Question

Find the average value of the function \(f(x) = \frac{1}{x}\) in the interval \({\rm{(1,4)}}\).

Short answer

The average value of the function \(f(x) = \frac{1}{x}\) in the interval \({\rm{(1,4)}}\) is \({\rm{0}}{\rm{.462}}\).

Step-by-step solution

Formula of Average Value.

Average value: The average value of the function\(f\)is defined on the interval\({\rm{(a, b)}}\)is\({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\).

Use the formula of the average value for calculation.

The given function is \(f(x) = \frac{1}{x}\) in the interval \({\rm{(1,4)}}\).

Obtain the average value of the function, \(f(x) = \frac{1}{x}\) as follows.

The limit values are got from the interval \({\rm{(1,4)}}\)as \({\rm{a = 1}}\) and \({\rm{b = 4}}\).

Substitute \({\rm{a = 1}}\), \({\rm{b = 4}}\) and \({\rm{f(x)}}\)in the formula \({f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\) and compute the average value of \({\rm{f(x)}}\) as follows.

\(\begin{aligned}{c}{f_{ave}} = \frac{1}{{b - a}}\int_a^b f (x)dx\\ = \frac{1}{{4 - 1}}\int_1^4 {\frac{1}{x}} dx\\ = \frac{1}{3}\int_1^4 {\frac{1}{x}} dx\end{aligned}\)

Further, simplify as follows.

\(\begin{aligned}{c}{f_{{\rm{ave }}}} &= \frac{1}{3}\int_1^4 {\frac{1}{x}} dx\\ &= \frac{1}{3}(\ln |x|)_1^4\end{aligned}\)

\(\begin{aligned}{c}{f_{{\rm{ave }}}} &= \frac{1}{3}(\ln |4| - \ln |1|)\\ &= 0.462\end{aligned}\)

Thus, the average value of the function \(f(x) = \frac{1}{x}\) in the interval \({\rm{(1,4)}}\) is \({\rm{0}}{\rm{.462}}\).