# Find the area between the curve f(x) = 1/(x+2) and the lines x=0  and x= 1.

hala718 | Certified Educator

f(x) = 1/(x+2)

The area under the curve is the integral of f(x)

F(x) = intg f(x) = intg (1/(x+2)

= ln (x+2)

Then the area between the lines x= 0 and x = 1 is:

Area = F(1) - F(0)

= ln (3) - ln (1)

= ln 3 - 0= lon3

Then the area = ln 3

giorgiana1976 | Student

For the beginning, the area which has to be found is located between the given curve f(x), the lines x = 0 and x = 1, also the x axis.

To calculate the area, we'll use the formula:

S = Integral (f(x) - ox)dx = Int f(x)dx = Int dx/(x+2)

We'll calculate the integral, using substitution technique.

We'll note x+2 = t.

We'll differentiate x+2.

(x+2)' = t'

dx = dt

Int dt/t = ln t = ln (x+2) + C

Now, we'll calculate the value of the area, using Leibnitz Newton formula::

S = F(1) - F(0), where

F(1) = ln (1+2) = ln 3

F(0) = ln (0+2) = ln 2

S = ln 3 - ln 2

We'll use the quotient property of logarithms:

S = ln (3/2)

neela | Student

The area A  enclosed by the curve f(x) and x axis between  x= a and x= b is given by:

A = Int f(x) dx from x = a to x = b'

Given functio is f(x) = 1/(1+x).

A = In (1/(1+x))dx  from x = 0 to x = 1.

Int (1/(1+x)) dx = ln (1+x).

A =  { ln(1+x) at x =1} - {ln(1+x) at x= 0}

A = ln(1+1) - ln (1+0)

A = ln2 - ln1

A = ln2 - 0

A = ln2.