Calculus: Analytic Geometry and Calculus, with Vectors

280 Integrals

which appears in (4.95) and is so important that it merits reproduction.
We recall that n must be even and that h = (b - a)/n. Whenever f is
Riemann integrable over the interval a < x < b, the error term a is near
zero when n is large. When f is continuous, and perhaps in some other
cases as well, experienced operators of pencils and slide rules and calcula-
tors and electronic computers neglect the e and habitually use the remain-
ing Simpson sum in the right member of (4.96) as an approximation to the
integral. A particular sum is often judged to be as accurate as desired
when this sum agrees to the desired number of decimal places with the
sum obtained by doubling n. In many practical applications, sur-
prisingly small values of n yield the desired accuracy.
Nearly everyone who understands the trapezoidal and Simpson formu-
las generates the following idea. It should be possible to derive still
better formulas by approximating f by polynomials of higher degree
having graphs passing through more of the points Po, P1, P2, P3,..
It turns out, however, that these formulas are more complicated than the
Simpson formula, and using them for a given h is not as satisfactory as
using the Simpson formula with a smaller h.

Problems 4.99

I Tables give

f

1 dx= log 2 = 0.69314 71806.

x

Show that the trapezoidal formula with n = 4 gives h = T, yo = 1, y, = 3

,and

f

2idx=e-}'T[

i x IV F } a+T } al=e+0.697024

and that use of the Simpson formula with n = 4 gives h = and

f -1dx=e+;'-+S[1+4+4+ +] =e+0.693254.

1 x

Show that the error terms are respectively -0.003877 and -0.000107. Observe

that it is almost equally easy to use the trapezoidal and Simpson formulas.

Remember that properly educated persons use the Simpson formula whenever

suitable occasions arise, but that they rarely if ever use the trapezoidal formula.

2 Tables give

log 2.5 = 0.91629 07319.

Using the Simpson formula with two subintervals, obtain the approximation

f

251dx=

1 1+ 4 + 11=0.223148.

2 X 12 L2 2.25 2.5

Show how this and the last numerical result of Problem 1 give the approximation

log 2.5 = 0.91640 2.