# Calculate the area enclosed by the x axis and the curve y = e^(2x) in the interval 0 less than/equal x less than/equal 2.0 < x < 2

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For the beginning, the area which has to be found is located between the given curve y = e^2x and the lines x = 0 and x = 2, also the x axis.

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

S = Integral (f(x) - ox)dx = Int f(x)dx = Int e^(2x) dx

Int e^(2x) dx = e^(2x)/2 + C

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

S = F(2) - F(0), where

F(2) = e^(2*2)/2 = e^4/2

F(0) = e^(2*0)/2 = e^0/2 = 1/2

S = e^4/2 - 1/2

S = (e^4 - 1)/2

We have a difference of squares, at numerator:

S = (e^2-1)(e^2+1)/2

**S = (e-1)(e+1)(e^2+1)/2**

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

A = Int f(x) dx from x=a to x = b. Given f(x) = e^(2x).

Therefore , A = Int e^(2x) dx from x= 0 to x = 2

Int e^2x dx = (e^2x)/2

Therefore A = {[e^2x)/2] at x =2} - {e^2x/2 at x= 0}

A = [e^(2*2)]/2 - [e^(2*0)]/2 = (e^4 - 1)/2