Question

Find the average value of the function on the given interval. f(x)=2 ln x; [1,e]

Answer 1

Answer:

$f_{avg}=\frac{1}{e-1}$

Step-by-step explanation:

We are given that a function

$f(x)=2lnx$

We have to find the average value of function on the given interval [1,e]

Average value of function on interval [a,b] is given by

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

Using the formula

$f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx$

By Parts integration formula

$\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx$

u=ln x and v=dx

Apply by parts integration

$f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))$

$f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})$

$f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}$

By using property lne=1,ln 1=0

$f_{avg}=\frac{1}{e-1}$