Q. 21 - Calculus

Question

Explain why the Mean Value Theorem for Integrals applies to the function $f\left(x\right)=x\left(x-6\right)$ on the interval $\left[1,5\right]$. Next state the conclusion of the Mean Value Theorem for Integrals in this particular case, and sketch a graph illustrating your conclusion. Then find all the values $c\in \left[1,5\right]$ for which $f\left(c\right)$ is equal to the average value of $f$ on $\left[1,5\right]$, and indicate these values on your graph. 

Step-by-Step Solution

Answer

The values are $c\approx 1.845;4.145$.

Step 1. Given Information

The function is ,

$f\left(x\right)=x\left(x-6\right)$

The interval is $\left[1,5\right]$.

Step 2. Graph

Consider the following figure of the function,

From above figure, it is clear that the function in continuous on $\left[1,5\right]$.

Therefore, the Mean value Theorem for Integrals is applicable to the above function.

Step 3. Calculation

The objective is to state the conclusion of the Mean Value Theorem for Integrals for the above function.

The conclusion is that there exists some point $c\in \left(a,b\right)$ such that

$f\left(c\right)=\frac{1}{b-a}{\int }_a^{b}f\left(x\right)dx$

The average value is,

$$\begin{gathered} f\left(c\right)=\frac{1}{5-1}{\int }_1^{5}x\left(x-6\right)dx \\ =\frac{-23}{3} \end{gathered}$$

So, there exists such that $f\left(c\right)=\frac{-23}{3}$

The objective is to find the values of $c$.

Now,

           $$\begin{gathered} f\left(c\right)=\left(c-6\right) \\ \frac{-23}{3}=c^2-6c \\ -23=3c^2-18c \\ 3c^2-18c+23=0 \\ c\approx 1.845;4.145 \end{gathered}$$  

Therefore, the values are $c\approx 1.845;4.145$