Answer
$$L = \int_{0}^{1} \sqrt{2 + \frac 1 {2t}} \space dt \approx 2.0915$$
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 $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(t + \sqrt t)}{dt} = 1 + \frac{1}{2 \sqrt t}$
$\frac{dy}{dt} = \frac{d(t - \sqrt t)}{dt} = 1 - \frac{1}{2 \sqrt t}$
3. Substitute the values into the definite integral:
$L = \int_{0}^{1} \sqrt{(1 + \frac{1}{2 \sqrt t})^2 + (1 - \frac{1}{2 \sqrt t})^2} \space dt$
$L = \int_{0}^{1} \sqrt{(1 + 2(\frac{1}{2 \sqrt t})+ \frac 1 {4t}) + (1 - 2(\frac{1}{2 \sqrt t})+ \frac 1 {4t})} \space dt$
$L = \int_{0}^{1} \sqrt{(1 + \frac 1 {4t}) + (1 + \frac 1 {4t})} \space dt$
$L = \int_{0}^{1} \sqrt{2 + \frac 2 {4t}} \space dt$
$L = \int_{0}^{1} \sqrt{2 + \frac 1 {2t}} \space dt$
4. Write this integral on your calculator (you can use online integral calculators), and get the result.
$L \approx 2.0915$
