Difference Equations to Differential Equations: Section 4.2 — Numerical Approximations of Definite Integrals
Section 4.2
Difference Equations
to
Differential Equations
Numerical Approximations
of Definite Integrals
Computing a definite integral of a function f over an interval [a, b] using upper and lower
sums, or even as the limit of Riemann sums, is, for all but the simplest cases, a difficult task.
As a result, definite integrals are almost never computed in that manner. For the most
part, definite integrals are evaluated either using the Fundamental Theorem of Calculus
or using numerical approximation techniques. We will take up the Fundamental Theorem
of Calculus approach in the next section; in this section we consider several methods for
numerical approximation.
The left-hand and right-hand rules
Recall that for an integrable function f on an interval [a, b], the left-hand rule approximaRb
tion for a f (x)dx, using n intervals, is given by
AL = h
n−1
X
f (a + ih)
(4.2.1)
f (a + ih),
(4.2.2)
i=0
and the right-hand rule approximation by
AR = h
n
X
i=1
where
h=
b−a
.
n
We now look at the accuracy of these approximations. Let xi = a+ih, i = 0, 1, 2, . . . , n,
the endpoints for a partition of [a, b] using n intervals of equal length h. Assume f is
continuous on [a, b] and differentiable on (a, b) and that x is a point in the ith interval,
that is, xi−1 ≤ x ≤ xi . Then the Mean Value Theorem tells us that there exists a point ci
in the interval (xi−1 , xi ) such that
f 0 (ci ) =
f (x) − f (xi−1 )
.
x − xi−1
(4.2.3)
Solving for f (x) in (4.2.3), we have
f (x) = f (xi−1 ) + f 0 (ci )(x − xi−1 ).
1
(4.2.4)
Copyright c by Dan Sloughter 2000
2
Numerical Approximations of Definite Integrals
Section 4.2
Integrating both sides of (4.2.4) over the interval [xi−1 , xi ] gives us
Z xi
Z xi
Z xi
f (x)dx =
f (xi−1 )dx +
f 0 (ci )(x − xi−1 )dx
xi−1
xi−1
xi−1
Z xi
= f (xi−1 )(xi − xi−1 ) +
f 0 (ci )(x − xi−1 )dx
(4.2.5)
xi−1
Z xi
f 0 (ci )(x − xi−1 )dx,
= f (xi−1 )h +
xi−1
where we have used the fact that the integral of a constant equals the constant multiplied
by the length of the interval. Hence
Z b
n Z xi
X
f (x)dx =
f (x)dx
a
=
i=1
n
X
xi−1
n Z xi
X
f (xi−1 )h +
i=1
= AL +
i=1
Thus we have
Z b
f (x)dx − AL =
a
≤
(4.2.6)
xi−1
i=1
n Z xi
X
f 0 (ci )(x − xi−1 )dx
f 0 (ci )(x − xi−1 )dx.
xi−1
n Z xi
X
f 0 (ci )(x − xi−1 )dx
i=1 xi−1
n Z xi
X
i=1
(4.2.7)
f 0 (ci )(x − xi−1 )dx .
xi−1
Now
Z xi
Z xi
0
f (ci )(x − xi−1 )dx ≤
xi−1
|f 0 (ci )(x − xi−1 )|dx,
(4.2.8)
xi−1
(see Problem 11 in Section 4.1), so
Z b
n Z xi
X
f (x)dx − AL ≤
|f 0 (ci )(x − xi−1 )|dx
a
=
i=1 xi−1
n Z xi
X
i=1
(4.2.9)
0
|f (ci )|(x − xi−1 )dx,
xi−1
where the last equality follows from the fact that x − xi−1 ≥ 0 for all x in [xi−1 , xi ]. Now
suppose f 0 is defined and continuous on [a, b] and let M be the maximum value of |f 0 (x)|
for x in [a, b]. Then
Z xi
Z xi
0
|f (ci )|(x − xi−1 )|dx ≤
M (x − xi−1 )dx
xi−1
xi−1
Z xi
M h2
,
(x − xi−1 )dx =
=M
2
xi−1
(4.2.10)
Section 4.2
Numerical Approximations of Definite Integrals
xi - 1
3
xi
Figure 4.2.1 Graph of y = x − xi−1 over the interval [xi−1 , xi ]
since the region beneath the graph of y = x − xi−1 over the interval [xi−1 , xi ] is a triangle
with base and height both of length h = xi − xi−1 (see Figure 4.2.1). Substituting (4.2.10)
into (4.2.9), and recalling that h = b−a
n , we have
Z b
f (x)dx − AL ≤
a
n
X
M h2
i=1
2
=
nM h2
nM (b − a)2
M (b − a)2
=
=
.
