Answer
Length=$612.3$
Work Step by Step
$\frac{d x}{d t}{=}\mathbf{1}-e^{t}$
${\frac{d y}{d t}}=1+e^{t}$
$\begin{aligned} \int_{-6}^6 \sqrt{\left[\frac{d x}{d t}\right]^2+\left[{\frac{d y}{d t}}\right]^2} d t & =\int_{-6}^6 \sqrt{\left(1-e^t\right)^2+\left(1+e^t\right)^2} d t \\ & =\int_{-6}^6 \sqrt{1-2 e^t+e^{2 t}+1+2 e^t+e^{2 t}} d t \\ & =\int_{-6}^6 \sqrt{2+2 e^{2 t}} d t .\end{aligned}$
$\operatorname{Length}\!=\!\int_{-6}^{6}f(t)d t$
$\begin{aligned} & =\frac{6-(-6)}{3(6)}[f(-6)+4 f(-4)+2 f(-2)+4 f(0)+2 f(2)+4 f(4)+f(6)] \\ & =\frac{2}{3}[f(-6)+4 f(-4)+2 f(-2)+4 f(0)+2 f(2)+4 f(4)+f(6)] \\ & =\frac{2}{3}\left(\sqrt{2+2 e^{-12}}+4 \sqrt{2+2 e^{-8}}+2 \sqrt{2+2 e^{-4}}+4 \sqrt{2+2 e^0}+2 \sqrt{2+2 e^4}+4 \sqrt{2+2 e^8}+\sqrt{2+2 e^{12}}\right. \\ & \approx 612.3 .\end{aligned}$
