   Chapter 5.2, Problem 16E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use a calculator or computer to make a table of values of left and right Riemann sums Ln and Rn for the integral ∫ 0 2 e − x 2 d x with n = 5, 10, 50, and 100. Between what two numbers must the value of the integral lie? Can you make a similar statement for the integral ∫ − 1 2 e − x 2 d x ? Explain.

To determine

To Calculate: The left Riemann sum Ln and right Riemann sums Rn for the integral 02ex2dx for n=5,10,50,and100 using the calculator.

To Find: The two numbers which the integral 02ex2dx lies between that two numbers.

To Explain: the statement for the integral 12ex2dx.

Explanation

Given:

The integral function as 02ex2dx.

Number of rectangles n=5,10,50and100.

Calculation:

Show the Equation of the integral as follows:

02ex2dx (1)

Consider the value of the function f(x)=ex2 (2)

Calculate the value of the function f(x)=ex2 within limit [0,2].

Substitute 0 for x in Equation (6).

f(1)=e(0)2=1

Substitute 2 for x in Equation (6).

f(2)=e(2)2=e4=0.183

The function f(x)=ex2 is a decreasing function within limits [0,2].

For n=5

Find the width (Δx) using the relation:

Δx=ban (3)

Here, the upper limit is b, the lower limit is a, and the number of rectangles is n.

The limits of the integral 02ex2dx, a=0 and b=2.

Substitute 2 for b, 5 for n, and 0 for a in Equation (3).

Δx=205=25

Calculate the right Riemann Sum for n=5,10,50and100.

The right endpoints are x1=25, x2=45, x3=65, x4=85 and x5=2.

The expression to find the right Riemann sum Rn as shown below:

Rn=i=1nf(xi)Δx=f(x1)Δx+f(x2)Δx+...+f(xn)Δx (4)

Here, the right endpoint height of first rectangle is f(x1), the width is Δx, height of right endpoint of second rectangle is f(x2), and left endpoint height of nth rectangle is f(xn).

Calculate the value of f(x1), f(x2), f(x3), f(x4), and f(x5) using the Equation (2).

Substitute 25 for x1 in Equation (2).

f(25)=e(25)2=e425=0.8521

Substitute 45 for x2 in Equation (2).

f(45)=e(45)2=e1625=0.5272

Substitute 65 for x3 in Equation (2).

f(65)=e(65)2=e3625=0.2369

Substitute 85 for x4 in Equation (2).

f(85)=e(85)2=e6425=0.0773

Substitute 105 for x5 in Equation (2).

f(2)=e(105)2=e10025=0.0183

Calculate the right Riemann sum using calculator.

Substitute 25 for Δx, 5 for n, and values of right end points in Equation (4).

R5=i=15f(xi)Δx=f(x1)Δx+f(x2)Δx+...+f(x5)Δx=f(25)×25+f(45)×25+...+f(2)×25=[(0.8521×0.4)+(0.5272×0.4)+(0.2369×0.4)+(0.0773×0.4)+(0.0183×0.4)]

=0.68479

The value of right Riemann Sum for n=5 is 0.68479_.

Similarly calculate the right Riemann sum for n=10,50,100.

Tabulate the values of right Riemann sum for n=5,10,50and100 as shown in table 1.

 n Rn 5 0.68479 10 0.68115 50 0.19652 100 0.09956

Table 1

Calculate the left Riemann Sum for n=5,10,50and100

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