Q. 60 - 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.  

$f\left(x\right)={(x+2)}^2-5, [-5, 0]$

Step-by-Step Solution

Answer

The exact average value of f is$-2.6667,$and it is verified from the graph of f.

The graph is

Step 1. Given Information.

The given function and interval is $f\left(x\right)={(x+2)}^2-5, [-5, 0].$

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,

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{0-(-5)}{\int }_{-5}^0{(x+2)}^2-5 dx \\ =\frac{1}{5}{\int }_{-5}^0\left(x^2+4x+4-5\right) dx \\ =\frac{1}{5}{\int }_{-5}^0\left(x^2+4x-1\right)dx \\ =\frac{1}{5}{\left[\frac{x^3}{3}+\frac{4x^2}{2}-x\right]}_{-5}^0 \\ =\frac{1}{5}\left[\frac{0}{3}+\frac{4\left(0\right)}{2}-0-\left(\frac{{\left(-5\right)}^3}{3}+\frac{4{\left(-5\right)}^2}{2}-\left(-5\right)\right)\right] \\ =\frac{1}{5}\left[0-\left(-\frac{125}{3}+50+5\right)\right] \\ =\frac{1}{5}\left[\frac{125}{3}-55\right] \\ =\frac{1}{5}\left[\frac{125-165}{3}\right] \\ =\frac{1}{5}\left(-\frac{40}{3}\right) \\ =-\frac{8}{3} \\ =-2.6667 \end{gathered}$$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that the average value is $-2.6667.$ Thus, the answer is right.