2
2n2
2n
(4.2.11)
In other words, the absolute value of the error of the left-hand rule approximation is
bounded by a constant multiplied by n1 . This explains the behavior of the example in
Section 4.1 where we saw that doubling the number of intervals would decrease the error
by a factor of 21 . The same techniques yield a similar result for the right-hand rule.
The trapezoidal rule
For a decreasing function, the left-hand rule is an upper sum and the right-hand rule is a
lower sum; for an increasing function, the left-hand rule is a lower sum and the right-hand
rule is an upper sum. Hence, for such functions, it would seem that the average of the
left-hand and right-hand rules, that is,
AL + AR
,
2
Rb
should provide a better approximation to a f (x)dx than either AL or AR . We will now
show that this is true in general.
Suppose f , f 0 , and f 00 are all defined and continuous on [a, b]. From (4.2.4) we know
that for any x in the interval [xi−1 , xi ] there exists a point ci in (xi−1 , xi ) such that
f (x) = f (xi−1 ) + f 0 (ci )(x − xi−1 ).
(4.2.12)
Similarly, there exists a point di in (xi−1 , xi ) such that
f (x) = f (xi ) + f 0 (di )(x − xi ).
(4.2.13)
4
Numerical Approximations of Definite Integrals
Section 4.2
Using f 0 in place of f in (4.2.4), there exists a point pi in (xi−1 , ci ) such that
f 0 (ci ) = f 0 (xi−1 ) + f 00 (pi )(ci − xi−1 )
(4.2.14)
f 0 (di ) = f 0 (xi−1 ) + f 00 (qi )(di − xi−1 )
(4.2.15)
and a point qi such that
Substituting (4.2.14) into (4.2.12) and (4.2.15) into (4.2.13), we have
f (x) = f (xi−1 ) + (f 0 (xi−1 ) + f 00 (pi )(ci − xi−1 ))(x − xi−1 )
= f (xi−1 ) + f 0 (xi−1 )(x − xi−1 ) + f 00 (pi )(ci − xi−1 )(x − xi−1 )
(4.2.16)
f (x) = f (xi ) + (f 0 (xi−1 ) + f 00 (qi )(di − xi−1 ))(x − xi )
= f (xi ) + f 0 (xi−1 )(x − xi ) + f 00 (qi )(di − xi−1 )(x − xi ).
(4.2.17)
and
Taking the average of (4.2.16) and (4.2.17) gives us
f (x) + f (x)
2
f (xi−1 ) + f (xi ) f 0 (xi−1 )((x − xi−1 ) + (x − xi ))
+
=
2
2
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
.
+
2
f (x) =
(4.2.18)
Now
Z xi
f (xi−1 ) + f (xi )
f (xi−1 ) + f (xi )
f (xi−1 ) + f (xi )
dx =
(xi − xi−1 ) =
h
2
2
2
xi−1
(4.2.19)
and
Z xi
f 0 (xi−1 )((x − xi−1 ) + (x − xi ))
dx = f 0 (xi−1 )
2
xi−1
Z xi
xi−1
xi−1 + xi
x−
2
dx
= 0,
where the final equality follows from the fact that region between the graph of
y =x−
xi−1 + xi
2
(4.2.20)
Section 4.2
Numerical Approximations of Definite Integrals
5
xi - 1
xi
Figure 4.2.2 Graph of y = x −
xi−1 + xi
over the interval [xi−1 , xi ]
2
and the interval [xi−1 , xi ] forms two triangles of equal area, one above the x-axis and one
below (see Figure 4.2.2). Moreover, if K is the maximum value of |f 00 (x)| for x in [a, b],
then
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
K
≤ (|ci − xi−1 ||x − xi−1 |
2
2
+ |di − xi−1 ||x − xi |).
Since the points ci , di , and x are all in [xi−1 , xi ], it follows that
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
≤ Kh2 .
