Answer
$$L = (e^3 - e^{-3})$$
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 $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(e^t + e^{-t})}{dt} = e^t - e^{-t}$
$\frac{dy}{dt} = \frac{d(5 - 2t)}{dt} = -2$
3. Substitute the values into the definite integral:
$L = \int_{\alpha}^{\beta} \sqrt{(e^t - e^{-t})^2 + (-2)^2} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(e^{2t} - 2e^te^{-t} + e^{-2t}) + 4} \space dt$
** Notice : $e^t e^{-t} = 1$
$L = \int_{\alpha}^{\beta} \sqrt{(e^{2t} - 2e^te^{-t} + e^{-2t}) + 4(e^t e^{-t})} \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(e^{2t} + 2e^te^{-t} + e^{-2t}) } \space dt$
$L = \int_{\alpha}^{\beta} \sqrt{(e^t + e^{-t})^2} \space dt$
$L = \int_{\alpha}^{\beta} {(e^t + e^{-t})} \space dt$
Since t goes from 0 to 3:
$L = \int_{0}^{3} {(e^t + e^{-t})} \space dt$
$L = (e^t - e^{-t})]^3_0$
$L = (e^3 - e^{-3}) - (e^0 - e^{-0})$
$L = e^3 - e^{-3} - (1 - 1)$
$L = (e^3 - e^{-3})$
