Q. 63 - 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.  

$f\left(x\right)={\left(e^x\right)}^2, [a, b]=[-1, 1]$

Step-by-Step Solution

Answer

The exact average value of f is$1.81,$and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)={\left(e^x\right)}^2, [a, b]=[-1, 1].$

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,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{1-(-1)}{\int }_{-1}^1{\left(e^x\right)}^2 dx \\ =\frac{1}{2} {\int }_{-1}^1\left(e^{2x}\right) dx \\ =\frac{1}{2}{\left[\frac{e^{2x}}{2}\right]}_{-1}^1 \\ =\frac{1}{2}\left[\frac{e^{2\left(1\right)}}{2}-\frac{e^{2\left(-1\right)}}{2}\right] \\ =\frac{1}{2}\left[\frac{e^2}{2}-\frac{e^{-2}}{2}\right] \\ =\frac{1}{4}\left(e^2-e^{-2}\right) \\ =1.81 \end{gathered}$$

Step 3. Verify the obtained result.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that the average value is $1.81.$ Thus, the answer is right.