2
(4.2.21)
Hence
Z xi
xi−1
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
dx ≤
2
Z xi
Kh2 dx.
xi−1
Now
Z xi
Kh2 dx = Kh2 (xi − xi−1 ) = Kh3 ,
(4.2.22)
xi−1
so we have
Z xi
xi−1
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
dx ≤ Kh3 .
2
(4.2.23)
6
Numerical Approximations of Definite Integrals
Section 4.2
Putting this all together, we have
Z b
f (x)dx =
a
n Z xi
X
i=1
f (x)dx
xi−1
n Z xi
X
f (xi−1 ) + f (xi )
=
dx
2
x
i−1
i=1
n Z xi
X
f 0 (xi−1 )((x − xi−1 ) + (x − xi ))
+
dx
2
i=1 xi−1
n Z xi
X
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
+
dx
2
i=1 xi−1
=
n
X
f (xi−1 ) + f (xi )
i=1
2
h+0
n Z xi
X
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
+
dx
2
i=1 xi−1
Pn
Pn
h i=1 f (xi−1 ) + h i=1 f (xi )
=
2
Z
n
x
i
X
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
+
dx
2
x
i−1
i=1
=
AL + AR
2
n Z xi
X
f 00 (pi )(ci − xi−1 )(x − xi−1 ) + f 00 (qi )(di − xi−1 )(x − xi )
dx,
+
2
i=1 xi−1
from which it follows, using (4.2.23), that
n
Z b
X
AL + AR
f (x)dx −
≤
Kh3 = nKh3 = nK
2
a
i=1
That is, if we approximate
Rb
a
b−a
n
3
=
K(b − a)3
.
n2
(4.2.24)
f (x)dx by
AL + AR
,
2
then the absolute value of the error is bounded by a constant multiplied by n12 . In particular, if we double the number of intervals, we should expect the error to decrease by a
factor of
2
1
1
= .
2
4
Section 4.2
Numerical Approximations of Definite Integrals
We call
AT =
AL + AR
2
7
(4.2.25)
Rb
a trapezoidal rule approximation for a f (x)dx. The name comes from the fact that (4.2.25)
may also be derived by replacing the areas of rectangles by the areas of trapezoids in the
Riemann sum approximations (see Problem 6).
Example
R 2 2 In Section 4.1 we saw that the left-hand and right-hand rule approximations
for −1 (x + 1)dx using n = 6 intervals are AL = 5.375 and AR = 6.875. Hence the
corresponding trapezoidal rule approximation is
AT =
5.375 + 6.875
= 6.125.
2
With n = 100, the left-hand and right-hand rules give usAL = 5.95545 and AR = 6.04545,
yielding a trapezoidal rule approximation of
AT =
Example
5.95545 + 6.04545
= 6.00045.
2
In a Section 4.1 we approximated
Z 10
A=
1
1
dt
t
using both left-hand and right-hand rules and, after noting that to 6 decimal places A =
2.302585, obtained the following table of values:
n
10
20
40
80
160
320
AR
1.960214
2.116477
2.205491
2.253003
2.277534
2.289994
AL
2.770214
2.521477
2.407991
2.354253
2.328159
2.315307
|A − AR |
0.342371
0.186108
0.097094
0.049582
0.025052
0.012591
|A − AL |
0.467629
0.218892
0.105406
0.051668
0.025574
0.012722
Using these results, we may compute the following trapezoidal rule approximations:
n
10
20
40
80
160
320
AT
2.365214
2.318977
2.306741
2.303628
2.302846
2.302650
|A − AT |
0.062629
0.016392
0.004156
0.001043
0.000265
0.000065
We see that the errors in the trapezoidal rule approximations are significantly smaller than
the corresponding errors for the left-hand and right-hand rule approximations. Moreover,
8
Numerical Approximations of Definite Integrals
Section 4.2
in agreement with our work above, the errors in the trapezoidal rule approximations decrease by a factor of, roughly, 14 when we double the number of intervals. Hence we see
the trapezoidal rule approximations converging to the value of the definite integral at a
significantly faster rate than do the left-hand and right-hand rule approximations.
The midpoint rule
As above, let f be an integrable function on an interval [a, b], n a positive integer, h = b−a
n ,
and, for i = 0, 1, 2, . . . , n, xi = a + ih, (the endpoints of a partition of [a, b] with n intervals
of equal length h). We may think of the trapezoidal rule as improving on the left-hand
and right-hand rules by approximating the area of the region beneath the graph of f and
above the interval [xi−1 , xi ] using a rectangle with height equal to the average of f (xi−1 )
and f (xi ). Another approach is to average xi−1 and xi before evaluating f . Since each
interval is of length h, we may find the midpoint by adding h2 to the left-hand endpoint.
