The work of a spring

How much work does it take to stretch a spring? You will make the program add up thin strips under the force-position graph, and check the answer against $\tfrac12 kx^2$.

The idea. The work of a changing force is the area under its graph: $W = \int F\,dx$. For a spring, $F = kx$, and the area from $x_1$ to $x_2$ is $\tfrac12 k(x_2^2 - x_1^2)$.

Your tasks.

1. Work out the force $F = kx$ at every position in x.
2. Use area(x, F) to find the work to stretch the spring from $0\ \text{m}$ to $0.25\ \text{m}$.
3. Find the work from $0.10\ \text{m}$ to $0.20\ \text{m}$ the same way, using only the positions in that range.