Answer
$2450$ $J$
Work Step by Step
A slice of water $Δx$ $m$ thick and lying at a depth of $x_i$ $m$ (where $0{\leq}x{\leq}1/2$) has volume $(2\times1\timesΔx)$ $m^{3}$, a mass of $2000Δx$ $kg$, weighs about $(9.8)(2000Δx)$ = $19,600Δx)$ $N$, and thus requires about $19,600x_iΔx$ $J$ of work for its removal.
So
$$\begin{align*}
W& = \lim\limits_{n \to \infty}\Sigma_{i=1}^{n}{19,600x_iΔx}\\
& = \int_0^{\frac{1}{2}}19,600xdx\\
& = [9800x^{2}]_0^{1/2}\\
& = 2450 J
\end{align*}$$
