Application of Integrals

By Hook's law we have
\[F=kx\]
where $k$ is the spring constant and $x$ is the distance form the natural length. So we have
\[5=k \cdot (0.4) \Rightarrow  k=\frac{5}{0.4}=\frac{50}{4}=\frac{25}{2}.\]
Hence we have
\[F=\frac{25}{2} x.\]
The total work done is given by 
\[W=\int_0^{1.1} F dx=\int_0^{1.1} \frac{25}{2} x dx= \frac{25}{4}x^2 \big|_0^{1.1}=\frac{25}{4}(1.1)^2-0=7.5625.\]
So the answer is $7.5625$.