Namely, if we let
h
,
2
h
h
3
c2 = x1 + = a + h + = a + h,
2
2
2
h
h
5
c3 = x2 + = a + 2h + = a + h,
2
2
2
..
.
c1 = a +
(4.2.26)
h
1
h
= a + (i − 1)h + = a + i −
h,
2
2
2
..
.
h
h
1
cn = xn−1 + = a + (n − 1)h + = a + n −
h,
2
2
2
ci = xi−1 +
then ci is the midpoint of the ith interval [xi−1 , xi ]. We call
n
X
n
X
1
AM =
f (ci )h = h
f a+ i−
h ,
2
i=1
i=1
(4.2.27)
the Riemann sum formed by evaluating f at the points c1 , c2 , c3 , . . . , cn , a midpoint rule
Rb
approximation of the definite integral a f (x)dx.
R2
Example To find the midpoint rule approximation for −1 (x2 + 1)dx using n = 6 intervals, we would have
2 − (−1)
h=
= 0.5.
6
Then the interval endpoints are x0 = −1, x1 = −0.5, x2 = 0, x3 = 0.5, x4 = 1, x5 = 1.5,
and x6 = 2, from which we find the midpoints c1 = −0.75, c2 = −0.25, c3 = 0.25, c4 = 0.75,
Section 4.2
Numerical Approximations of Definite Integrals
5
4
3
2
1
1
-1
2
Z 2
Figure 4.2.3 Midpoint rule approximation for
(x2 + 1)dx
−1
c5 = 1.25, and c6 = 1.75 (see Figure 4.2.3). Thus, letting f (x) = x2 + 1, we have
AM = 0.5(f (−0.75) + f (−0.25) + f (0.25) + f (0.75) + f (0.125) + f (1.75))
= 0.5(1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625)
= 5.9375.
With n = 100 intervals, using (4.2.27) with a computer, we have AM = 5.999775.
Example
Applying the midpoint rule to the problem of approximating
Z 10
A=
1
1
dt,
t
we obtain the following table (again rounded to 6 decimal places):
9
10
Numerical Approximations of Definite Integrals
Section 4.2
1
0.8
0.6
0.4
0.2
2
4
6
8
10
Z 10
Figure 4.2.4 Midpoint rule approximation for
1
n
AM
|A − AM |
10
20
40
80
160
320
2.272740
2.294504
2.300515
2.302064
2.302455
2.302552
0.029845
0.008081
0.002070
0.000521
0.000130
0.000033
1
dt
t
Notice that, as with the trapezoidal rule, doubling the number of intervals decreases the
error by a factor of about 41 . Moreover, note that the error in each approximation is
approximately 21 of the corresponding error for the trapezoidal rule.
An analysis of the error in the midpoint rule, similar to that which we did above
for the left-hand, right-hand, and trapezoidal rules, would show that the absolute value
of the error is bounded by a constant multiplied by n12 . Hence doubling the number of
intervals will decrease the error by, roughly, a factor of 41 , as was evidenced in the previous
example. Moreover, a more careful examination of the error (one requiring the use of
Taylor polynomials, which we will discuss in Chapter 5) would show that there is a sense
in which it is typically on the order of 12 the size of the error of the trapezoidal rule.
Simpson’s rule
We saw above that averaging the left-hand and right-hand rules, two approximation methods with errors bounded by a constant multiple of n1 , resulted in an approximation method,
the trapezoidal rule, with an error bounded by a constant multiple of n12 . We might now
think that we could improve on the trapezoidal and midpoint rules, two rules with errors
bounded by a constant multiple of n12 , by computing their average. However, it turns out
that the relationship between these two rules is not as simple as with the left-hand and
right-hand rules; in fact, one really needs to use Taylor polynomials in order to understand
the error terms fully. At the same time, there is a hint in our previous example. Given
Section 4.2
Numerical Approximations of Definite Integrals
11
that the error from the midpoint rule was about 12 of the error of the trapezoidal rule,
it would be reasonable to guess that perhaps an average of the two which gives twice as
much weight to the midpoint rule would be appropriate. This in fact turns out to be the
right mixture, and we define
1
2
AT + 2AM
AT + AM =
.
