Q. 62 - 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)= 4x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}, [a, b]=[0, 2]$

Step-by-Step Solution

Answer

The exact average value of f is$4.5255,$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)= 4x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}, [a, b]=[0, 2].$

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,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-0}{\int }_0^{2}4x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} dx \\ =\frac{1}{2}{\left[\frac{4x^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{{\displaystyle \frac{5}{2}}}\right]}_0^2 \\ =\frac{1}{2}{\left[\frac{2}{5}\times 4x^{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^2 \\ =\frac{1}{2}\left[\frac{8{\left(2\right)}^{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{5}-0\right] \\ =\frac{1}{2}\left[\frac{8}{5}\sqrt{32}\right] \\ =\frac{4}{5}\sqrt{32} \\ =\frac{16}{5}\sqrt{2} \\ =3.2\sqrt{2} \\ \approx 4.5255 \end{gathered}$$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that the average value is $4.5255.$Thus, the answer is right.