254 Appendix E: A Detailed Look at f
In many cases you will be familiar enough with the function you want to
integrate that you’ll know whether the function has any quick wiggles
relative to the interval of integration. If you’re not familiar with the
function, and you have reason to suspect that it may cause problems, you
can quickly plot a few points by evaluating the function using the
subroutine you wrote for that purpose.
If for any reason, after obtaining an approximation to an integral, you
have reason to suspect its validity, there’s a very simple procedure you
can use to verify it: subdivide the interval of integration into two or more
adjacent subintervals, integrate the function over each subinterval, then
add the resulting approximations. This causes the function to be sampled
at a brand new set of sample points, thereby more likely revealing any
previously hidden spikes. If the initial approximation was valid, it will
equal the sum of the approximations over the subintervals.
Conditions That Prolong Calculation Time
In the preceding example (page 251), you saw that the algorithm gave an
incorrect answer because it never detected the spike in the function. This
happened because the variation in the function was too quick relative to
the width of the interval of integration. If the width of the interval were
smaller, you would get the correct answer; but it would take a very long
time if the interval were still too wide.
For certain integrals such as the one in that example, calculating the
integral may be unduly prolonged because the width of the interval of
integration is too large relative to certain features of the functions being
integrated. Consider an integral where the interval of integration is wide
enough to require excessive calculation time but not so wide that it would
be calculated incorrectly. Note that because f (x) = xe
−x
approaches zero
very quickly as x approaches ∞, the contribution to the integral of the
function at large values of x is negligible. Therefore, you can evaluate the
integral by replacing ∞, the upper limit of integration, by a number not so
large as 10
99
, say 10
3
.
To Next Page
Loading...
Need help?
General
