Solve Integral Problem: Work Done By Gas in Barrel

[alucrid]

Homework Statement

A 80 g bullet is fired from a rifle having a barrel 0.550 m long. Assuming the origin is placed where the bullet begins to move, the force (in Newtons) exerted by the expanding gas on the bullet is 13000 + 5000x - 26000x2, where x is in meters.

(a) Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.

(b) If the barrel is 1.15 m long, how much work is done, and how does this value compare to the work calculated in part (a)? (work done with the 1.15 m barrel / work done with the 0.550 m barrel)

Homework Equations

13000 + 5000x - 26000x2

13000x + (5000/2)x² - (26000/3)x³

The Attempt at a Solution

I know I have to get the integral. So that would be 13000x + (5000/2)x² - (26000/3)x³. So when I substitute .550 m in I get 3580.5 J which is then 3.5805 Kj. Then for 1.5 m i get -62625 J, which is -6.2625 Kj. Did I do this right? I have a feeling I didn't, since I know almost nothing about integrals.

 

[unscientific]

Complete Solution Removed

 

[Hootenanny]

unscientific, please check your PM's