(4.2.28)
3
3
3
Rb
We call AS a Simpson’s rule approximation for a f (x)dx. This method of approximating
definite integrals is named for Thomas Simpson (1710-1761). Simpson developed this rule
in 1743 as a method for approximating the area under a curve after first approximating
the curve with a number of parabolic arcs.
AS =
Example
R 2 2 Using n = 6 intervals, we saw above that the trapezoidal rule approximation
for −1 (x + 1)dx is AT = 6.1250 and the midpoint rule approximation is AM = 5.9375.
Thus the corresponding Simpson’s rule approximation is
AS =
6.1250 + (2)(5.9375)
18
=
= 6.
3
3
With n = 100 intervals, we have AT = 6.000450 and AM = 5.999775, giving us
AS =
18
6.000450 + (2)(5.999775)
=
= 6.
3
3
It may seem surprising in this example that we get the same result with 100 intervals as
we do with 6, but in fact this is the exact answer. It may be shown, either by careful
examination of the error or by deriving the rule from parabolic approximations, that
Simpson’s rule will find the exact value for the definite integral of any quadratic polynomial.
What is even more surprising, careful examination of the error using Taylor polynomials
shows that Simpson’s rule is exact for cubic polynomials as well.
Example Using the values for the trapezoidal and midpoint rule approximations obtained above, we have the following approximations for
Z 10
A=
1
1
dt
t
using Simpson’s rule:
n
10
20
40
80
AS
2.303565
2.302662
2.302590
2.302585
|A − AS |
0.000980
0.000077
0.000005
0.000000
We have stopped the table at 80 intervals because at that point the approximation is
accurate to 6 decimal places.
12
Numerical Approximations of Definite Integrals
Section 4.2
1
0.8
0.6
0.4
0.2
0.5
1
1.5
2
2.5
3
Figure 4.2.5 Region beneath y = sin(x) over the interval [0, π]
It may be shown that the absolute value of the error using Simpson’s rule is bounded
by a constant multiple of n14 , resulting in a dramatic improvement over both the trapezoidal and midpoint rules. For Simpson’s rule, doubling the number of intervals typically
decreases the error by a factor of
4
1
1
=
,
2
16
a general fact for which we can see some evidence in the preceding example. To be fair,
since Simpson’s rule makes use of both the trapezoidal rule and the midpoint rule approximations, the function being integrated must be evaluated both at the endpoints and at the
midpoint of each interval. This requires evaluating the function at 2n + 1 points, whereas
the trapezoidal rule evaluates the function at n + 1 points and the midpoint rule evaluates the function at n points. Thus, for direct comparison of errors, one should compare,
for example, Simpson’s rule with 10 subdivisions to the other rules using 20 subdivisions.
Nevertheless, Simpson’s rule converges to the value of the integral much faster than the
other methods and, hence, is the method of preference among the ones we have discussed.
Even faster methods exist, but we will leave them for a more advanced course.
When approximating the value of an integral, there is in general no practical way to
know how many intervals are necessary in order to obtain an approximation to a desired
level of accuracy. Analogous to the way in which we applied Newton’s method, we normally
compute a sequence of approximations, perhaps starting with only two intervals and then
doubling the number of intervals from one approximation to the next, stopping when we
obtain two successive approximations whose difference, in absolute value, is less than the
desired level of accuracy. The next example illustrates this procedure.
Example Suppose we wish to approximate, with an error less than 0.0001, the area A
of the region between one arch of the curve y = sin(x) and the x-axis, as shown in Figure
4.2.5. That is, we want to find
Z π
A=
sin(x)dx.
0
Section 4.2
Numerical Approximations of Definite Integrals
13
Starting with n = 2 intervals and using Simpson’s rule, we generate the following table of
approximations, rounding to 6 decimal places:
n
AS
2
4
8
16
2.004560
2.000269
2.000017
2.000001
Since the absolute value of the difference between the last two approximations is less than
0.0001, we stop at this point and use 2.0000 as our approximation for A.
Problems
1. Approximate each of the following integrals using the trapezoidal and midpoint rules
with n = 4 intervals.
