# Question 03038

Aug 3, 2017

${f}_{a v e} = 1$
$c = 6$
$c = 8$

#### Explanation:

The average value of a function is found by:

${f}_{a v e} = \frac{1}{b - a} \cdot {\int}_{a}^{b} f \left(x\right) \mathrm{dx}$

on some interval $\left[a , b\right]$

We have $f \left(x\right) = {\left(x - 7\right)}^{2} , a = 6 ,$ and $b = 9$.

Therefore:

${f}_{a v e} = \frac{1}{3} {\int}_{6}^{9} {\left(x - 7\right)}^{2} \mathrm{dx}$

We can use a basic substitution to solve, where $u = x - 7$

$= \frac{1}{3} {\int}_{- 1}^{2} {u}^{2} \mathrm{du}$

=>=1/3*1/3u^3]_(-1)^2#

$= \frac{1}{9} \left(8 + 1\right)$

$= 1$

To find $c$ such that ${f}_{a v e} = f \left(c\right)$, we can use the mean value theorem for integrals, which says that if you have a function which is continuous on a closed interval $\left[a , b\right]$, then there is some number $c$ in $\left[a , b\right]$ such that $f \left(c\right) = {f}_{a v e}$.

We found ${f}_{a v e} = 1 ,$ so $f \left(c\right) = 1$.

$\implies {\left(c - 7\right)}^{2} = 1$

$\implies {c}^{2} - 14 c + 49 = 1$

$\implies {c}^{2} - 14 c + 48 = 0$

We find that $c = 8$ or $c = 6$. Both of these numbers are contained within the given interval.

$c = 6 : f \left(6\right) = {\left(6 - 7\right)}^{2} = 1$

$c = 8 : f \left(8\right) = {\left(8 - 7\right)}^{2} = 1$

The area of the rectangle is given by $\left(b - a\right) f \left(c\right)$ where $\left(b - a\right)$ is the width and $f \left(c\right)$ is the height. This gives the rectangle a width of $3$ and a height of $1$. You would therefore want the top right graph.