the most important
2


integrals in all of mathematics, for it is the Gaussian probability distribution integral
that plays such an important role in statistics and probability.
Thus we need other methods for getting our hands on the value of an integral.
One method would be to return to the original de¬nition, that is to the Riemann
sums. If we need to know the value of
1
e’x dx
2

0

then we can approximate this value by a Riemann sum
1
e’x dx ≈ e’(0.25) · 0.25 + e’(0.5) · 0.25 + e’(0.75) · 0.25 + e’1 · 0.25.
2 2 2 2 2

0

A more accurate approximation could be attained with a ¬ner approximation:
10
1
’x 2
e’(j ·0.1) · 0.1
2
dx ≈ (——)
e
0 j =1
253
CHAPTER 8 Applications of the Integral

or
100
1
’x 2
e’(j ·0.01) · 0.01
2
dx ≈ ()
e
0 j =1
The trouble with these “numerical approximations” is that they are calcula-
tionally expensive: the degree of accuracy achieved compared to the number of
calculations required is not attractive.
Fortunately, there are more accurate and more rapidly converging methods for
calculating integrals with numerical techniques. We shall explore some of these in
the present section.
It should be noted, and it is nearly obvious to say so, that the techniques of
this section require the use of a computer. While the Riemann sum (——) could be
computed by hand with some considerable effort, the Riemann sum ( ) is all but
infeasible to do by hand. Many times one wishes to approximate an integral by the
sum of a thousand terms (if, perhaps, ¬ve decimal places of accuracy are needed).
In such an instance, use of a high-speed digital computer is virtually mandatory.

8.7.1 THE TRAPEZOID RULE
The method of using Riemann sums to approximate an integral is sometimes called
“the method of rectangles.” It is adequate, but it does not converge very quickly
and it begs more ef¬cient methods. In this subsection we consider the method of
approximating by trapezoids.
Let f be a continuous function on an interval [a, b] and consider a partition
P = {x0 , x1 , . . . , xk } of the interval. As usual, we take x0 = a and xk = b. We also
assume that the partition is uniform.




Fig. 8.44
In the method of rectangles we consider a sum of the areas of rectangles.
Figure 8.44 shows one rectangle, how it approximates the curve, and what error
254 CHAPTER 8 Applications of the Integral

is made in this particular approximation. The rectangle gives rise to a “triangular”
error region (the difference between the true area under the curve and the area of the
rectangle). We put quotation marks around the word “triangular” since the region
in question is not a true triangle but instead is a sort of curvilinear triangle. If we
instead approximate by trapezoids, as in Fig. 8.45 (which, again, shows just one
region), then at least visually the errors seem to be much smaller.




Fig. 8.45
In fact, letting x = xj ’ xj ’1 as usual, we see that the ¬rst trapezoid in the
¬gure has area [f (x0 )+f (x1 )]· x/2. The second has area [f (x1 )+f (x2 )]· x/2,
and so forth. In sum, the aggregate of the areas of all the trapezoids is
1 1
· {f (x0 ) + f (x1 )} · x + · {f (x1 ) + f (x2 )} · x + ···
2 2
1
+ · {f (xk’1 ) + f (xk )} · x
2
x
= · {f (x0 ) + 2f (x1 ) + 2f (x2 )
2
+ · · · + 2f (xk’1 ) + f (xk )}. (†)
It is known that, if the second derivative of f on the interval [a, b] does not exceed
M then the approximation given by the sum (†) is accurate to within
M · (b ’ a)3
.
12k 2
[By contrast, the accuracy of the method of rectangles is generally not better than
N · (b ’ a)2
,
2k
where N is an upper bound for the ¬rst derivative of f . We see that the
method of trapezoids introduces an extra power of (b ’ a) in the numerator
255
CHAPTER 8 Applications of the Integral

of the error estimate and, perhaps more importantly, an extra factor of k in the
denominator.]
EXAMPLE 8.29
Calculate the integral
1
e’x dx
2

0
to two decimal places of accuracy.
SOLUTION
We ¬rst calculate that if f (x) = e’x then f (x) = (4x 2 ’ 2)e’x and
2 2


therefore |f (x)| ¤ 2 = M for 0 ¤ x ¤ 1. In order to control the error, and to
have two decimal places of accuracy, we need to have
M · (b ’ a)3
< 0.005
12k 2
or
2 · 13
< 0.005.
12k 2
Rearranging this inequality gives
100
< k2.
3
Obviously k = 6 will do.
So we will use the partition P = {0, 1/6, 1/3, 1/2, 2/3, 5/6, 1}. The corres-
ponding trapezoidal sum is
1/6 ’02
+ 2e’(1/6) + 2e’(1/3) + 2e’(1/2)
2 2 2
S= ·e
2
+ 2e’(2/3) + 2e’(5/6) + e’1 .
2 2 2



Some tedious but feasible calculation yields then that
1
S= · {1 + 2 · .9726 + 2 · .8948 + 2 · .7880
12
+ 2 · .6412 + 2 · .4994 + .3679}
8.9599
= = .7451.
12
We may use a computer algebra utility like Mathematica or Maple to
calculate the integral exactly (to six decimal places) to equal 0.746824. We thus
see that the answer we obtained with the Trapezoid Rule is certainly accurate
to two decimal places. It is not accurate to three decimal places.
256 CHAPTER 8 Applications of the Integral

It should be noted that Maple and Mathematica both use numerical tech-
niques, like the ones being developed in this section, to calculate integrals. So our
calculations merely emulate what these computer algebra utilities do so swiftly and
so well.

