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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