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Math Help - Integrals

  1. #1
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    Integrals

    I was told the best way to solve this problem dx/x^2-x with the limits b=3
    and a=2, is to used parital fractions. I dont understand this. Please help.

    The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.


    I sincerely thank you for your help.
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  2. #2
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    Quote Originally Posted by roseh
    I was told the best way to solve this problem dx/x^2-x with the limits b=3
    and a=2, is to used parital fractions. I dont understand this. Please help.

    The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.


    I sincerely thank you for your help.
    I understand the first problem to be:
    \int_2^3 dx \, \frac{1}{x^2-x}

    So. You want to decompose the fraction into linear (or nonfactorable quadratic) parts.
    \frac{1}{x^2-x}=\frac{1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}
    Where A and B are some constants.

    There are various and sundry ways to solve for A and B. My favorite is simply to add the two fractions and compare the numerator with the numerator of the original expression. So:
    \frac{A}{x}+\frac{B}{x-1}=\frac{A(x-1)+B(x)}{x(x-1)}

    So we know that A(x-1)+B(x)=1.

    By expanding the LHS and matching like coefficients we get that A+B=0 and -A=1. So A=-1 and B=1. Finally:

    \int_2^3 dx \, \frac{1}{x^2-x}=\int_2^3 dx \, \frac{-1}{x} + \int_2^3 dx \, \frac{1}{x-1}

    =-ln(|x|) \right | _2^3 + ln(|x-1|) \right| _2^3

    =(-ln3+ln2)+(ln2-ln1) = -ln3+2ln2-0=ln(2^2/3)

    =ln(4/3)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by roseh
    I was told the best way to solve this problem dx/x^2-x with the limits b=3
    and a=2, is to used parital fractions. I dont understand this. Please help.

    The second problem is dx/x^2-4x+4 with the limits b=1 and a=-1.


    I sincerely thank you for your help.
    For your second problem:
    \int_{-1}^1 dx \, \frac{1}{x^2-4x+4}=\int_{-1}^1 dx \, \frac{1}{(x-2)^2}

    The fraction decomposition in this case is not
    \frac{1}{(x-2)^2} \neq \frac{A}{x-2}+\frac{B}{x-2}
    You will quickly see by adding the fractions on the RHS that this does not work. What we need for repeated linear factors is:
    \frac{1}{(x-2)^2}=\frac{A}{x-2}+\frac{Bx+C}{(x-2)^2}

    Again, I will add the fractions on the RHS and equate the numerator with the original expression.
    A(x-2)+Bx+C=1
    So A+B=0 and -2A+C=1. There is no single set of numbers A,B, and C that will work. The simplest is A=B=0, C=1 which gives you back your original problem. (This implies that partial fractions is NOT the simplest way to do the problem, which you may have already noticed! )

    In the spirit of partial fractions I will use A=-1, B=1, C=-1.

    \int_{-1}^1 dx \, \frac{1}{x^2-4x+4}=\int_{-1}^1 dx \, \frac{1}{(x-2)^2}

    =\int_{-1}^1 dx \, \frac{-1}{x-2} + \int_{-1}^1 dx \, \frac{x-1}{(x-2)^2}
    I will put y = x - 2 in the second integral, so x = y + 2 and dx = dy:

    =-ln(|x-2|) \right | _{-1}^1 + \int_{-3}^{-1} dy \,  \frac{y+1}{y^2}

    =(-ln1+ln3)+ \int_{-3}^{-1} dy \, \frac{1}{y} + \int_{-3}^{-1} dy \, \frac{1}{y^2}

    =-0+ln3+ln(|y|) \right | _{-3}^{-1} - \frac{1}{y} \right | _{-3}^{-1}

    = ln3+(ln1-ln3)-(\frac{1}{-1} - \frac{1}{-3})

    =0-(-1+\frac{1}{3})=\frac{2}{3}

    Of course, this would have been a whole lot simpler just to set y = x - 2 in the original integral.

    Note: I didn't mention it explicitly... Remember to check that the integral doesn't cross any "improper" points. (ie. Make sure the integrand doesn't go either to zero or infinity.)

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    The only case of partial fractions I have not mentioned (because it didn't come up) is the case of non-factorable quadratic factors. By "non-factorable" I mean the quadratic has no real zeros. (You CAN use complex factors, but I, as a Physicist, usually don't.) So how would you decompose:
    \frac{x^2+1}{2x^3+4x^2+3x+1}

    I happen to know (because I built the problem!) that 2x^3+4x^2+3x+1=(x+1)(2x^2+2x+1). So we can break the original fraction into:

    \frac{x^2+1}{2x^3+4x^2+3x+1}=\frac{A}{x+1}+\frac{B  x+C}{2x^2+2x+1}

    We then add the fractions on the RHS and compare numerators:

    A(2x^2+2x+1)+(Bx+C)(x+1)=x^2+1
    Note that the quadratic factor always has a linear term over it.

    So we get that 2A+B=1, 2A+(B+C)=0, and A+C=1. I get A=2, B=-3, and C=-1.

    \frac{x^2+1}{2x^3+4x^2+3x+1}=\frac{2}{x+1}+\frac{-3x+-1}{2x^2+2x+1}
    as you can check by adding fractions.

    In general for repeated factors, we always up the degree of the polynomial in the numerator for every factor we have, so for:
    \frac{1}{x^4}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{D  x^2+Ex+F}{x^3}+\frac{Gx^3+Hx^2+Ix+J}{x^4}

    So if you have something like:
    \frac{1}{x^2(x^2-3x+1)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^  2-3x+1}

    -Dan
    Last edited by topsquark; March 28th 2006 at 08:19 AM. Reason: Can't tell my left hand from my right!
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    Thanks soooooooo much! The explanation/breakdown was terrific.
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by roseh
    Thanks soooooooo much! The explanation/breakdown was terrific.
    We aim to please!

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    Humbug!

    Something was bugging me about a previous post so I looked it up.

    My method for the repeated linear factors is not correct. (Well, technically it will work, but it isn't the simplest form!)

    If you have:
    \frac{x}{(x-2)^2}
    The partial fraction expansion is:
     \frac{x}{(x-2)^2}= \frac{A}{x-2}+ \frac{B}{(x-2)^2}

    Previously I had told you the numerator of the second term would be Bx+C, which will work I suppose, since you can always just set B=0, but the new expansion I just gave you will work better. In other words, for repeated linear factors the numerators are just constants.

    Sorry for the confusion!

    -Dan
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