Answer
$$W \approx 1919.145{\text{ ft - lb}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}F\left( x \right) = 2000x{e^{ - x}},{\text{ }}\left( {0 \leqslant x \leqslant 5} \right) \cr
& {\text{The work is given by }} \cr
& W = \int_a^b {F\left( x \right)} dx \cr
& W = \int_0^5 {2000x{e^{ - x}}} dx \cr
& {\text{Integrate by tables }}\int {u{e^u}du} = \left( {u - 1} \right){e^u} + C \cr
& W = 2000\int_0^5 {\overbrace {\left( { - x} \right)}^u\overbrace {{e^{ - x}}}^{{e^u}}\overbrace {\left( { - 1} \right)dx}^{du}} \cr
& W = 2000\left[ {\left( { - x - 1} \right){e^{ - x}}} \right]_0^5 \cr
& W = - 2000\left[ {\left( {x + 1} \right){e^{ - x}}} \right]_0^5 \cr
& W = - 2000\left[ {\left( {5 + 1} \right){e^{ - 5}} - \left( {0 + 1} \right){e^{ - 0}}} \right] \cr
& W = - 2000\left[ {6{e^{ - 5}} - 1} \right] \cr
& W = 2000\left( {1 - \frac{6}{{{e^5}}}} \right) \cr
& W \approx 1919.145{\text{ ft - lb}} \cr} $$
