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Math Help - [SOLVED] Definite Integral, & Partial fractions

  1. #16
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    Okay, yea, once I got the division right everything went well, I know what to do now.
    So I now have:
    \int_{-1}^{3} \frac {85}{x+7}- \frac {6}{x+3}dx Could I split the integral, integrate and do the limits individually?
    Thanks!
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  2. #17
    Super Member 11rdc11's Avatar
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    Yes and here is a hint \int\frac{du}{u} = ln(u)
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  3. #18
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    Unhappy

    Okay, so I did that and I get 36.8285 for the answer, I then calculated it out on my calculator and it got came out to 85ln(5)-91ln(3)-36 which is .8285 only, so it's just without the 36. I guess I did something wrong with A and B? I also know that .8285 is the correct answer. I just checked.
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  4. #19
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    Oh wait, what do I need to do with the (x-10) on the outside, that must equal -36 right?
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  5. #20
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    \int_{-1}^{3} \bigg(\frac {85}{x+7}- \frac {6}{x+3}\bigg)dx

    [85ln(x+7) - 6ln(x+3)]^3_{-1}

    so

    85ln(10) - 6ln(6) - 85ln(6) + 6ln(2) = ?
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  6. #21
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    Quote Originally Posted by 11rdc11 View Post
    Yes and here is a hint \int\frac{du}{u} = ln(u)
    No, it's \ln |u| .. it doesn't matter here, but it's not very correct to write it
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  7. #22
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Moo View Post
    No, it's \ln |u| .. it doesn't matter here, but it's not very correct to write it
    Lol you too picky but you are right!
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  8. #23
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    I don't mean to be Mr. Picky, but:

    \left(x-10 + \frac{79x+213}{x^2+10x+21}\right) \times (x^2+10x+21)

    = (x-10)(x^2+10x+21) + 79x + 213

    = x^3 + 10x^2 + 21x - 10x^2 - 100x - 210 + 79x + 213

    = x^3 + 3

    \therefore \frac{x^3+3}{x^2+10x+21} = x-10 + \frac{79x+213}{x^2+10x+21}

    No need to change the integral answers though. The OP should be able to solve them on his own by now...right Matt3D?
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  9. #24
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    Quote Originally Posted by 11rdc11 View Post
    \int_{-1}^{3} \bigg(\frac {85}{x+7}- \frac {6}{x+3}\bigg)dx

    [85ln(x+7) - 6ln(x+3)]^3_{-1}

    so

    85ln(10) - 6ln(6) - 85ln(6) + 6ln(2) = ?
    Yea, that is what I got which is 36.8285
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  10. #25
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by 11rdc11 View Post
    Yep the first term after doing long division can't be x^2. It would have to be x because x times x^2 equals x^3. Now if the first term was x^2 we would have x^2 times x^2 which would be x^4

    Oops stand corrected! I didn't notice matt3d corrected his long division and he is correct it 213 not 210. Well everything Jameson said to do after with the partial fractions is correct. We all made minor mistakes lol.
    Lol yea we had a couple of mistakes but it all worked out in the end
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  11. #26
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    Quote Originally Posted by Chop Suey View Post
    I don't mean to be Mr. Picky, but:

    \left(x-10 + \frac{79x+213}{x^2+10x+21}\right) \times (x^2+10x+21)

    = (x-10)(x^2+10x+21) + 79x + 213

    = x^3 + 10x^2 + 21x - 10x^2 - 100x - 210 + 79x + 213

    = x^3 + 3

    \therefore \frac{x^3+3}{x^2+10x+21} = x-10 + \frac{79x+213}{x^2+10x+21}

    No need to change the integral answers though. The OP should be able to solve them on his own by now...right Matt3D?
    Yea, I know what you mean, it's all good though, I just thought that I should point it out... I'm still stuck on the answer though.
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  12. #27
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by matt3D View Post
    Yea, that is what I got which is 36.8285
    Hmm your right lets backtrack

    Ok I figured out why it wrong we forgot the other part of the original integral lol

    \int_{-1}^3 \bigg(x - 10 + \frac{85}{x+7} - \frac{6}{x+3}\bigg)dx

    so

    \frac{x^2}{2} -10x + 85ln(x+7) - 6ln(x+3)\bigg|^3_{-1}
    Last edited by 11rdc11; September 16th 2008 at 01:07 AM.
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  13. #28
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    Haha, awesome, I thought it was from the (x-10) too. Yep, so its 8/2-(30+10) which is -36.
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