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.
Expert Answer
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:
\[ -1 = c – 1\]
\[ c = 0\]
\[ +1= c – 1\]
\[ c = 2\]