Q. 58 - Calculus

Question

For each function f and interval [a, b] in Exercises 56–67, use definite integrals and the Fundamental Theorem of Calculus to find the exact average value of f from x = a to x = b. Then use a graph of f to verify that your answer is reasonable.  

$f\left(x\right)=4-x^2, \left[-2,2\right]$

Step-by-Step Solution

Answer

The exact average value of f  is $2.66,$ and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)=4-x^2, \left[-2,2\right].$

Step 2. Finding the exact average value.

To find the exact average value of f  from x = a to x= b, we will use the formula:  $$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-(-2)}{\int }_{-2}^{2}4-x^2 dx \\ =\frac{1}{4}{\left[4x-\frac{x^3}{3}\right]}_{-2}^2 \\ =\frac{1}{4}\left[4\left(2\right)-\frac{{\left(2\right)}^3}{3}-\left(4\left(-2\right)-\left(\frac{{\left(-2\right)}^3}{3}\right)\right)\right] \\ =\frac{1}{4}\left(8-\frac{8}{3}+8-\frac{8}{3}\right) \\ =\frac{1}{4}\left(\frac{24-8+24-8}{3}\right) \\ =\frac{1}{4}\left(\frac{32}{3}\right) \\ =\frac{8}{3} \\ \approx 2.66 \end{gathered}$$

Step 3. Verification.

By using the graphing utility the graph of f  is 

From the graph, we can depict that the average value is $2.66.$ Thus, the answer is right.