Z 1
Z π
2
(a)
x dx
(b)
sin(x)dx
0
0
Z 5
Z 1
1
dt
t
(c)
1
Z 4
(e)
(t2 + t)dt
(d)
z 3 dz
−1
(f)
0
Z 2p
4 − x2 dx
0
2. Use your results from Problem 1 to compute the corresponding Simpson’s rule approximation for each integral.
3. Approximate each of the following integrals using the trapezoidal and midpoint rules
with n = 50 intervals.
Z 1
Z π
2
(a)
x dx
(b)
sin(x)dx
0
0
Z 5
Z 2
(c)
(4x2 + 3x − 6)dx
−2
Z π
(e)
(g)
x2 cos(x)dx
sin(x)
dx
x
1
Z 2π q
(f)
1 − sin2 (t) dt
(d)
0
0
Z 5
Z π
1
dx
2
−5 x + 1
(h)
sin(3x)cos(x)dx
−π
4. Use your results from Problem 3 to compute the corresponding Simpson’s rule approximation for each integral.
5. Approximate the following definite integrals using Simpson’s rule. Starting with n = 2
intervals, compute a sequence of approximations by doubling the number of intervals
from one approximation to the next. Stop when the absolute value of the difference
between two successive approximations is less than 0.00001.
14
Numerical Approximations of Definite Integrals
Z 4
(a)
Z 6
2
x dx
(b)
(3x2 + 4x − 3)dx
1
0
Z π
Z 2π
(c)
sin2 (x)dx
(d)
0
Section 4.2
sin2 (x) cos2 (x)dx
0
Z π p
(e)
1 + cos2 (θ) dθ
Z 2
(f)
−π
sin(x)
dx
x
0.1
6. Suppose f is integrable on the interval [a, b]. Divide [a, b] into n equal intervals of
length h = b−a
be the endpoints of these intervals. Let AT
n and let x0 , x1 , x2 , . . . , xn R
b
be the trapezoidal rule approximation for a f (x)dx.
(a) Show that
AT =
h
(f (x0 ) + 2f (x1 ) + 2f (x2 ) + · · · + 2f (xn−1 ) + f (xn )).
2
(b) Assume f (x) ≥ 0 for all x in [a, b]. For i = 1, 2, 3, . . . , n, let Ai be the area of the
trapezoid with vertices at (xi−1 , 0), (xi−1 , f (xi−1 )), (xi , f (xi )), and (xi , 0) (that is,
Ai is the area of a trapezoid with one side being the interval [xi−1 , xi ] and parallel
sides extending from the x-axis up to the graph of f ). Then we could approximate
Rb
f (x)dx by A1 + A2 + A3 + · · · + An . Show that
a
AT = A1 + A2 + A3 + · · · + An .
7. Suppose f is integrable on the interval [a, b]. Divide [a, b] into 2n equal intervals of
length h = b−a
Let
2n and let x0 ,, x1 , x2 , . . . , x2n be the endpoints of these intervals.
Rb
AT and AM be the trapezoidal rule and midpoint rule approximations for a f (x)dx
using the n intervals with endpoints x0 , x2 , x4 , . . . , x2n . Let AS be the corresponding
Simpson’s rule approximation. Show that
AS =
h
(f (x0 ) + 4f (x1 ) + 2f (x2 ) + 4f (x3 ) + · · · + 2f (x2n−2 ) + 4f (x2n−1 ) + f (x2n )).
3
8. Let T (t) be the temperature at t hours after midnight at the Kalispell airport and
suppose the following values for T were recorded on March 15, 1955:
Time (t)
0.0
Temperature (T ) 40
0.5
38
1.0
37
1.5
36
2.0
33
2.5
30
3.0
28
3.5
28
4.0
27
4.5
26
Time (t)
5.0
Temperature (T ) 24
5.5
26
6.0
30
6.5
30
7.0
32
7.5
35
8.0
37
8.5
38
9.0
40
9.5
45
Time (t)
10.0 10.5 11.0 11.5 12.0
Temperature (T )
47
47
48
49
50
R 12
(a) Approximate 0 T (t)dt using Simpson’s rule. You may wish to use the formula in
Problem 7.
Section 4.2
Numerical Approximations of Definite Integrals
15
(b) What does
1
A=
12
Z 12
T (t)dt
0
represent?
24
1 X
(c) How does A compare with
T
25 t=0
t
?
2
9. Find the area beneath one arch of the curve y = sin2 (x).
10. Let R be the region in the plane bounded by the curves y = x2 and y = (x − 2)2 and
the x-axis. Find the area of R.