What is the area of the region enclosed by the curves y=lnx and y=ln^2x ?
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hala718 | High School Teacher | (Level 1) Educator Emeritus
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.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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.