BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 5, Problem 38RE
To determine

To Draw: the graph for the function f(x)=cos2xsinx.

To Evaluate: The value of the integral function 02πf(x)dx

Expert Solution

Answer to Problem 38RE

The value of the integral function is 0.

Explanation of Solution

Given information:

The function is f(x)=cos2xsinx.

The region lies between x=0 and x=2π.

Calculation:

Show the function as follows:

f(x)=cos2xsinx (1)

Substitute 0 for x in Equation (1).

f(0)=cos2(0)sin(0)=1×0=0

Substitute π6 for x in Equation (1).

f(π6)=cos2(π6)sin(π6)=0.75×0.5=0.38

Calculate the remaining values of the function as shown below.

Show the x and f(x) values as in Table (1).

xcos2xsinxf(x)
01.000.000.00
π60.750.500.38
π30.250.870.22
π20.001.000.00
2π30.250.870.22
5π60.750.500.38
π1.000.000.00
7π60.75-0.50-0.38
4π30.25-0.87-0.22
3π20.00-1.000.00
5π30.25-0.87-0.22
11π60.75-0.50-0.38
2π1.000.000.00

Table 1

Sketch the function as shown in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5, Problem 38RE

Refer to Figure 1.

The value of the integral function 02πf(x)dx is 0.

Consider u=cosx (2)

Differentiate both sides of the Equation (2).

du=sinxdx (3)

Calculate the lower limit value of u using Equation (2).

Substitute 0 for x in Equation (2).

u=cos(0)=1

Calculate the upper limit value of u using Equation (2).

Substitute 2π for x in Equation (2).

u=cos(2π)=1

Substitute u for cosx and du for sinxdx in Equation (1) as shown below.

f(u)=u2(du)=u2du (4)

Find the integral value of the function as shown below.

02πf(x)dx=11(u2du)=(u33)11=(133133)=0

Therefore, the integral value of the function is 0.

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