posted by Jane .
Find the average value of the function on the given interval. f(x)=6x lnx; [1,e^2]
average value is area under the curve divided by the interval length
what do you get?
Use integration by parts for x lnx
am I on the right track?
Integrate the function over that interval and divide the integral by e^2 - 1.
Try integration by parts for the integral.
Let lnx = u and dv = 6 x dx
du = dx/x and v = 3 x^2
Integral of f(x) dx =Integral u dv
= uv - Integral v du
= 3 x^2 lnx - Integral of 3x dx
= 3x^2 lnx - (3/2)x^2
so my answer would be 3/2 (3e^4+1)/ (e^2-1)?
That is correct.