# The largest value of the integral function F ( x ) = ∫ 2 x f ( t ) d t among the following: (A) F ( 0 ) (B) F ( 1 ) (C) F ( 2 ) (D) F ( 3 ) (E) F ( 4 ) ### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805 ### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 5.2, Problem 48E
To determine

## To find: The largest value of the integral function F(x)=∫2xf(t)dt among the following:(A)F(0)(B)F(1)(C)F(2)(D)F(3)(E)F(4)

Expert Solution

The largest integral function is F(2) and the correct option is (C).

### Explanation of Solution

Given information:

The graph of the function f.

The integral function is F(x)=2xf(t)dt (1)

Calculation:

Show the Property of definite integral

abf(x)dx=baf(x)dx (2)

Part (A):

Calculate the value of the integral function at x=0 using Equation (1).

Substitute 0 for x in Equation (1)

F(0)=20f(t)dt (3)

Apply the Property of Definite integral.

Modify Equation (3) using Equation (2).

F(0)=02f(t)dt

Thus, the value of F(0) is negative.

Part (B):

Calculate the value of the integral function at x=1 using Equation (1).

Substitute 1 for x in Equation (1).

F(1)=21f(t)dt (4)

Apply the Property of the Definite integral.

Modify Equation (4) using Equation (2).

F(1)=12f(t)dt

Thus, the value of F(1) is negative.

Part (C):

Calculate the value of the integral function at x=2 using Equation (1).

Substitute 2 for x in Equation (2).

F(2)=22f(t)dt (5)

The value of F(2) is a integral with width of the interval Δx=0.

Thus, the value of F(2) is 0.

Part (D):

Calculate the value of the integral function at x=3 using Equation (1).

Substitute 3 for x in Equation (1).

F(1)=23f(t)dt

Refer to graph.

The area of the function within limits [2,3] is negative as it is below the x-axis.

Thus, the value of F(3)=23f(t)dt and is negative.

Part (E):

Calculate the value of the integral function at x=4 using Equation (1).

Substitute 4 for x in Equation (1)

F(4)=24f(t)dt

Refer to graph of function f.

The area of the function within limits [2,4] is negative as it is below the x-axis.

Thus, the value of F(4) is negative.

Conclusion:

Compare the value of F(0), F(1), F(2), F(3) and F(4). The values of F(0), F(1), F(3), and F(4) is negative.

The value of F(2) is zero and non negative.

Thus, the value of F(2) is the largest among the given functions.

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