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 5.2, Problem 49E
To determine

To calculate: The value of the integral function 42[f(x)+2x+5]dx.

Expert Solution

Answer to Problem 49E

The value of the integral function 42[f(x)+2x+5]dx is 15.

Explanation of Solution

Given information:

The integral function is 42[f(x)+2x+5]dx (1)

The region lies between x=4 and x=2.

Calculation:

The integral property 2 is shown below:

ab[f(x)+g(x)]dx=abf(x)dx+abg(x)dx (2)

The property is valid only, if the functions f(x) and g(x) are continuous.

Apply the integral property 2 in Equation (1).

42[f(x)+2x+5]dx=42f(x)dx+242xdx+425dx (3)

Rearrange Equation (3) as shown below.

42[f(x)+2x+5]dx=I1+2I2+I3 (4)

Consider the regions A, B, and C bounded by the graph of f as in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.2, Problem 49E , additional homework tip  1

Refer to Figure 1.

The value of the region which lies in the integral 42f(x)dx is 3.

The value of the region which lies in the integral 20f(x)dx is 3.

The value of the region which lies in the integral 02f(x)dx is 3.

Calculate the integral I1 by rearranging the integral function 42f(x)dx as shown below:

I1=42f(x)dx+20f(x)dx+02f(x)dx=3+33=3

Calculate I2 using the relation:

I2=42xdx (5)

Consider the function g(x)=x.

The function g(x) represents a straight line with slope 1.

Plot the function g(x) within the limits [4,2] as shown in Figure 2.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 5.2, Problem 49E , additional homework tip  2

Refer to Figure 2.

Take the area bounded by the function g(x) within the limits [4,2] as A1 and A2

Modify Equation (5) using the Figure 2.

I2=42xdx=A1+A2=(12×2×2)(12×4×4)=6

Calculate I3 using the relation:

I2=425dx=5[x]42=5(2(4))=30

Calculate the integral function using Equation (4).

Substitute 3 for I1, 6 for I2 and 30 for I3 in Equation (4).

42[f(x)+2x+5]dx=I1+2I2+I3=3+2(6)+30=312+30=15

Therefore, the value of the integral function is 15.

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