A force of 13 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length?
Using Hooke's Law, f(x)=kx , I set it up like this:
f(x) = 13
I plug that in to the to integral of 13x on the interval from 4 to 6. Then I got 65/2 in-lbs. I divide by 12 becuase the problem requires it to be ft-lb. Thus I got I got 65/24 ft-lbs, but the site says it is wrong. What did I do wrong?