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anonymous

  • 4 years ago

Under certain circumstances, the number of generations of "good" bacteria needed to increase the frequency of a particular gene from is n=4.572 integral 1/((x^2)(1-x)) dx

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  1. anonymous
    • 4 years ago
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    \[n=4.572\int\limits_{0.1}^{0.6}1/(x^2(1-x))dx\]

  2. anonymous
    • 4 years ago
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    find n, rounded to the nearest whole

  3. anonymous
    • 4 years ago
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    \[\int \frac{1}{x^2(1-x)} dx\] Hmm I would like to try partial fraction \[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\]

  4. anonymous
    • 4 years ago
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    partial fraction been a long time... i think its the denominator your working on right?

  5. anonymous
    • 4 years ago
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    \[1 = Ax^2 + Bx^2 + Bx + Cx + C \] Comparing co-efficient C = 1; (A+B) = 0; (B+C) = 0 Now you can integrate it! :-D

  6. anonymous
    • 4 years ago
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    Umm the Numerator \[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} = 1 = Ax^2 + Bx^2 + Bx + Cx + C\] Denominators cancel out.

  7. anonymous
    • 4 years ago
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    \[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} \implies 1 = Ax^2 + Bx^2 + Bx + Cx + C\] This version is much better

  8. anonymous
    • 4 years ago
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    your 1/x^2(1-x) becomes 1/x^2(1+x) typo?

  9. anonymous
    • 4 years ago
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    Oh Crap, yeah typo but the concept is same and I also messed up my calculations but the concept remains the same. Instead of \[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\] you will have to do \[\frac{1}{x^2(1-x)} = \frac{A}{1-x} + \frac{Bx +C}{x^2}\]

  10. anonymous
    • 4 years ago
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    so its gonna be 1=\[x^2(A-B)+x(B-C)+C\]

  11. anonymous
    • 4 years ago
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    C= 1, (A-B) = 0, (B-C)=0

  12. anonymous
    • 4 years ago
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    yeah exactly!

  13. anonymous
    • 4 years ago
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    my answer is 50 =) is that right?

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