Answer
$$L = \int_0^2 {\sqrt {1 + 4{e^{ - 4x}}} } dx$$
Work Step by Step
$$\eqalign{
& y = {e^{ - 2x}}{\text{ on }}\left[ {0,2} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = {e^{ - 2x}}{\text{ and }}\left[ {0,2} \right] \to a = 0{\text{ and }}b = 2.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - 2x}}} \right] \cr
& f'\left( x \right) = - 2{e^{ - 2x}} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_0^2 {\sqrt {1 + {{\left( { - 2{e^{ - 2x}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_0^2 {\sqrt {1 + 4{e^{ - 4x}}} } dx \cr} $$
