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Homework Help: What is the distance a particle has moved if I know the work done

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    I was asked to find the distance a particle had traveled and was given its integral solved, except the upper limit was an equation.

    [itex](x^2+2x)[/itex] is the force in pounds that acts on the particle

    2. Relevant equations
    [itex]\int_{0}^{x^2+2}(x^2+2x)dx = 108[/itex]

    3. The attempt at a solution
    I know I must find what number the top limit equation equals, but I've been inserting random numbers with no success. Also, I don't know if it is possible to find the value of x and plug it in. This is ridiculous, why couldn't they just give me the integral with the limits and I solve it. :grumpy:
  2. jcsd
  3. Feb 26, 2012 #2
    What is the integral of x2 + 2x?
  4. Feb 26, 2012 #3
    Its [itex]\frac{1}{3}x^3+x^2[/itex]
  5. Feb 26, 2012 #4

    Ray Vickson

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    Maybe the problem is poor notation; this problem's notation is terrible--it double-uses the symbol x. Much better: write [tex] \int_{0}^{x^2+2}(x'^2+2x')dx' = 108,[/tex] which makes it clear what is the integration variable (x') and what is the unknown (x).

  6. Feb 26, 2012 #5
    I see what you did there, but how does that makes it easier to solve? I still don't know what to do. :frown:

    Rewinding everything I know:

    The force exerted on the particle: [tex](x'^2+2x')[/tex]

    The initial point of the particle: 0

    The distance traveled by the particle (that's my upper limit) given by: [tex](x^2+2)[/tex]

    And the total work done by the particle after traveling the distance I'm looking for: 108

    Still not helps writing all that. How can I find the upper limit value or the x value to plug it in and get 108 as answer?
    Last edited: Feb 26, 2012
  7. Feb 26, 2012 #6
    I agree with RGV. It may be easier to think about using a different variable name for the integral than the x you're trying to solve for.

    Try plugging in the limits of integration as if you are evaluating the integral.
  8. Feb 26, 2012 #7
    I kept trying putting in random values and found that the upper value must be 6. In other words:
    [tex] \int_{0}^{6}(x'^2+2x')dx' = 108,[/tex]
    However, isn't there a process to reach that? I did it by brute force.
  9. Feb 26, 2012 #8

    Ray Vickson

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    You have already evaluated the indefinite integral [itex] F(x') = \int(x'^2 + 2x')\, dx'.[/itex] Using your formula for F, what would be [itex] \int_0^a (x'^2 + 2x') \,dx'[/itex]? What is preventing you from substituting in [itex] a = x^2 + 2[/itex]? Mind you, the final result will not be pretty, but it gives you a place to start.

  10. Feb 26, 2012 #9
    Here's what I did, perhaps you may find the process of some use.

    1) I computed the integral and applied the limits.
    2) I know that the integral with the limit substitutions = 108
    3) Now it's mostly a matter of expanding the terms and factoring to find solutions.
    4) The only applicable solution I found was x=2
    5) The upper limit with the x=2 substitution is indeed equal to 6.

    Sorry for the short reply, I did it by hand and did the expansions and factoring by calculator since they were quite long. If you would like, I could write it out or post a picture of my solution if you did not understand my steps above.

    Hope that helps,

  11. Feb 26, 2012 #10
    It gives me x = 2, but I wonder if there is a rule or something for me to find that the value was 6 instead of doing it by guessing like I did.

    Yes please, an image would help a lot. Thanks. Its just that its weird to have to solve it by guessing, but you seem to have some process there.
    Last edited: Feb 26, 2012
  12. Feb 26, 2012 #11
    I hope you can read what I wrote. The site kills photo quality apparently and I didn't have enough time to copy it with my scanner.


    Hope that helps,

  13. Feb 26, 2012 #12
    Oh, I understand what you did there. Thank you very much. I feel dumb now because you make it look so easy and it took me so long.
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