Q. 56 - 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.

$f\left(x\right)=x-1, \left[-1,3\right]$

Step-by-Step Solution

Answer

The exact average value of f  is $0$ and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)=x-1, \left[-1,3\right].$

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

Thus,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{3-(-1)}{\int }_{-1}^3(x-1) dx \\ =\frac{1}{4}{\left[\frac{x^2}{2}-x\right]}_{-1}^3 \\ =\frac{1}{4}\left[\frac{3^2}{2}-3-\left(\frac{{\left(-1\right)}^2}{2}-(-1)\right)\right] \\ =\frac{1}{4}\left[\frac{9}{2}-3-\frac{1}{2}-1\right] \\ =\frac{1}{4}\left[\frac{9-6-1-2}{2}\right] \\ =\frac{1}{4}\left[0\right] \\ =0 \end{gathered}$$

Step 3. Verification.

The graph of f is

From the graph, we can depict that the average value is $0.$ Thus, the answer is right.