Page 268 - Calculus Demystified

Applications of the Integral
CHAPTER 8
of the error estimate and, perhaps more importantly, an extra factor of k in the 255
denominator.]
EXAMPLE 8.29
Calculate the integral

1 2
e −x dx
0
to two decimal placesof accuracy.
SOLUTION
2

We first calculate that if f(x) = e −x 2 then f (x) = (4x − 2)e −x 2 and

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 · 1 3
< 0.005.
12k 2
Rearranging this inequality gives
100 2
<k .
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 $ 2 2 2 2
S = · e −0 + 2e −(1/6) + 2e −(1/3) + 2e −(1/2)
2
%
2
2
2
+ 2e −(2/3) + 2e −(5/6) + e −1 .
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.