f(x) = e^2x

Let F(x) = intg f(x)

==> F(x) = intg e^2x dx

= e^2x/2 + c

==> F(x) = 2^2x/2 + C

Now the difinite integral from x= 0 to x= 2 is the area between the f(x) and x= 0 and x= 2:

A = F(2) - F(0)

=(e^2*2/2 + c ) - (e^2*0 / 2 + c)

= (e^4)/2 - e^0/2

= (e^4)/2 - 1/2

= (e^4 - 1)/2

**==> The definite integral is (e^4 - 1)/2**

The definite integral is the area which has to be found, that 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**