Answer
$$\int_{0}^{1} \int_{0}^{1} \sqrt{s+t} \ d s \ d t=\frac{8}{15} \left(2\sqrt 2 -1\right)$$
Work Step by Step
Given $$\int_{0}^{1} \int_{0}^{1} \sqrt{s+t} \ d s \ d t$$So we have
\begin{aligned} I&=\int_{0}^{1} \int_{0}^{1} \left(s+t\right)^{\frac{1}{2}} d s d t\\
&=\frac{2}{3}\int_{0}^{1} \left(s+t\right)^{\frac{3}{2}}| _{0}^{1}\ \ d t\\
&=\frac{2}{3}\int_{0}^{1} \left(1+t\right)^{\frac{3}{2}} \ d t- \frac{2}{3}\int_{0}^{1} \left(t\right)^{\frac{3}{2}} \ \ d t\\\\
&=\frac{4}{15} \left(1+t\right)^{\frac{5}{2}} |_{0}^{1}- \frac{4}{15} \left(t\right)^{\frac{5}{2}} |_{0}^{1}\\\\
&=\frac{4}{15} \left(1+1\right)^{\frac{5}{2}} - \frac{4}{15} \left(1\right)^{\frac{5}{2}}- \frac{4}{15} \left(1\right)^{\frac{5}{2}}+0\\\\
&=\frac{4}{15} \left((\sqrt 2)^{5} -2\right)\\
&=\frac{4}{15} \left(4\sqrt 2 -2\right)\\
&=\frac{8}{15} \left(2\sqrt 2 -1\right)\\
\end{aligned}
