 # (a) Find the average value $f$ on the given interval. (b) Find c such that $f_{ave} = f(c)$. Equation given below

This problem aims to find the average value of a function on a given interval and also find the slope of that function. This problem requires knowledge of the fundamental theorem of calculus and basic integration techniques.

To find the average value of a function on a given interval, we will integrate and divide the function by the length of the interval, so the formula becomes:

$f_{ave} = \dfrac{1}{b-a} \int_{a}^{b} f(x) \,dx$

To find $c$, we are going to use the mean value theorem, which states that there exists a point $c$ on the interval such that $f(c)$ equals the average value of the function.

We are given a function along with it’s limits:

$f(x) = (x – 3)^2 , [2, 5]$

##### Part a:

Formula for calculating $f_{ave}$ is:

$\dfrac{1}{b-a} \int_{a}^{b} f(x) \,dx$

where $a$ and $b$ are the distinct limits of the integral that are $2$ and $5$, respectively, and $f(x)$ is the function with respect to $x$, given as $(x-3)^2$.

Plugging in values in the formula, we get:

$\dfrac{1}{5-2} \int_{2}^{5} (x-3)^2 \,dx$

Substituting $u = x – 3$

and then taking their derivative: $du = dx$

Changing the upper limit $u = 5 – 3$, that is $u = 2$

As well as the lower limit $u = 2 – 3$, that is $u = -1$

Further solving the problem:

$=\dfrac{1}{3} \int_{-1}^{2} u^2 \,du$

$=\dfrac{1}{3} \left[\dfrac{u^3}{3} \right]_{-1}^{2}$

$= \dfrac{1}{3} \left[\dfrac{2^3}{3} – \dfrac{-1^3}{3} \right]$

$= \dfrac{1}{3} \left[\dfrac{8}{3} + \dfrac{1}{3} \right]$

$= \dfrac{1}{3} \times \dfrac{9}{3}$

$f_{ave}= 1$

This is the average of the function.

##### Part b:

$f(c) = (c – 3)^2$

As given in the problem, $f_{ave} = f(c)$, and since $f_{ave}$ equal to $1$ as computed in part $a$, our equation becomes:

$1 = (c – 3)^2$

solving for $c$:

$\pm 1 = c -3$

solving for $-1$ and $+1$ separately:

$-1 = c – 3$

$c = 2$

$+1 = c – 3$

$c = 4$

## Numerical Results

Part a: $f_{ave} = 1$

Part b: $c =2 , c = 4$

## Example

Given Equation:

$f(x) = (x – 1) , [1, 3]$

##### Part a:

Putting the values in the formula to compute $f_{ave}$

$\dfrac{1}{3-1} \int_{1}^{3} (x-1) \,dx$

Substituting $u = x – 1$

Then derivating $du = dx$

Upper limit $u = 3 – 1$, that is $u = 2$

Lower limit $u = 1 – 1$, that is $u = 0$

$=\dfrac{1}{2} \int_{0}^{2} u \,du$

$=\dfrac{1}{2} \left[\dfrac{u^2}{2} \right]_{0}^{2}$

$=\dfrac{1}{2} \left[\dfrac{4}{2} – \dfrac{0}{2} \right]$

$=\dfrac{1}{2} \left[2 \right]$

$= 1$

##### Part b:

$f(c) = (c – 1)$

As in question $f_{ave} = f(c)$, and $f_{ave}$ is equals to $1$ as computed in part $a$.

$1 = (c – 1)$

solving for $c$:

$\pm 1 = c -1$

solving for $-1$ and $+1$ separately:

$-1 = c – 1$

$c = 0$

$+1= c – 1$

$c = 2$