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January 30, 2015

January 30, 2015

Posted by **Jane** on Saturday, November 26, 2011 at 5:51pm.

- Calculus -
**Steve**, Saturday, November 26, 2011 at 5:57pmaverage value is area under the curve divided by the interval length

what do you get?

Use integration by parts for x lnx

- Calculus -
**Jane**, Saturday, November 26, 2011 at 6:07pm(1/e^2-1)(6x^2 lnx/2-x^2/4)

am I on the right track?

- Calculus -
**drwls**, Saturday, November 26, 2011 at 6:14pmIntegrate 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

- Calculus -
**Jane**, Saturday, November 26, 2011 at 6:26pmso my answer would be 3/2 (3e^4+1)/ (e^2-1)?

- Calculus -
**Steve**, Sunday, November 27, 2011 at 12:57pmThat is correct.

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