## Calculus: Early Transcendentals 8th Edition

The formula for the arc length of a curve on the interval (a,b) is $_{a}\int^{b}(\sqrt {1+(dy/dx)^2}$. The derivative of the function $x\times e^{-x}$ with respect to x is (using the product rule): $(x)'e^{-x}+x(e^{-x})'$=$(1)e^{-x}+x*-e^{-x}$=$(1-x)\times e^{-x}$. Plugging this in, we get: $_{0}\int^{2}(\sqrt {1+(1-x)^2\times e^{-2x}}$. A calculator provides the final answer: 2.1024.