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 Knight Paul. 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 Knight Paul. in the United States

Mar 22,2025

UpStudy AI · Accepted Answer

The average value $$ A $$ of a function $$ f(x) $$ over the interval $$[a, b]$$ is given by

$$
A = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx.
$$

For the function

$$
f(x) = \frac{1}{x}
$$

on the interval $$[1,6]$$, we have:

1. Compute the integral:

$$
\int_{1}^{6} \frac{1}{x} \, dx = \ln x \Big|_{1}^{6} = \ln 6 - \ln 1 = \ln 6.
$$

2. Calculate the average value:

$$
A = \frac{1}{6-1} \cdot \ln 6 = \frac{\ln 6}{5}.
$$

3. By the Mean Value Theorem for Integration, there exists a point $$ c $$ in $$[1,6]$$ such that

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

4. Solve for $$ c $$:

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

5. Rounding to the nearest thousandth:

$$
\ln 6 \approx 1.791759,
$$

$$
c \approx \frac{5}{1.791759} \approx 2.792.
$$

Thus, the value of $$ c $$ is

$$
c \approx 2.792.
$$
$$ c \approx 2.792 $$