Answer:
[tex]f_{avg}=\frac{1}{e-1}[/tex]
Step-by-step explanation:
We are given that a function
[tex]f(x)=2lnx[/tex]
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
[tex]\frac{1}{b-a}\int_{a}^{b}f(x)dx[/tex]
Using the formula
[tex]f_{avg}=\frac{1}{e-1}\int_{1}^{e}lnx dx[/tex]
By Parts integration formula
[tex]\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx[/tex]
u=ln x and v=dx
Apply by parts integration
[tex]f_{avg}=\frac{1}{e-1}([xlnx]^{e}_{1}-\int_{1}^{e}(\frac{1}{x}\times xdx))[/tex]
[tex]f_{avg}=\frac{1}{e-1}(elne-ln1-[x]^{e}_{1})[/tex]
[tex]f_{avg}=\frac{1}{e-1}(e-0-e+1)=\frac{1}{e-1}[/tex]
By using property lne=1,ln 1=0
[tex]f_{avg}=\frac{1}{e-1}[/tex]
