Posted by Zia on Thursday, January 3, 2013 at 4:43am.
you will need the intersection of
y = lnx and y = 2-lnx
2lnx= 2
lnx = 1
x = e
so you will have a region shaped like an "arrow head" from x=e to 4
Area = ∫(2 - 2lnx) dx from x = e to 4
(remember the integral of lnx is
xlnx - x , which should be in your repertoire of common integrals )
= [-2xlnx + 4x] from e to 4
= -8ln4 + 16 - (-2e + 4e)
= .....
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