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

First, we'll determine the limits of integration. These
limits are represented by the intercepting points of the given
curves.

We'll equate:

(ln x)^2
= ln x

We'll subtract ln
x:

(ln x)^2 - ln x = 0

We'll
factorize by ln x:

ln x*(ln x - 1) =
0

We'll cancel each factor:

ln
x = 0 => x = e^0 = 1

ln x - 1 = 0 => ln x = 1
=> x = e

The intercepting points are (1 , 0) and (e
, 1).

The lower limit of integration is x = 0 and the upper
limit of integration is x = e.

To determine what curve is
above of the other, we'll determine the monotony of the first derivative of the function
f(x) = (ln x)^2 - ln x.

f'(x) = 2lnx/x -
1/x

f'(x) = (2ln x - 1)/x

If x
= 1 => f'(1) = -1

If x = e => f'(e) =
1/e

The curve that is above is ln x and the area of the
region is the definite integral of the function: ln x - (ln
x)^2.

Int [ln x - (ln x)^2]dx = Int ln x dx - Int (ln x)^2
dx

We'll calculate Int ln x dx by
parts:

Int udv = uv - INt
vdu

Let u = ln x => du =
dx/x

Let dv = dx => v =
x

Int ln x dx = x*ln x - Int
dx

Int ln x dx = x*(ln x -
1)

We'll apply Leibniz Newton to determine the definite
integral:

Int ln x dx = F(e) -
F(1)

F(e) = e*(ln e - 1) = e*(1-1) =
0

F(1) = -1

F(e) - F(1) =
1

We'll calculate Int (ln x)^2 dx by
parts:

Let u = (ln x)^2 => du = 2ln
xdx/x

Let dv = dx => v =
x

Int (ln x)^2dx = x*(ln x)^2 - Int 2ln
xdx

Int (ln x)^2dx = x*(ln x)^2 - 2*Int ln
xdx

But Int ln x dx = 1

Int
(ln x)^2dx = x*(ln x)^2 - 2

We'll apply Leibniz Newton to
determine the definite integral:

F(e) - F(1) = e -
2

Int [ln x - (ln x)^2]dx = 1 - e +
2

Int [ln x - (ln x)^2]dx = 3 -
e

The area of the region bounded by the given
curves is of (3 - e) square units.