Solve Integral Problem: Work Done By Gas in Barrel

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The discussion revolves around calculating the work done by gas on a bullet fired from a rifle with a specified force equation. For a barrel length of 0.550 m, the work done is calculated to be 3580.5 J, or 3.5805 kJ. When extending the barrel to 1.15 m, the work done is found to be -62625 J, or -6.2625 kJ. The significant difference in work values raises concerns about the accuracy of the calculations, particularly regarding the use of integrals. The participant expresses uncertainty about their understanding of integrals and the correctness of their results.
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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.
 
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