Q. 69 - 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.

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

Step-by-Step Solution

Answer

The real number is $1$ and the average value is $3$ and it is verified from the graph.

Step 1. Given Information.

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

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

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{2-\left(-1\right)}{\int }_{-1}^2\left(2x^2+1\right)dx \\ =\frac{1}{3}{\left[\frac{2x^3}{3}+x\right]}_{-1}^2 \\ =\frac{1}{3}\left[\frac{2{\left(2\right)}^3}{3}+2-\left(\frac{2{\left(-1\right)}^3}{3}+(-1)\right)\right] \\ =\frac{1}{3}\left[\frac{16}{3}+2+\frac{5}{3}\right] \\ =\frac{1}{3}\left[\frac{16+6+5}{3}\right] \\ =\frac{1}{3}\left[\frac{27}{3}\right] \\ =3 \end{gathered}$$

Thus, the average value is $3.$

Now,

$$\begin{gathered} f\left(c\right)=3 \\ 2{\left(c\right)}^2+1=3 \\ 2c^2=2 \\ {c}^2=1 \\ c=\pm 1 \end{gathered}$$

The real number will be $1, 1\in \left[-1,2\right].$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that $f\left(c\right)=3.$ Thus, the real number is $1, 1\in \left[-1,2\right].$ Hence, the answer is right.