$\begin{array}{1 1}6061.83J\\6051.83J\\6000J\\2699.29J\end{array} $

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Answer : 6061.83J

The work is found by integrating the force over the distance

$W=\int\limits_{x_i}^{x_f} F.ds$

For the barrel of length $l_n=0.441m$,we need only change the limits of integration,

$W=\int\limits_0^{l_n} (a+bx-cx^2)dx$

$\;\;\;\;=ax+\large\frac{1}{2}$$bx^2-\large\frac{1}{3}$$cx^3\big|_0^{l_n}$

$\;\;\;\;=(13500N)(0.441m)+\large\frac{1}{2}$$(8170N/m)(0.441m)^2-\large\frac{1}{3}$$(24000N/m^2)(0.441m)^3$

$\;\;\;\;=6061.83J$

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