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Riemann Sums in MATLAB
Definition. Let f (x) be a function on an interval [a, b], and suppose this interval is parti- tioned by the values a = x 0 < x 1 < ... < xn− 1 < xn = b. Any sum of the form
R =
∑^ n
k=
f (ck)△xk,
where △xk = xk − xk− 1 and ck ∈ [xk− 1 , xk] is referred to as a Riemann sum of f. If we let P denote our choice of partition (the choice of values x 0 , x 1 , ..., xn), and we let
‖P ‖ := max(△x 1 , △x 2 , ..., △xn)
denote the norm of this partition, then we say f is Riemann integrable if
lim ‖P ‖→ 0
∑^ n
k=
f (ck)△xk
exists. Following Leibniz, our notation for this limit is
∫ (^) b
a
f (x)dx = lim ‖P ‖→ 0
∑^ n
k=
f (ck)△xk.
Example 1. As our first example, we will consider the case in which ck is always chosen as the right endpoint of the interval [xk− 1 , xk]. If we take a regular partition with n intervals,
then each interval has length △x = b− na , and the kth endpoint is
xk = a + k△x.
The Riemann sum becomes
R =
∑^ n
k=
f (a + k△x)△x.
Suppose we would like to approximate the integral
∫ (^2)
0
e−x
2 dx
with n = 4. We have △x = 2 − 4 0 = .5 and the values
x 0 = 0 x 1 =. 5 x 2 = 1 x 3 = 1. 5 x 4 = 2. 0.
The Riemann sum is
R =
∑^4
k=
f (0 +. 5 k).5 = .5(e−.^5
2
2
2
2 ) =. 6352.
More generally, we can write a MATLAB function M-file that carries out this calculation for any function f (defined as an inline function), endpoints a and b and regular partition with n points. See rsum1.m.
function value=rsum1(f,a,b,n) %RSUM1: Computes a Riemann Sum for the function f on %the interval [a,b] with a regular partition of n points. %The points on the intervals are chosen as the right endpoints. value = 0; dx = (b-a)/n; for k=1:n c = a+kdx; value = value + f(c); end value = dxvalue;
We run this in MATLAB with the following lines in the Command Window.
f=inline(’exp(-xˆ2)’) f = Inline function: f(x) = exp(-xˆ2) rsum1(f,0,2,4) ans =
rsum1(f,0,2,10) ans =
rsum1(f,0,2,100) ans =
rsum1(f,0,2,1000) ans =
To four decimal places, the correct value is .8821. △ One interesting aspect of the Riemann sum is that the points ck need not be chosen in the same place on each interval. That is, suppose we partition the interval [0, 1] with 0 = x 0 < x 1 = 12 < x 2 = 1. In this case, a possible Riemann sum is
f (0)
Assignments
- Alter the M-file rsum1.m so that it computes Riemann sums of the given function by taking the values ck as the left endpoints of each interval. Use your M-file to estimate ∫ (^2)
0
e−x
2 dx
for regular partitions with n = 10, 100 , 1000.
- Alter the M-file rsum1.m so that it computes Riemann sums of the given function by taking the values ck as the midpoints of each interval. Use your M-file to estimate ∫ (^2)
0
e−x
2 dx
for regular partitions with n = 10, 100 , 1000.
