Page 270 - Calculus Demystified

Applications of the Integral
CHAPTER 8
We have a continuous function f on the interval [a, b] and we have a partition 257
P ={x 0 ,x 1 ,...,x k } of our partition as usual. It is convenient in this technique to
assume that we have an even number of intervals in the partition.

Fig. 8.46

Now each rectangle, over each segment of the partition, is capped off by an
arc of a parabola. Figure 8.46 shows just one such rectangle. In fact, for each pair
of intervals [x 2j−2 ,x 2j−1 ], [x 2j−1 ,x 2j ], we consider the unique parabola passing
through the endpoints

(x 2j−2 ,f(x 2j−2 )), (x 2j−1 ,f(x 2j−1 )), (x 2j ,f(x 2j )). (∗)
2
Note that a parabola y = Ax + Bx + C has three undetermined coefficients, so
three points as in (∗) will determine A, B, C and pin down the parabola.
In fact (pictorially) the difference between the parabola and the graph of f is
so small that the error is almost indiscernible. This should therefore give rise to a
startling accurate approximation, and it does.
Summing up the areas under all the approximating parabolas (we shall not
perform the calculations) gives the following approximation to the integral:

b x

f(x) dx ≈ {f(x 0 ) + 4f(x 1 ) + 2f(x 2 ) + 4f(x 3 )
a 3
+ 2f(x 4 ) + ··· + 2f(x k−2 ) + 4f(x k−1 ) + f(x k )}.

If it is known that the fourth derivative f (iv) (x) satisfies |f (iv) (x)|≤ M on [a, b],
then the error resulting from Simpson’s method does not exceed

M · (b − a) 5
.
180 · k 4