rieman sum

7/30/2019 rieman sum

1/6

Math 242 Lab Riemann Sums & Numerical Integration

1 Part I: Riemann Sums

1.1 Sigma notation

One strategy for calculating the area of a region is to cut the region into simple shapes, calculatethe area of each simple shape, and then add these smaller areas together to get the area of the

whole region. We will use that approach, but it is useful to have a notation for adding a lot ofvalues together: the sigma () notation.

The function to the right of the sigma is called the summand, and the numbers below andabove the sigma are called the lower and upper limits of the summation.

Summation A way to read Sigmanotation the sigma notation notation

12 + 22 + 32 + 42 + 52 the sum of k squared

5

k=1 k2

for k equals 1 to k equals 51

3+ 1

4+ 1

5+ 1

6+ 1

7the sum of 1 over k

7

k=3 k1

for k equals 3 to k equals 7

20 + 21 + 22 + 23 + 24 + 25 the sum of 2 to the j-th power

5

j=0 2j

for j equals 0 to j equals 5a2 + a3 + a4 + a5 + a6 + a7 the sum of a sub i

7

i=2 aifor i equals 2 to i equals 7

The variable (typically i, j, or k) used in the summation is called the counter or index variable.

Problem 1: Write the summation denoted by each of the following:

(a)

5

k=1 k3 =

(b)

7j=2(1)j 1j =

(c)

4

m=0(2m + 1) =

1


2/6


x f(x) g(x) h(x)

1 2 4 32 3 1 33 1 -2 34 0 3 3

5 3 5 3

Figure 1: Table for Example 1.

In practice, the sigma notation is frequently used with the standard function notation:

3

k=1

f(k + 2) = f(1 + 2) + f(2 + 2) + f(3 + 2) = f(3) + f(4) + f(5)

and4

k=1

f(xi) = f(x1) + f(x2) + f(x3) + f(x4).

Example 1. Use the values in Table 1 to evaluate

5

k=2 2f(k) and

5

j=3(5 + f(j 2)).

Solution:

5

k=2 2f(k) = 2f(2)+2f(3)+2f(4)+2f(5) = 2(3)+2(1)+2(0)+2(3) = 14.

5

j=3(5 + f(j2)) =(5+f(32))+(5+f(42))+(5+f(52)) = (5+f(1))+(5+f(2))+(5+f(3)) = (5+ 2)+(5+3)+(5+1) = 21.

Problem 2: Use the values of f, g and h in Table 1 to evaluate the following:

(a)

5

k=2 g(k) =

(b)

5i=3[f(i 1) + h(i)] =

Since the sigma notation is simply a notation for addition, it has all of the familiar propertiesof addition.

Theorem 1. (Summation Properties)

Sum of Constants:n

k=1 C = C+ C+ C+ + C (n terms) = nC.

Addition:n

k=1(ak + bk) =n

k=1 ak +n

k=1 bk.

Subtraction:n

k=1(ak bk) =n

k=1 ak n

k=1 bk.

Constant Multiple:n

k=1 Cak = Cn

k=1 ak.

Preserves positivity: if bk ak for all k thenn

k=1 bk n

k=1 ak. In particular, if ak 0for all k then

nk=1 ak 0.

Additivity of ranges: if 1 m n thenm

k=1 ak +n

k=m+1 ak =n

k=1 ak.

2


3/6


1.2 Sums of areas of rectangles

Later, we will approximate the areas under curves by building rectangles as high as the curve,calculating the area of each rectangle, and then adding the rectangular areas together.

Example 2. Evaluate the sum of the rectangular areas in Figure 2 and write the sum using the sigma notation.Solution:

sum of the rectangular areas = sum of (base) (height) for each rectangle= (1)(1/3) + (1)(1/4) + (1)(1/5) = 47/60.

Using the sigma notation,

(1)(1/3) + (1)(1/4) + (1)(1/5) =5

k=3

1

k.

Problem 3: Evaluate the sum of the rectangular areas in Figure 3 and write the sum using thesigma notation.

Solution:

3


4/6


Example 3. Write the sum of the areas of the rectangles in Figure 5 using the sigma notation.Solution: The area of each rectangle is (base)(height).

rectangle base height area1 x1 x0 f(x1) (x1 x0)f(x1)2 x2 x1 f(x2) (x2 x1)f(x2)

3 x3 x2 f(x3) (x3 x2)f(x3)

The area of the k-th rectangle is (xk xk1)f(xk), and the total area of the rectangles is the sum3

k=1(xk xk1)f(xk).

1.3 Area under a curve by Riemann sums

Suppose we want to calculate the area between the graph of a positive function f and the interval[a, b] on the xaxis (Fig. 7). The Riemann Sum method is to build several rectangles with y = f(x)bases on the interval [a, b] and sides that reach up to the graph of f (Fig. 8). Then the areas ofthe rectangles can be calculated and added together to get a number called a Riemann sum of fon [a, b]. The area of the region formed by the rectangles is an approximationof the area we want.

Example 4. Approximate the area in Figure 2(a) between the graph of f and the interval [2, 5] on the xaxis

by summing the areas of the rectangles in Figure 2(b).Solution: The total area of rectangles is (2)(3) + (1)(5) = 11 square units.

Figure 2: Illustration for Example 4.

In order to effectively describe this process, some new vocabulary is helpful: a partition of aninterval and the mesh of the partition. A partition P of a closed interval [a, b] into n subintervalsis a set ofn + 1 points {x0 = a, x1, x2, x3,...,xn1, xn = b} in increasing order, a = x0 < x1 < x2

rieman sum

Documents