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Math Help - integral

  1. #1
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    integral

    Hi, I have a problem with that:
    <br />
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} dx<br />
    Last edited by alex23; September 12th 2009 at 09:50 AM.
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  2. #2
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    Quote Originally Posted by alex23 View Post
    Hi, I have a problem with that:
    <br />
\int\limits_{0}^{1}\frac{3x - 4}{x^2 - 3x + 2}<br />
    Slight problem, the integrand has a nonintegrable singularity at one end of the range of integration!

    CB
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  3. #3
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    oh - sry in previous integral was a mistake.

    <br />
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} = (3*\frac{1}{2}*x^2 - 4x)  |_{-1}^{0} - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)<br />
    <br />
*(-(x^2-3x+2)^{-2}) =<br />

    good for now ?
    Last edited by alex23; September 12th 2009 at 07:15 AM.
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  4. #4
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    Hi alex23
    Quote Originally Posted by alex23 View Post
    oh - sry in previous integral was a mistake.

    <br />
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} = (3*\frac{1}{2}*x^2 - 4x)  |_{-1}^{0} - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)<br />
    <br />
*(-(x^2-3x+2)^{-2}) =<br />

    good for now ?
    I don't think your work is right. Do you know partial fraction ?
    And you forgot "dx" in your question
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  5. #5
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    <br />
\int\limits_{-1}^{0}\frac{3x - 4}{x^2 - 3x + 2} dx = \int\limits_{-1}^{0}(3x - 4)*(x^2 - 3x + 2)^{-1} dx = <br />
    <br />
(-5\frac{1}{2}) - \int\limits_{-1}^{0}(3*\frac{1}{2}*x^2 - 4x)*(-(x^2-3x+2)^{-2}) dx =<br />

    what is wrong ?
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  6. #6
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    \frac{3x-4}{x^2-3x+2}<br />

    \frac{3x-4}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2}<br />

    3x-4 = A(x-2) + B(x-1)

    let x = 2 ... B = 2

    let x = 1 ... A = 1

    \int \frac{1}{x-1} + \frac{2}{x-2} \, dx = \ln|x-1| + 2\ln|x-2| + C<br />
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  7. #7
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    I remember that - this is definite integral.

    But I think I have not definite integral. I have tried to solve this problem using this pattern:

    <br />
\int\limits_{a}^{b} f'(x)g(x) dx = f(x)g(x) |_{a}^{b}  - \int\limits_{a}^{b} f(x)g'(x) dx<br />

    But You said mine solutions from previous posts based on this pattern were wrong.

    So now I have solution by skeeter - what next ? As I understand this is some other method based on definite integral and now I can't use above pattern ? So what should I do now ?
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  8. #8
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    \int_{-1}^0 \frac{1}{x-1} + \frac{2}{x-2} \, dx =

    \left[\ln|x-1| + 2\ln|x-2|\right]_{-1}^0 =<br />

    [0 + 2\ln{2}] - [\ln{2} + 2\ln{3}] = \ln{2} - \ln{9} = \ln\left(\frac{2}{9}\right)<br />
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  9. #9
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    thx skeeter - Your solution is different than solution which I know and Your solution is easier

    As I understand in Your solution we must count definite integral and then  |_{a}^{b}

    ? Only that ?

    So maybe mine pattern is for count area under curve ? I had integrals 3. years ago and I remember nothing.... :/
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  10. #10
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    Hi alex23
    Quote Originally Posted by alex23 View Post
    <br />
\int\limits_{a}^{b} f'(x)g(x) dx = f(x)g(x) |_{a}^{b}  - \int\limits_{a}^{b} f(x)g'(x) dx<br />

    But You said mine solutions from previous posts based on this pattern were wrong.
    Your formula is right. It' integration by part. But your work is not right.
    What are your f(x), f('x), g(x), and g'(x) ?
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