A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

This is what I did...

F=10

x=4

4 going to 6

F=kx

10=k(4)

k=2.5

f=2.5x

I then took the integral of 2.5x from 4 to 6 and my answer was 25 ft-lbs. The correct answer is 15/4 ft-lbs.

Can someone explain?