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Solving an equation after integrating

  1. Jun 30, 2011 #1
    1. The problem statement, all variables and given/known data

    i have an equation that i want to solve for b:
    [itex]\int _0^1 (a+b x)^2 x^c dx=2[/itex]

    2. Relevant equations
    \int _0^1 (a+b x)^2 x^c dx=2
    given: c>0

    3. The attempt at a solution

    To evaluate the integral on the left hand side, I expanded the bracket as:
    (a+b x)^2=a^2+2abx+(bx)^2.

    Then i evaluated the integral and simplifed the result to get:

    [itex]\frac{(b^2+2ab+a^2)c^2+(3b^2+8ab+5a^2)c+2b^2+6ab+6a^2}{c^3+6c^2+11c+6}=2[/itex]

    and now i'm stuck.
    Since we have linear and quadratic powers of b, i don't know how to make b the subject. any ideas will be very much appreciated.
    Thank you
     
  2. jcsd
  3. Jun 30, 2011 #2

    micromass

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    What is the integral

    [tex]\int_0^1{(a^2x^c+2abx^{c+1}+b^2x^{c+2})dx}[/tex]

    Don't "simplify" anything, just calculate the integral, nothing more.
     
  4. Jun 30, 2011 #3
    after evaluating the integral, we have:

    (a^2/(c+1)) + (2ab/(c+3))+(b^2/(c+3))=2

    (this looks much cleaner :))

    but still, how would i make b the subject?

    thank you in advance
     
  5. Jun 30, 2011 #4

    micromass

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    Well, you have

    [tex]\frac{1}{c+3}b^2+\frac{2a}{c+2}b+\frac{a^2-2c-2}{c+1}=0[/tex]

    this is a quadratic equation in b and can be solved by the quadratic formula.
     
  6. Jun 30, 2011 #5
    so, i will have 2 solutions?
     
  7. Jun 30, 2011 #6

    micromass

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    Yes, unless the solutions coincide. But I doubt that will happen.
     
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