Question

How do you find the point c in the interval $0 \leqslant x \leqslant 3$ such that $f(c)$ is equal to the average value of $f(x) = 3{x^2}$ ?

Answer 1

In order to find the solution, we need to first find the value of ${f_{ave}}$ and then place the value and doing some simplification we get the required answer.

Formula used: ${f_{ave}} = \dfrac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx}$ ,
$\int\limits_a^b {{x^2}dx} = \dfrac{{{x^{2 + 1}}}}{{2 + 1}}$

Complete step-by-step solution:
In the given, the interval is given as: $0 \leqslant x \leqslant 3$ which is equal to $\left[ {0,3} \right]$ where $0$ is the lower limit known as $a$ and $3$ is the upper limit known as $b$.
Now first we need to find the value of ${f_{ave}}$
To find the value of ${f_{ave}}$, we need to place the value of the intervals in the given formula:
Therefore, ${f_{ave}} = \dfrac{1}{{b - a}}\int\limits_a^b {f\left( x \right)dx}$
Where, $a = 0$, $b = 3$ and $f\left( x \right) = 3{x^2}$
Thus, we get: ${f_{ave}} = \dfrac{1}{{3 - 0}}\int\limits_0^3 {f\left( {3{x^2}} \right)dx}$
$\Rightarrow {f_{ave}} = \dfrac{1}{3}\int\limits_0^3 {f\left( {3{x^2}} \right)dx}$
Now, we will integrate to the right of the integration symbol using the formula: $\int {{x^n}dx = } \dfrac{{{x^{n + 1}}}}{{n + 1}}$ , where n is the power to which $x$ is raised
First we take out the constant $3$ from the integration side in order to apply the formula further:
$\Rightarrow {f_{ave}} = \dfrac{1}{3} \times 3\int\limits_0^3 {f\left( {{x^2}} \right)dx}$
On integrating and solving further, we get:
$\Rightarrow {f_{ave}} = {\left[ {\dfrac{{{x^{2 + 1}}}}{{2 + 1}}} \right]^3}_0$
On simplification we get the required answer.
$\Rightarrow {f_{ave}} = \left[ {{{\dfrac{x}{3}}^3}} \right]_0^3$
Now, let’s bring the constant term outside the bracket which is = $\dfrac{1}{3}$
Therefore, ${f_{ave}} = \dfrac{1}{3}\left[ {{x^3}} \right]_0^3$
We will place the value of the intervals in $x$, to find the value of${f_{ave}}$ :
${f_{ave}} = \dfrac{1}{3}\left[ {{3^3} - 0} \right]$
On solving it further, we get:
${f_{ave}} = \dfrac{1}{3}\left[ {27 - 0} \right]$
On simplification we get,
${f_{ave}} = \dfrac{1}{3} \times 27 = 9$
Therefore, ${f_{ave}} = 9$
Now that we have found the value of ${f_{ave}}$ , we need to put this value in $f\left( x \right)$ to get our required solution. Here $f\left( x \right)$= $f\left( {3{x^2}} \right)$
Thus, $f\left( x \right) = {f_{ave}}$
$\Rightarrow 3{x^2} = 9$
Dividing both sides with$3$, we get:
$\Rightarrow {x^2} = 3$
In order to remove the square, we take the square root of both the sides:
$\sqrt {{x^2}}$ $= \sqrt 3$
$\Rightarrow x = \pm \sqrt 3$
Since the interval is $\left[ {0,3} \right]$ , hence $- \sqrt 3$ is not included and thus our answer is:
$x = \sqrt 3$
Thus, $c = \sqrt 3$

The value of c is equal to $\sqrt 3$.

Note: Integration is simply a process of adding up infinitesimally small parts of any object to give a whole. It can be used to find area, volume, displacement etc. the inverse of integration is differentiation. Integration and differentiation are two main methods of calculus. Some common formulas of integration are as follows:
$\int\limits_a^a {f\left( x \right)dx = 0}$ , where the limits are same
$\int\limits_a^b {f\left( x \right)dx = - \int\limits_b^a {f\left( x \right)dx} }$ , where the limits are interchanged
$\int\limits_a^b f \left( x \right)dx = \int\limits_a^b {f\left( t \right)dt}$ , as long as the limits remain the same, the change in variables do not affect as such.