anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[n=4.572\int\limits_{0.1}^{0.6}1/(x^2(1-x))dx\]
anonymous
  • anonymous
find n, rounded to the nearest whole
anonymous
  • anonymous
\[\int \frac{1}{x^2(1-x)} dx\] Hmm I would like to try partial fraction \[\frac{1}{x^2(1+x)} = \frac{A}{1+x} + \frac{Bx +C}{x^2}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
partial fraction been a long time... i think its the denominator your working on right?
anonymous
  • anonymous
\[1 = Ax^2 + Bx^2 + Bx + Cx + C \] Comparing co-efficient C = 1; (A+B) = 0; (B+C) = 0 Now you can integrate it! :-D
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[\frac{1}{x^2(1+x)} = \frac{Ax^2 + Bx^2 + Bx + Cx + C}{x^2 (x+1)} \implies 1 = Ax^2 + Bx^2 + Bx + Cx + C\] This version is much better
anonymous
  • anonymous
your 1/x^2(1-x) becomes 1/x^2(1+x) typo?
anonymous
  • anonymous
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 \[\frac{1}{x^2(1-x)} = \frac{A}{1-x} + \frac{Bx +C}{x^2}\]
anonymous
  • anonymous
so its gonna be 1=\[x^2(A-B)+x(B-C)+C\]
anonymous
  • anonymous
C= 1, (A-B) = 0, (B-C)=0
anonymous
  • anonymous
yeah exactly!
anonymous
  • anonymous
my answer is 50 =) is that right?

Looking for something else?

Not the answer you are looking for? Search for more explanations.