A spring of stiffness k = 200 N/m is stretched from 0 to 0.2 m. The work done is the integral of kx with respect to x, which equals one half k x squared. How many joules of work are done?

4

$W = \tfrac{1}{2}(200)(0.2^{2}) = 100 \times 0.04 = 4$ J. Using $W = Fd$ with the FINAL force of 40 N would give 8 J, double the truth, because the force grew from zero, and only the integral accounts for that.