What is the area of the region enclosed by the curves ylnx a

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

Solution

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.