Chapter 5.2, Problem 48E

If , $F(x)={\displaystyle {\int }_2^{x}f(t)dt}$, where f is the function whose graph is given, which of the following values is largest?

(A) F(0)

(B) F(1)

(C) F(2)

(D) F(3)

(E) F(4)

To determine

To find: The largest value of the integral function $F\left(x\right)={\displaystyle {\int }_2^{x}f\left(t\right)dt}$ among the following:

(A)$F\left(0\right)$

(B)$F\left(1\right)$

(C)$F\left(2\right)$

(D)$F\left(3\right)$

(E)$F\left(4\right)$

Answer to Problem 48E

The largest integral function is $F\left(2\right)$ and the correct option is (C).

Explanation of Solution

Given information:

The graph of the function $f$.

The integral function is $F\left(x\right)={\displaystyle {\int }_2^{x}f\left(t\right)dt}$ (1)

Calculation:

Show the Property of definite integral

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=-{\displaystyle {\int }_b^{a}f\left(x\right)}dx$ (2)

Part (A):

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

Substitute 0 for x in Equation (1)

$F\left(0\right)={\displaystyle {\int }_2^{0}f\left(t\right)}dt$ (3)

Apply the Property of Definite integral.

Modify Equation (3) using Equation (2).

$F\left(0\right)=-{\displaystyle {\int }_0^{2}f\left(t\right)}dt$

Thus, the value of $F\left(0\right)$ is negative.

Part (B):

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

Substitute 1 for x in Equation (1).

$F\left(1\right)={\displaystyle {\int }_2^{1}f\left(t\right)}dt$ (4)

Apply the Property of the Definite integral.

Modify Equation (4) using Equation (2).

$F\left(1\right)=-{\displaystyle {\int }_1^{2}f\left(t\right)}dt$

Thus, the value of $F\left(1\right)$ is negative.

Part (C):

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

Substitute 2 for x in Equation (2).

$F\left(2\right)={\displaystyle {\int }_2^{2}f\left(t\right)}dt$ (5)

The value of $F\left(2\right)$ is a integral with width of the interval $\Delta x=0$.

Thus, the value of $F\left(2\right)$ is 0.

Part (D):

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

Substitute 3 for x in Equation (1).

$F\left(1\right)={\displaystyle {\int }_2^{3}f\left(t\right)}dt$

Refer to graph.

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

Thus, the value of $F\left(3\right)={\displaystyle {\int }_2^{3}f\left(t\right)}dt$ 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)

$F\left(4\right)={\displaystyle {\int }_2^{4}f\left(t\right)}dt$

Refer to graph of function f.

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

Thus, the value of $F\left(4\right)$ is negative.

Conclusion:

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

The value of $F\left(2\right)$ is zero and non negative.

Thus, the value of $F\left(2\right)$ is the largest among the given functions.