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Well, it's not exactly homework, but did I do this integral right?

  1. Jun 9, 2005 #1
    I had a test this morning in cal 3 *crosses fingers* and I'm pretty sure I did well, but one of the questions was(I REALLY need somone to tell me how to do all the pretty math font stuff with symbols)
    S=integral, for this problem the upper limit was y, lower limit was x)

    f(x,y)=S(t^2+2t+3)^17 dt and then it asked for Fx and Fy, but that wasn't my problem

    so u=t^2+2t+3, du = 2t+2dt

    so you get (1/(2t+2))(u^18)/18, plug everything back in for u, and you get that decent ball of poo, then plug in the limits so you get [ball of poo with y in place of t] - [ball of poo with x in place of t], then just differentiate, for Fx the first part is 0, so it's -[more poo] and Fy is just [some more poo]-0?

    Edit: Hah, I typed you instead of u
  2. jcsd
  3. Jun 9, 2005 #2


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    For the math font (LaTeX code), look here: https://www.physicsforums.com/showthread.php?t=8997

    So the function is:


    It doesn't look like substitution will work here, since the integrand doesn't have the form of a product of a function and its derivative.

    Since you integrate a function with the upper limit as a variable and you then have to differentiate it, the fundamental theorem of calculus springs to mind.
  4. Jun 9, 2005 #3
    Oh nut crappers, you're right.

    Wait, so the way I didn't doesn't work, even if it's cumbersome? And won't the fundamental theorem only help me with the y partial derivative?
  5. Jun 9, 2005 #4


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    Switch the integral to:

    Now you can take the partial derivative with respect to x. The only difference is the '-' sign.

    I'd write the entire function as:
    [tex]f(x,y)=\int_a^y(t^2+2t+3)^{17}dt- \int_a^x(t^2+2t+3)^{17}dt[/tex]

    where a is some constant, just to get the x and y in different integrals.
    Last edited: Jun 9, 2005
  6. Jun 9, 2005 #5
    Ok ok, I blew that obvious way, but are you sure what I did doesn't work? I end up with some weird thing for du, and have to divide by 2t+2 and all that wash, but I seem to recall doing that in the past
  7. Jun 9, 2005 #6


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    Actually calculating the integral would work for simple integrals. I get the feeling that your professor purpoely gave a derivative that would be extremely hard to calculate. This integral is huge!

    What you did here with the u and t is not right. It looks like you took (1/(2t+2)) out of the integral. t is not a constant so it can't be taken out of the integral.
  8. Jun 9, 2005 #7
    Yah, I did end up with a huge integral, so I pray for luck

    I'm not 100 percent sure I retyped what I did right, so here's hoping...oh well, I probably still made an A. I did take the (1/2t+2) out, but at that point you were differentiating with respect to u, or something like that, I thought. Oh well, partial credit
  9. Jun 9, 2005 #8


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    I've plugged it into Maple. A primitive of the integrand is:


    Not something you want to work with on an exam, especially not with pen and paper.
    Last edited: Jun 9, 2005
  10. Jun 9, 2005 #9
    Hey, that's what I got, with pencil!

    ...ok, not true
  11. Jun 9, 2005 #10


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    I'm getting a headache just looking at that integral.

    Anyway, don't worry about it schattenjaeger. I bet you did great overall.
  12. Jun 9, 2005 #11
    Thanks, I hope so, 10 questions, and I'm really confident on the other 9, and with a little luck I might even get some(well, a little)partial credit when he says what I did or was trying to do, though I did miss the point of the problem

    Thing is, if it had been a single variable case with 0 as the lower limit and the variable in the upper, I would've known what to do, but I'm not sure how to do it in this situation
  13. Jun 9, 2005 #12


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    Yeah, I think it helps to split the integral into two parts as I did in post #4. As long as the lower limit is a constant, you can differentiate with respect to the upper variable. You can choose 'a' to be any constant (makes no difference here), including 0.

    When you take the partial with respect to x, the first integral disappears and you get [tex]-(x^2+2x+3)^{17}[/tex]

    When you take the partial with respect to y, the second integral disappears and you get [tex](y^2+2y+3)^{17}[/tex]
  14. Jun 9, 2005 #13
    Well, that makes great sense now, thanks!
  15. Jun 17, 2005 #14
    Well, made an 86 on the test, so good enough, for sure. Of course I got that problem wrong and big long reminder about how you can't have two variables after a substitution scrawled in red over my answer:) He DID lament in class though that he had meant to pick an integral we couldn't solve, but had actually picked one we could, though no one who had tried had done it correctly. How WOULD you solve that integral, not taking into account the fundamental theorem of calculus?
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