Given the function defined on the interval
below, find the value of where
the average value.

on the interval

Round to the nearest thousandth.

Ask by Barnett Cruz. in the United States

Mar 22,2025

Question

Given the function defined on the interval
below, find the value of where
the average value.

on the interval

Round to the nearest thousandth.

Ask by Barnett Cruz. in the United States

Mar 22,2025

UpStudy AI · Accepted Answer

We begin with the formula for the average value of a function $$ f(x) $$ on the interval $$[a,b]$$:

$$
\text{Average} = \frac{1}{b-a} \int_a^b f(x) \, dx
$$

For the function $$ f(x)=\frac{1}{x} $$ on the interval $$[1,6]$$, we have:

$$
\text{Average} = \frac{1}{6-1} \int_1^6 \frac{1}{x} \, dx = \frac{1}{5} \int_1^6 \frac{1}{x} \, dx
$$

The integral of $$ \frac{1}{x} $$ is the natural logarithm:

$$
\int_1^6 \frac{1}{x} \, dx = \ln x \Big|_1^6 = \ln 6 - \ln 1 = \ln 6
$$

Thus, the average value of the function is:

$$
\text{Average} = \frac{1}{5} \ln 6
$$

We are looking for the $$ c $$ such that:

$$
f(c) = \frac{1}{c} = \frac{1}{5} \ln 6
$$

Solving for $$ c $$:

$$
\frac{1}{c} = \frac{\ln 6}{5} \quad \Rightarrow \quad c = \frac{5}{\ln 6}
$$

Now, we calculate $$ c $$ numerically. Using the approximation $$ \ln 6 \approx 1.792 $$:

$$
c \approx \frac{5}{1.792} \approx 2.792
$$

Thus, the value of $$ c $$ is approximately $$ 2.792 $$ when rounded to the nearest thousandth.
$$ c \approx 2.792 $$