   Chapter 5.1, Problem 7E ### 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

# Evaluate the upper and lower sums for f(x) = 2 + sin x, 0 ≤ x ≤ π, with n = 2, 4, and 8. Illustrate with diagrams like Figure 14.

To determine

To evaluate:

The upper and lower sums for the function with n=2.

The upper and lower sums for the function with n=4.

The upper and lower sums for the function with n=8.

Explanation

Given information:

The curve function is f(x)=2+sinx.

The region lies between x=0 and x=π. So the limits are a=0 and b=π.

Calculation:

Draw the graph for the function f(x)=2+sinx with two rectangles as shown in Figure (1).

Calculation:

Find the width (Δx) using the relation:

Δx=ban (1)

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

The expression to find the upper sum of areas of nth rectangle is shown below.

Uppersum=f(x1)Δx+f(x2)Δx+...+f(xn)Δx (2)

Here, the maximum height of the first rectangle is f(x1), the width is Δx, the maximum height of the second rectangle is f(x2), and the maximum height of nth rectangle is f(xn)

Find the upper sum of areas for two rectangles as shown below.

Substitute π for b, 0 for a, and 2 for n in Equation (1).

Δx=π02=π2

Refer to Figure (1).

The maximum values of the function occur at x=π2 on the both subintervals.

Take the maximum height of the first rectangle’s f(x1) value as 3 and the maximum height of the second rectangle’s f(x2) value as 3.

Substitute 2 for n, 3 for f(x1), π2 for Δx, and 3 for f(x2) in Equation (2).

Uppersum=(3×π2)+(3×π2)=3π+3π2=6π2=9.425

Therefore, the upper sum of the function for two rectangles is 9.425_.

The expression to find the lower sum of areas of nth rectangle is shown below.

Lowersum=f(x0)Δx+f(x1)Δx+...+f(xn1)Δx (3)

Here, the minimum height of the first rectangle is f(x0), the minimum height of the second rectangle is f(x1), and the minimum height of nth rectangle is f(xn1)

Find the lower sum of areas for two rectangles as shown below.

Refer to Figure (1).

The minimum values of the function occur at x=0 and x=π.

Take the minimum height of the first rectangle’s f(x0) value as 2 and the minimum height of the second rectangle’s f(x1) value as 2.

Substitute 2 for n, 2 for f(x0), π2 for Δx, and 2 for f(x1) in Equation (3).

Lowersum=(2×π2)+(2×π2)=2π+2π2=4π2=6.283

Therefore, the lower sum of the function for two rectangles is 6.283_.

Draw the graph for the function f(x)=2+sinx with four rectangles as shown in Figure (2).

Find the upper sum of areas for four rectangles as shown below.

Substitute π for b, 0 for a, and 4 for n in Equation (1).

Δx=π04=π4

Refer to Figure (2).

Take the maximum height of the first rectangle’s f(x1) value as 2.7, the maximum height of the second rectangle’s f(x2) value as 3, the maximum height of the third rectangle’s f(x3) value as 3, and the maximum height of the fourth rectangle’s f(x4) value as 2.7.

Substitute 4 for n, 2.7 for f(x1), π4 for Δx, 3 for f(x2), 3 for f(x3), and 2.7 for f(x4) in Equation (2).

Uppersum=(2.7×π4)+(3×π4)+(3×π4)+(2.7×π4)=2.7π+3π+3π+2.7π4=11.4π4=8.954

Therefore, the upper sum of the function for four rectangles is 8.954_.

Find the lower sum of areas for four rectangles as shown below.

Refer to Figure (2).

Take the minimum height of the first rectangle’s f(x0) value as 2, the minimum height of the second rectangle’s f(x1) value as 2.7, the minimum height of the third rectangle’s f(x2) value as 2.7, and the minimum height of the fourth rectangle’s f(x3) value as 2

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 