Q. 71 - Calculus

Question

For each function f and interval [a, b] given in Exercises 68–73, find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. Then use a graph of f to verify that your answer is reasonable.

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

Step-by-Step Solution

Answer

The real number is $\frac{-3+\sqrt{201}}{12},$ and the average value is $\frac{11}{3}$ and it is verified from the graph.

Step 1. Given Information.

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

Step 2. Finding a real number c ∈ (a, b).

To find a real number c ∈ (a, b) such that f(c) is the average value of f on [a, b]. We will use the average value formula: $\frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx.$

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-(-2)}{\int }_{-2}^2\left(1+x+2x^2\right) dx \\ =\frac{1}{4}{\left[x+\frac{x^2}{2}+\frac{2x^3}{3}\right]}_{-2}^2 \\ =\frac{1}{4}\left[2+\frac{2^2}{2}+\frac{2{\left(2\right)}^3}{3}-\left(-2+\frac{{\left(-2\right)}^2}{2}+\frac{2{\left(-2\right)}^3}{3}\right)\right] \\ =\frac{1}{4}\left[4+\frac{16}{3}+\frac{16}{3}\right] \\ =\frac{1}{4}\left(\frac{44}{3}\right) \\ =\frac{11}{3} \end{gathered}$$

Thus, the average value is $\frac{11}{3}\approx 3.66.$

Now,

$$\begin{gathered} f\left(c\right)=\frac{11}{3} \\ 1+c+2c^2=\frac{11}{3} \\ 2c^2+c-\frac{8}{3}=0 \\ 6c^2+3c-8=0 \\ c=\frac{-3\pm \sqrt{{\left(3\right)}^2-4\left(6\right)\left(-8\right)}}{2\left(6\right)} \\ c=\frac{-3\pm \sqrt{201}}{12} \\ c=\frac{-3+\sqrt{201}}{12},\frac{-3-\sqrt{201}}{12} \end{gathered}$$

Therefore, the real number will be $\frac{-3+\sqrt{201}}{12},\frac{-3+\sqrt{201}}{12}\in \left[-2,2\right].$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that $f\left(c\right)\approx 3.66.$ Thus, the real number is $\frac{-3+\sqrt{201}}{12},\frac{-3+\sqrt{201}}{12}\in \left[-2,2\right].$ Hence, the answer is right.