You Try It: How ¬ne a partition would we have needed to use if we wanted four
decimal places of accuracy in the last example? If you have some facility with a
computer, use the Trapezoid Rule with that partition and con¬rm that your answer
agrees with Mathematica™s answer to four decimal places.
EXAMPLE 8.30
Use the Trapezoid Rule with k = 4 to estimate
1
1
dx.
1 + x2
0

SOLUTION
Of course we could calculate this integral precisely by hand, but the point
here is to get some practice with the Trapezoid Rule. We calculate
± 
 
1/4  1 1
1 1 1
S= · +2· +2· +2· + .
2  1+02 1+12 
2 2 2
 
1 2 3
1+ 4 1+ 4 1+ 4

A bit of calculation reveals that
1 5323
S= · ≈ 0.782794 . . . .
8 850
Now if we take f (x) = 1/(1 + x 2 ) then f (x) = (6x 2 ’ 2)/(1 + x 2 )3 .
Thus, on the interval [0, 1], we have that |f (x)| ¤ 4 = M. Thus the error
estimate for the Trapezoid Rule predicts accuracy of
M · (b ’ a)3 4 · 13
= ≈ 0.020833 . . . .
12 · 42
12k 2
This suggests accuracy of one decimal place.
Now we know that the true and exact value of the integral is arctan 1 ≈
0.78539816 . . .. Thus our Trapezoid Rule approximation is good to one, and
nearly to two, decimal places”better than predicted.

8.7.2 SIMPSON™S RULE
Simpson™s Rule takes our philosophy another step: If rectangles are good, and
trapezoids better, then why not approximate by curves? In Simpson™s Rule, we
approximate by parabolas.
257
CHAPTER 8 Applications of the Integral

We have a continuous function f on the interval [a, b] and we have a partition
P = {x0 , x1 , . . . , xk } of our partition as usual. It is convenient in this technique to
assume that we have an even number of intervals in the partition.




Fig. 8.46

Now each rectangle, over each segment of the partition, is capped off by an
arc of a parabola. Figure 8.46 shows just one such rectangle. In fact, for each pair
of intervals [x2j ’2 , x2j ’1 ], [x2j ’1 , x2j ], we consider the unique parabola passing
through the endpoints

(—)
(x2j ’2 , f (x2j ’2 )), (x2j ’1 , f (x2j ’1 )), (x2j , f (x2j )).

Note that a parabola y = Ax 2 + Bx + C has three undetermined coef¬cients, so
three points as in (—) will determine A, B, C and pin down the parabola.
In fact (pictorially) the difference between the parabola and the graph of f is
so small that the error is almost indiscernible. This should therefore give rise to a
startling accurate approximation, and it does.
Summing up the areas under all the approximating parabolas (we shall not
perform the calculations) gives the following approximation to the integral:
b x
f (x) dx ≈ {f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 )
3
a
+ 2f (x4 ) + · · · + 2f (xk’2 ) + 4f (xk’1 ) + f (xk )}.

If it is known that the fourth derivative f (iv) (x) satis¬es |f (iv) (x)| ¤ M on [a, b],
then the error resulting from Simpson™s method does not exceed

M · (b ’ a)5
.
180 · k 4
258 CHAPTER 8 Applications of the Integral

EXAMPLE 8.31
e’x dx to two decimal places of
2
1
Use Simpson™s Rule to calculate 0
accuracy.
SOLUTION
If we set f (x) = e’x then it is easy to calculate that
2



f (iv) (x) = e’x · [12 ’ 72x 2 + 32x 4 ].
2



Thus |f (x)| ¤ 12 = M.
In order to achieve the desired degree of accuracy, we require that
M · (b ’ a)5
< 0.005
180 · k 4
or
12 · 15
< 0.005.
180 · k 4


Y
Simple manipulation yields
FL
200
< k4.
15
AM

This condition is satis¬ed when k = 2.
Thus our job is easy. We take the partition P = {0, 1/2, 1}. The sum arising
from Simpson™s Rule is then
TE



1/2
S= {f (0) + 4f (1/2) + f (1)}
3
1
= {e’0 + 4 · e’(1/2) + e’1 }
2 2 2

6
1
= {1 + 3.1152 + 0.3679}
6
1
≈ · 4.4831
6
≈ 0.7472
Comparing with the “exact value” 0.746824 for the integral that we noted in
Example 8.29, we ¬nd that we have achieved two decimal places of accuracy.
It is interesting to note that if we had chosen a partition with k = 6, as we did
in Example 8.29, then Simpson™s Rule would have guaranteed an accuracy of
M · (b ’ a)5 12 · 15
= ≈ 0.00005144,
180 · k 4 180 · 64
or nearly four decimal places of accuracy.
259
CHAPTER 8 Applications of the Integral

EXAMPLE 8.32
Estimate the integral
1
1
dx
1 + x2
0

using Simpson™s Rule with a partition having four intervals. What degree of
accuracy does this represent?

SOLUTION
Of course this example is parallel to Example 8.30, and you should compare
the two examples. Our function is f (x) = 1/(1 + x 2 ) and our partition is
P = {0, 1/4, 2/4, 3/4, 1}. The sum from Simpson™s Rule is

1/4
S= · {f (0) + 4f (1/4) + 2f (1/2) + 4f (3/4) + f (1)}
3
1 1 1
= · +4·
1 + 02 1 + (1/4)2
12
1 1 1

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