Answer
$L = \int_0^2 \sqrt{2 + 2e^{-2t} } \space dt \approx 3.1416$
Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(t + e^{-t})}{dt} = 1 - e^{-t}$
$\frac{dy}{dt} = \frac{d(t - e^{-t})}{dt} = 1 + e^{-t}$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(1 - e^{-t})^2 + (1 + e^{-t})^2} \space dt$
- Since $t$ goes from 0 to 2:
$L = \int_0^2 \sqrt{(1 - e^{-t})^2 + (1 + e^{-t})^2} \space dt$
$L = \int_0^2 \sqrt{(1 -2e^{-t} + e^{-2t}) + (1 +2e^{-t} + e^{-2t})} \space dt$
$L = \int_0^2 \sqrt{(1 -2e^{-t} + e^{-2t}) + (1 +2e^{-t} + e^{-2t})} \space dt$
$L = \int_0^2 \sqrt{2 + 2e^{-2t} } \space dt$
4. Write this integral on your calculator, and get the result.
$L \approx 3.1416$
