330
5.6. NUMERICAL INTEGRATION
tool for estimating definite integrals. 11
Summary
In this section, we encountered the following important ideas:
• For a definite integral such as
1
0
e −x 2 dx when we cannot use the First Fundamental
Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative,
we can estimate the integral’s value by using a sequence of Riemann sum approximations.
Typically, we start by computing L n , R n , and M n for one or more chosen values of n.
• The Trapezoid Rule, which estimates
b
a
f (x) dx by using trapezoids, rather than
rectangles, can also be viewed as the average of Left and Right Riemann sums. That is,
T n =
1
2 (L n + R n ).
• The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs
of the respective errors of these rules are opposites. Hence, by taking the weighted
average S n =
2M n +T n
3
, we can build a much more accurate approximation to
b
a
f (x) dx
by using approximations we have already computed. The rule for S n is known as
Simpson’s Rule, which can also be developed by approximating a given continuous
function with pieces of quadratic polynomials.
Exercises
1. Consider the definite integral
1
0
x tan(x) dx.
(a) Explain why this integral cannot be evaluated exactly by using either usubstitution or by integrating by parts.
(b) Using 4 subintervals, compute L 4 , R 4 , M 4 , T 4 , and S 4 .
(c) Which of the approximations in (b) is an over-estimate to the true value of
1
0
x tan(x) dx? Which is an under-estimate? How do you know?
2. For an unknown function f (x), the following information is known.
• f is continuous on [3, 6];
• f is either always increasing or always decreasing on [3, 6];
• f has the same concavity throughout the interval [3, 6];
• As approximations to
6
3
f (x) dx, L 4 = 7.23, R 4 = 6.75, and M 4 = 7.05.
11 One reason that Simpson’s Rule is so effective is that S 2n benefits from using 2n + 1 points of data.
Because it combines M n , which uses n midpoints, and T n , which uses the n + 1 endpoints of the chosen
subintervals, S 2n takes advantage of the maximum amount of information we have when we know function
values at the endpoints and midpoints of n subintervals.
5.6. NUMERICAL INTEGRATION
tool for estimating definite integrals. 11
Summary
In this section, we encountered the following important ideas:
• For a definite integral such as
1
0
e −x 2 dx when we cannot use the First Fundamental
Theorem of Calculus because the integrand lacks an elementary algebraic antiderivative,
we can estimate the integral’s value by using a sequence of Riemann sum approximations.
Typically, we start by computing L n , R n , and M n for one or more chosen values of n.
• The Trapezoid Rule, which estimates
b
a
f (x) dx by using trapezoids, rather than
rectangles, can also be viewed as the average of Left and Right Riemann sums. That is,
T n =
1
2 (L n + R n ).
• The Midpoint Rule is typically twice as accurate as the Trapezoid Rule, and the signs
of the respective errors of these rules are opposites. Hence, by taking the weighted
average S n =
2M n +T n
3
, we can build a much more accurate approximation to
b
a
f (x) dx
by using approximations we have already computed. The rule for S n is known as
Simpson’s Rule, which can also be developed by approximating a given continuous
function with pieces of quadratic polynomials.
Exercises
1. Consider the definite integral
1
0
x tan(x) dx.
(a) Explain why this integral cannot be evaluated exactly by using either usubstitution or by integrating by parts.
(b) Using 4 subintervals, compute L 4 , R 4 , M 4 , T 4 , and S 4 .
(c) Which of the approximations in (b) is an over-estimate to the true value of
1
0
x tan(x) dx? Which is an under-estimate? How do you know?
2. For an unknown function f (x), the following information is known.
• f is continuous on [3, 6];
• f is either always increasing or always decreasing on [3, 6];
• f has the same concavity throughout the interval [3, 6];
• As approximations to
6
3
f (x) dx, L 4 = 7.23, R 4 = 6.75, and M 4 = 7.05.
11 One reason that Simpson’s Rule is so effective is that S 2n benefits from using 2n + 1 points of data.
Because it combines M n , which uses n midpoints, and T n , which uses the n + 1 endpoints of the chosen
subintervals, S 2n takes advantage of the maximum amount of information we have when we know function
values at the endpoints and midpoints of n subintervals.