Answer
$\displaystyle \frac{16\sqrt{2}-8}{15}\approx 0.9752$
Work Step by Step
$\displaystyle \int_{0}^{1}\int_{0}^{1}\sqrt{s+t}ds dt=\displaystyle \int_{0}^{1}[\int_{0}^{1}\sqrt{s+t}ds ]dt=\left[\begin{array}{ll}
u=t+s & \\
du=ds & \\
t=0\Rightarrow u=t, & t=1\Rightarrow u=t+1
\end{array}\right]$
... treat t as a constant in the inner integral.
$=\displaystyle \int_{0}^{1}[\int_{t}^{t+1}u^{1/2}du ]dt$
$= =\displaystyle \int_{0}^{1}[\frac{u^{3/2}}{3/2} ]_{t}^{t+1}dt$
$=\displaystyle \frac{2}{3}\int_{0}^{1}[(t+1)^{3/2}-t^{3/2}]dt=$
$=\displaystyle \frac{2}{3}[\frac{(t+1)^{5/2}-t^{5/2}}{5/2}]_{0}^{1}$
$=\displaystyle \frac{4}{15}[(2^{5/2}-1)-(1-0)]$
$=\displaystyle \frac{4}{15}(2^{5/2}-2)$
$=\displaystyle \frac{4}{15}(4\sqrt{2}-2)$
$= \displaystyle \frac{16\sqrt{2}-8}{15}\approx 0.9752$
