Answer
$62.5$ $ft-lb$
Work Step by Step
The chain's weight density is
$\frac{25lb}{10ft}$ = $2.5$ $lb/ft$
The part of the chain $x$ $ft$ below the ceiling (for $5{\leq}x{\leq}10$) has ot be lifted $2(x-5)$ $ft$, so the work needed to lift the $i$th subinterval of the chain is
$2({x_i}-5)(2.5Δx)$
The total work needed is
$$\begin{align*}
W& = \lim\limits_{n \to \infty}{\Sigma_{i=1}^{n}}2(x_i-5)(2.5)Δx\\
& = \int_5^{10}[2(x-5)(2.5)]dx\\
& = 5\left[\frac{1}{2}x^{2}-5x\right]_5^{10}\\
& = 62.5 ft-lb
\end{align*}$$
