Page 269 - Calculus Demystified

Applications of the Integral
CHAPTER 8
256
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 fine 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 confirm 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.
0 1 + x 2
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
2 1+0 2 1 2 2 2 3 2 1+1 
1+ 1+ 1+
 
4 4 4
A bit of calculation reveals that
1 5323
S = · ≈ 0.782794 ....
8 850
2
2 3
2
Now if we take f(x) = 1/(1 + x ) then f (x) = (6x − 2)/(1 + x ) .

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 · 1 3
= ≈ 0.020833 ....
12k 2 12 · 4 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.