Q. 72 - 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.

$f\left(x\right)=100(1-x), [a, b] = [0, 10]$

Step-by-Step Solution

Answer

The real number is $5,$ and the average value is $-400$ and it is verified from the graph.

Step 1. Given Information.

The given function and the interval is $f\left(x\right)=100(1-x), [a, b] = [0, 10].$

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

$$\begin{gathered} \frac{1}{b-a}{\int }_a^{b}f\left(x\right) dx \\ =\frac{1}{10-0}{\int }_0^{10}100(1-x) dx \\ =\frac{1}{10}{\int }_0^{10}(100-100x) dx \\ =\frac{1}{10}{\left[100x-\frac{100x^2}{2}\right]}_0^{10} \\ =\frac{1}{10}\left[100\left(10\right)-\frac{100{\left(10\right)}^2}{2}-\left(100\left(0\right)-\frac{100\left(0\right)}{2}\right)\right] \\ =\frac{1}{10}\left(1000-\frac{10000}{2}-0\right) \\ =\frac{1}{10}\left(-\frac{8000}{2}\right) \\ =-400 \end{gathered}$$

Thus, the average value is $-400.$

Now,

$$\begin{gathered} f\left(c\right)=-400 \\ 100(1-c)=-400 \\ 100-100c=-400 \\ -100c=-500 \\ c=5 \end{gathered}$$

Therefore, the real number is $5.$

Step 3. Verification.

By using the graphing utility, the graph of f  is 

From the graph, we can depict that $f\left(c\right)=-400.$ Thus, the real number is $5, 5\in \left[0,10\right].$ Hence, the answer is right.