Question

Find the average value of the function g(x) = lnx/x over the interval [1, e].

Answer 1

Answer:

$\displaystyle \bar g=\frac{1}{2(e-1)}$

Step-by-step explanation:

Average Value of a Function

Given a function g(x), the average value of g in a given interval (xo,x1) is given by

$\displaystyle \bar g=\frac{1}{x_1-x_0}\int_{x_0}^{x_1} g(x)dx$

Plugging in the given data

$\displaystyle \bar g=\frac{1}{e-1}\int_{1}^{e} \frac{lnx}{x}dx$

Let's compute the indefinite integral

$\displaystyle I=\int \frac{lnx}{x}dx$

We'll use the substitution u=lnx, du=dx/x. Then

$\displaystyle I=\int u.du$

$\displaystyle I=\frac{u^2}{2}$

Taking back the substitution

$\displaystyle I=\frac{ln^2x}{2}$

The average value is

$\displaystyle \bar g=\frac{1}{e-1}\frac{ln^2x}{2} \ | _1^e$

$\displaystyle \bar g=\frac{1}{e-1}\left(\frac{ln^2e}{2}-\frac{ln^21}{2} \right )$

$\displaystyle \bar g=\frac{1}{e-1}\left(\frac{1}{2}-0 \right )$

$\displaystyle \bar g=\frac{1}{2(e-1)}$