Find the exact area enclosed between the curve y=e^(2x) and the lines y=1 and x=2

You need to evaluate the definite integral, hence, you need to find the limits of integration, such that:

`{(y = e^(2x)),(y = 1):} => e^(2x) = 1 => e^(2x) = e^0 => 2x = 0 => x = 0`

You need to notice that `y = e^(2x)` is above `y...

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You need to evaluate the definite integral, hence, you need to find the limits of integration, such that:

`{(y = e^(2x)),(y = 1):} => e^(2x) = 1 => e^(2x) = e^0 => 2x = 0 => x = 0`

You need to notice that `y = e^(2x)` is above `y = 1` for `x in [0,2]` , hence, you need to evaluate the following definite integral to find the area of the region enclosed by the given cuves, such that:

`int_0^2 (e^(2x) - 1) dx`

You need to use the property of linearity of integrals, such that:

`int_0^2 (e^(2x) - 1) dx = int_0^2 (e^(2x)) dx -int_0^2 dx`

`int_0^2 (e^(2x) - 1) dx = ((e^(2x))/2 - x)|_0^2`

You need to use the fundamental theorem of calculus, such that:

`int_0^2 (e^(2x) - 1) dx = (e^4/2 - 2 - e^0/2 + 0)`

`int_0^2 (e^(2x) - 1) dx = (e^4/2 - 2 - 1/2)`

`int_0^2 (e^(2x) - 1) dx = (e^4 - 5)/2`

Hence, evaluating the area of the region enclosed by the curve `y = e^(2x)` and the line `y = 1` , for `x in [1,2]` yields `int_0^2 (e^(2x) - 1) dx = (e^4 - 5)/2.`

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