5.6. NUMERICAL INTEGRATION
329
(b) Compute M 1 for each function to approximate
1
0
f (x) dx,
1
0
g(x) dx, and
1
0
h(x) dx, respectively.
(c) Compute T 1 for each of the three functions, and hence compute S 1 for each of
the three functions.
(d) Evaluate each of the integrals
1
0
f (x) dx,
1
0
g(x) dx, and
1
0
h(x) dx exactly
using the First FTC.
(e) For each of the three functions f , g, and h, compare the results of L 1 , R 1 ,
M 1 , T 1 , and S 2 to the true value of the corresponding definite integral. What
patterns do you observe?
1
2
1
2
1
2
Figure 5.19: Axes for plotting the functions in Activity 5.17.
⊳
The results seen in the examples in Activity 5.17 generalize nicely. For instance, for any
function f that is decreasing on [a, b], L n will over-estimate the exact value of
b
a
f (x) dx,
and for any function f that is concave down on [a, b], M n will over-estimate the exact
value of the integral. An excellent exercise is to write a collection of scenarios of possible
function behavior, and then categorize whether each of L n , R n , T n , and M n is an over- or
under-estimate.
Finally, we make two important notes about Simpson’s Rule. When T. Simpson first
developed this rule, his idea was to replace the function f on a given interval with a
quadratic function that shared three values with the function f . In so doing, he guaranteed
that this new approximation rule would be exact for the definite integral of any quadratic
polynomial. In one of the pleasant surprises of numerical analysis, it turns out that
even though it was designed to be exact for quadratic polynomials, Simpson’s Rule
is exact for any cubic polynomial: that is, if we are interested in an integral such as
5
2
(5x 3 − 2x 2 + 7x − 4) dx, S 2n will always be exact, regardless of the value of n. This is just
one more piece of evidence that shows how effective Simpson’s Rule is as an approximation
329
(b) Compute M 1 for each function to approximate
1
0
f (x) dx,
1
0
g(x) dx, and
1
0
h(x) dx, respectively.
(c) Compute T 1 for each of the three functions, and hence compute S 1 for each of
the three functions.
(d) Evaluate each of the integrals
1
0
f (x) dx,
1
0
g(x) dx, and
1
0
h(x) dx exactly
using the First FTC.
(e) For each of the three functions f , g, and h, compare the results of L 1 , R 1 ,
M 1 , T 1 , and S 2 to the true value of the corresponding definite integral. What
patterns do you observe?
1
2
1
2
1
2
Figure 5.19: Axes for plotting the functions in Activity 5.17.
⊳
The results seen in the examples in Activity 5.17 generalize nicely. For instance, for any
function f that is decreasing on [a, b], L n will over-estimate the exact value of
b
a
f (x) dx,
and for any function f that is concave down on [a, b], M n will over-estimate the exact
value of the integral. An excellent exercise is to write a collection of scenarios of possible
function behavior, and then categorize whether each of L n , R n , T n , and M n is an over- or
under-estimate.
Finally, we make two important notes about Simpson’s Rule. When T. Simpson first
developed this rule, his idea was to replace the function f on a given interval with a
quadratic function that shared three values with the function f . In so doing, he guaranteed
that this new approximation rule would be exact for the definite integral of any quadratic
polynomial. In one of the pleasant surprises of numerical analysis, it turns out that
even though it was designed to be exact for quadratic polynomials, Simpson’s Rule
is exact for any cubic polynomial: that is, if we are interested in an integral such as
5
2
(5x 3 − 2x 2 + 7x − 4) dx, S 2n will always be exact, regardless of the value of n. This is just
one more piece of evidence that shows how effective Simpson’s Rule is as an approximation