Answer
\[\int_{0}^{1}\sqrt{2+\frac{1}{2t}}\;\;dt\;\;\approx 1.49923\]
Work Step by Step
\[x=t+\sqrt{t}\;,\;y=t-\sqrt{t}\;\;\,0\leq t\leq 1\]
Differentiate $x$ with respect to $t$:
\[\frac{dx}{dt}=1+\frac{1}{2\sqrt{t}}\]
Differentiate $y$ with respect to $t$:
\[\frac{dy}{dt}=1-\frac{1}{2\sqrt{t}}\]
Therefore the arc length is given by:
\[L=\int_{0}^{1}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\]
\[L=\int_{0}^{1}\sqrt{\left(1+\frac{1}{2\sqrt{t}}\right)^2+\left(1-\frac{1}{2\sqrt{t}}\right)^2}dt\]
\[L=\int_{0}^{1}\sqrt{\left(1+\frac{1}{4t}+\frac{2}{2\sqrt{t}}\right)+\left(1+\frac{1}{4t}-\frac{2}{2\sqrt{t}}\right)}dt\]
\[L=\int_{0}^{1}\sqrt{2+\frac{1}{2t}}\;\;dt\]
By using calculator,
\[L\approx 1.49923\]
