Question

30 Points! I need this fast! Thank you

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

Answer 1

Answer:

Step-by-step explanation:

I don't often see calculus problems here!  I like it when I do!

The formula for the average value of a function is

$\frac{1}{b-a}\int\limits^a_b f({x}) \, dx$

Since e is greater than 1, our lower bound is 1 and the upper is e, and filling in our function we have this integral:

$\frac{1}{e-1}\int\limits^e_1 {\frac{ln(x)}{x} } \, dx$

We'll use a u substitution to simplify before we integrate.  I teach my calculus students to rewrite the integral identifying what the dx is and therefore, what our du has to "match".  For example, the rewrite is:

$\frac{1}{e-1}\int\limits^e_1 {ln(x)}*\frac{1}{x} \, dx$

Let u = ln(x), then

$\frac{du}{dx}=\frac{1}{x}$ and

$du=\frac{1}{x}dx$

What our du is equal to is the same as what we have designated as our dx in the integral.  They "match". So I know I chose the right u.  In terms of u, our integral is

$\frac{1}{e-1}\int\limits^e_1 {u} \, du$

which integrates to $\frac{u^2}{2}$

Making the back substitution of ln(x) for u gives us:

$\frac{1}{e-1}[\frac{(lnx)^2}{2}(1,e)$

Using the First Fundamental Theorem of Calculus:

$\frac{1}{1.718281828}[\frac{(ln(e))^2}{2}-\frac{(ln(1))^2}{2}]$ which simplifies down to

$\frac{1}{1.718281828}[\frac{1}{2}-0]$ which finally becomes

$\frac{1}{3.436563657}=.2909883534$

Round to however many places you need.

Good luck in your calculus class!