The integral is the area under a curve if we let x vary? What about Work and Force?

Question

If Force = 6.25 - 5x + x^2 then work = 6.25x - 2.5x^2-(1/3)x^3

Is this next statement correct.

"Work is the area under the curve of Force, if we let the upper limit of integration vary and assume the lower limit of integration to be zero"

Its been 2 weeks since FTC

Also, I asked a question 30 minutes ago here....

http://answers.yahoo.com/question/index;_ylt=AkQEXBOdvTuQoXZOGKd6XMTsy6IX;_ylv=3?qid=20080717234932AAp0gYc

the question is simply solve ∫ e^(-4)x^2 dx from .25 to 1.25

So answer both if you feel willing.

Answer 1

The limits of integration are simply the inital and finish positions. If the object that the force is acting on starts at zero, then that statement is correct. Also, work does not depend on direction,at least with a conservative force, so that means if you start at zero move it to x=100, back to x=-100, and finish at zero, you still integrate from 0 to 0 giving you no net work.