Answer
$$W = 160{\text{in - lb}}$$
Work Step by Step
$$\eqalign{
& {\text{From the information we have:}} \cr
& F\left( x \right) = 20{\text{pounds}},{\text{ }}\left( {{\text{Variable force}}} \right) \cr
& x = {\text{9inches}},{\text{ }}\left( {{\text{Variable distance}}} \right) \cr
& \cr
& {\text{Using the Hooke's Law to find the spring constant }}k \cr
& F\left( x \right) = kx \cr
& 20 = k\left( 9 \right) \cr
& k = \frac{{20}}{9} \cr
& \cr
& F\left( x \right) = kx \cr
& F\left( x \right) = \frac{{20}}{9}x \cr
& \cr
& {\text{The work required is }} \cr
& W = \int_a^b {F\left( x \right)} dx \cr
& {\text{Stretching the spring 0ft to 1ft, }} \cr
& 1{\text{ft}}\left( {\frac{{12{\text{in}}}}{{1{\text{ft}}}}} \right) = 12{\text{in}} \cr
& {\text{Interval }}\underbrace {\left[ {0{\text{in}},12{\text{in}}} \right]}_{\left[ {a,b} \right]},{\text{ then}} \cr
& W = \int_0^{12} {\frac{{20}}{9}x} dx \cr
& {\text{Integrating}} \cr
& W = \frac{{20}}{9}\left[ {\frac{1}{2}{x^2}} \right]_0^{12} \cr
& W = \frac{{10}}{9}\left[ {{x^2}} \right]_0^{12} \cr
& W = \frac{{10}}{9}\left[ {{{\left( {12} \right)}^2} - {{\left( 0 \right)}^2}} \right] \cr
& W = 160{\text{in - lb}} \cr} $$
