What is the area of the region enclosed by the curves y=lnx and y=ln^2x ?

Expert Answers

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First we need to determine the points of intersection between lnx and ln^2 x

==> ln x = ln^2 x

==> ln^2 x - ln x = 0

==> lnx ( lnx -1) =0

==> lnx = 0 ==> x = 1

==> lnx-1 = 0 ==> lnx = 1 ==> x = e

Then we need to find the integral between x = 1 and x= e

==> Int ln^2 x - lnx = Int lnx ( lnx-1) dx

Int lnx dx

 v= lnx ==> dv = 1/x  dx

du = dx ==> u = x

==> Int vdu = v*u - Int u dv

==> Int lnx dx = xlnx - Int x*1/x dx

==> int lnx dx = xlnx - x + C........(1)

==> Int ln^2 x

u= ln^2x ==> du = 2lnx 1/x dx

dv = dx ==> v = x

==> Int u dv = u*v - Int v du

==> Int ln^2 x dx = xln^2 x - Int 2lnx dx

==> Int ln^2 x = xln^2 x - 2* (1)

==> Int ln^2 x = xln^2 x - 2( xlnx -x) + C

==> Int ln^2 x = xln^2 x - 2xlnx + 2x + C .............(2)

Now we will find the definite integral for both (1) and (2).

==> In lnx dx ( x= 1 tp x= e) = xlnx -x ( x=1 tp x= e)

==> elne - e - ( 1ln 1 - 1)  = e - e - 0 + 1 = 1

Then the area under y= lnx is 1 square unit.

==> Int ln^2 x dx ( x= 1 tp x = e)

==> xln^2 x -2xlnx + 2x

==> ( eln^2 e - 2elne + 2e - ( 1ln^2 1 - 2ln1 + 2)

==> e - 2e + 2e - 2 = e-2

Then the area under the curve ln^2 x is e-2 square units.

Then the area between lnx and ln^2 x is e-2-1 = l e-3l = 3- e

Then the area between the curves is (3-e) square units.

 

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