Question

Find the average value of the function g(x) = $\frac{lnx}{x}$ over the interval [1, e].

Answer 1

Answer:

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

Step-by-step explanation:

Average Value of a Function

Given a function g(x), we can compute the average value of g in a given interval (a,b) with the equation:

$\displaystyle average=\frac{1}{b-a}\int_{a}^{b} g(x)dx$

We use the given data

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

We now compute the indefinite integral with a u-substitution

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

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

$\displaystyle I=\int u.du$

Integrating

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

Since u=lnx

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

The average value is

$\displaystyle average=\frac{1}{e-1}\left|\frac{ln^2x}{2} \right|_1^e$

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

Since lne=1, and ln1=0

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

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