Answer
$$
x=t+e^{-t}, y=t-e^{-t}, \quad 0 \leq t \leq 2
$$
$L \approx 3.1416$
Work Step by Step
$$
x=t+e^{-t}, y=t-e^{-t}, \quad 0 \leq t \leq 2
$$
then
$$
d x / d t=1-e^{-t}, \quad d y / d t=1+e^{-t},
$$
so
$$
\begin{aligned}
(d x / d t)^{2}+(d y / d t)^{2}& =\left(1-e^{-t}\right)^{2}+\left(1+e^{-t}\right)^{2} \\
&=1-2 e^{-t}+e^{-2 t}+1+2 e^{-t}+e^{-2 t} \\
&=2+2 e^{-2 t}.
\end{aligned}
$$
Thus the length of the given curve is
$$
\begin{aligned}
L &=\int_{a}^{b} \sqrt{(d x / d t)^{2}+(d y / d t)^{2}} d t \\
&=\int_{0}^{2} \sqrt{2+2 e^{-2 t}} d t\\
& \approx 3.1416
\end{aligned}
$$
