Q. 70 - Calculus

Question

For each function f and interval [a, b] given in Exercises 68–73, find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. Then use a graph of f to verify that your answer is reasonable.

$f\left(x\right)=9-x^2, [a, b] = [0, 3]$

Step-by-Step Solution

Answer

The real number is $\sqrt{3},$ and the average value is $6$ and it is verified from the graph.

Step 1. Given Information.

The given function and the interval is $f\left(x\right)=9-x^2, [a, b] = [0, 3].$

Step 2. Finding a real number c ∈ (a, b).

To find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. We will use the average value formula: $\frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx.$

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{3-0}{\int }_0^3\left(9-x^2\right)dx \\ =\frac{1}{3}{\left[9x-\frac{x^3}{3}\right]}_0^3 \\ =\frac{1}{3}\left[9\left(3\right)-\frac{3^3}{3}-\left(9\left(0\right)-\frac{0}{3}\right)\right] \\ =\frac{1}{3}\left(27-9-0\right) \\ =\frac{1}{3}\left(18\right) \\ =6 \end{gathered}$$

Thus, the average value is $6.$

Now,

$$\begin{gathered} f\left(c\right)=6 \\ 9-c^2=6 \\ -c^2=-3 \\ c=\pm \sqrt{3} \end{gathered}$$

The real value will be $\sqrt{3}, \sqrt{3}\in \left[0,3\right].$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that $f\left(c\right)=6.$ Thus, the real number is $\sqrt{3}, \sqrt{3}\in \left[0,3\right].$Hence, the answer is right.