Calculus
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 
(1/e^21)(6x^2 lnx/2x^2/4)
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^21)?

That is correct.
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