Chapter 6.5, Problem 7E

(a) Find the average value of f on the given interval.

(b) Find c such that f_ave = f(c).

(c) Sketch the graph off and a rectangle whose area is the same as the area under the graph of f.

f(x) = (x – 3)², [2, 5]

To determine

To find: The average value of the function $f\left(x\right)={\left(x-3\right)}^2$.

Answer to Problem 7E

The average value of the function is 1.

Explanation of Solution

Given information:

The function is $f\left(x\right)={\left(x-3\right)}^2,\left[2,5\right]$.

Calculation:

Show the function as below:

$f\left(x\right)={\left(x-3\right)}^2$ (1)

The expression to find the average value of the function $\left(f_{\text{avg}}\right)$ is shown below:

$f_{\text{avg}}=\frac{1}{b-a}{\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx$ (2)

Here, the upper limit is b, the lower limit is a, and the function is $f\left(x\right)$.

Substitute 2 for a, 5 for b, and ${\left(x-3\right)}^2$ for $f\left(x\right)$ in Equation (2).

$\begin{array}{c}{f}_{\text{ave}}=\frac{1}{5-2}{\displaystyle \underset{2}{\overset{5}{\int }}{\left(x-3\right)}^{2}dx}\\ =\frac{1}{3}{\displaystyle \underset{2}{\overset{5}{\int }}{\left(x-3\right)}^{2}dx}\end{array}$ (3)

Integrate Equation (3) with respect to x and apply the limits.

$\begin{array}{c}{f}_{\text{ave}}=\frac{1}{3}{\displaystyle \underset{2}{\overset{5}{\int }}\left(x^2-6x+9\right)dx}\\ =\frac{1}{3}{\left[\frac{x^3}{3}-\frac{6x^2}{2}+9x\right]}_2^5\\ =\frac{1}{3}\left\{\left[\frac{5^3}{3}-\left(3\times 5^2\right)+\left(9\times 5\right)\right]-\left[\frac{2^3}{3}-\left(3\times 2^2\right)+\left(9\times 2\right)\right]\right\}\\ =\frac{1}{3}\left(3\right)\end{array}$

$=1$

Thus, the average value of the function is 1.

To determine

To find: The value of count c.

Answer to Problem 7E

The value of c is 2 or 4.

Explanation of Solution

Given information:

The function is $f\left(x\right)={\left(x-3\right)}^2,\left[2,5\right]$.

Calculation:

Find the value of c using the formula:

$f\left(c\right)=f_{\text{ave}}$ (4)

Substitute 1 for $f_{\text{ave}}$ in Equation (4).

$f\left(c\right)=1$ (5)

Substitute c for x in Equation (1)

$f\left(c\right)={\left(c-3\right)}^2$ (6)

Substitute 1 for $f\left(c\right)$ in Equation (6).

$\begin{array}{l}{\left(c-3\right)}^2=1\\ c-3=\sqrt{1}\\ c-3=\pm 1\end{array}$ (7)

Find c value using Equation (7).

$\begin{array}{l}c-3=1\\ c=4\end{array}$

Find the another value of c.

$\begin{array}{l}c-3=-1\\ c=-1+3\end{array}$

$c=2$

Thus, the value of c is 2 or 4.

To determine

To draw: The graph of the function f.

Explanation of Solution

Substitute 2 for x in Equation (1)

$\begin{array}{c}y={\left(2-3\right)}^2\\ =1\end{array}$

Substitute 4 for x in Equation (1)

$\begin{array}{c}y={\left(4-3\right)}^2\\ =1\end{array}$

Draw the graph for the curve function $y={\left(x-3\right)}^2$ and mark the coordinates (2,1), (4,1) as shown in Figure 1.