Q. 59 - 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. 

$f\left(x\right)=4, [-37.2, 103.75]$

Step-by-Step Solution

Answer

The exact average value of f is $4,$and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)=4, [-37.2, 103.75].$

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

Thus,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{103.75-(-37.2)}{\int }_{-37.2}^{103.75}4 dx \\ =\frac{1}{140.95}\times 4{\int }_{-37.2}^{103.75} dx \\ =\frac{4}{140.95}{\left[x\right]}_{-37.2}^{103.75} \\ =\frac{4}{140.95}\left[103.75-(-37.2)\right] \\ =\frac{4}{140.95}\left(140.95\right) \\ =4 \end{gathered}$$

Step 3. Verification.

By the graphing utility, the graph of f  is 

From the graph, we can depict that the average value is $4.$ Thus, the answer is right.