Antidifferentiation for a larger problem

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Homework Statement


I'm trying to antidifferentiate a function in order to find the volume of a solid. The function comes out to be (2x+1)/(x^2+8x+16)


Homework Equations



N/A

The Attempt at a Solution


I've tried a multitude of things. I figured that partial fractions would work, but it turns out that since the denominator is a square that doesn't work. I'm not asking you to solve, just a nudge in the right direction
 

Answers and Replies

  • #2
cristo
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I figured that partial fractions would work, but it turns out that since the denominator is a square that doesn't work.
What do you mean by this? Try the partial fraction decomposition [tex]\frac{A}{(x+4)}+\frac{B}{(x+4)^2}[/tex]
 
  • #3
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ALright. I don't understand though. Where did the third (x+4) come from. The denominator only factors to (x+4)(x+4)
 
  • #4
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Okay even when i try that I get

A(x+4)^2+B(x+4) = 2x+1

When I plug in to eliminate one of the variables (x=-4) then I get 0+0 = -7. I'm probably doing something wrong.
 
  • #5
cristo
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ALright. I don't understand though. Where did the third (x+4) come from. The denominator only factors to (x+4)(x+4)
We know that the number of unknown constants must be equal to the degree of the denominator, and so the easiest way to set it up is to take the two fractions as A/(x+4) and B/(x+4)2

Okay even when i try that I get

A(x+4)^2+B(x+4) = 2x+1

When I plug in to eliminate one of the variables (x=-4) then I get 0+0 = -7. I'm probably doing something wrong.
This is not correct: You should have the following expression: [tex]\frac{2x+1}{(x+4)^2}\equiv \frac{A}{(x+4)}+\frac{B}{(x+4)^2}[/tex] Multiplying this expression by (x+4)2 does not give what you have above.
 
  • #6
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Ok. I'll try that. Why did you use those denominators though, for the partials. Intuitively I would use x+4 and x+4 because they multiply together to get x^2 +8x+16

Thanks though
 
  • #7
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Alright. So I solve that equation and get 2x+1 = A(x+4) + B

A = 1 and B = -7

WHen I antidifferentiate I get ln(x+4) + 7/(x+4) Seems that I'm missing a factor of 2 in front of the first term. Sorry for the problems.
 
  • #8
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nevermind on that last part, i found a mistake. but the previous question is still troubling me
 
  • #9
cristo
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Ok. I'll try that. Why did you use those denominators though, for the partials. Intuitively I would use x+4 and x+4 because they multiply together to get x^2 +8x+16

Thanks though
Well, if we use (x+4) as the only denominator we obtain something like [tex]\frac{2x+1}{(x+4)^2}\equiv \frac{A}{x+4}+\frac{B}{x+4}=\frac{A+B}{x+4}[/tex]; notice that this can never be in the required form (since the denominator will always be linear). Thus, we take the two fractions as (x+4) and (x+4)2, so that the fraction looks like[tex]\frac{A}{x+4}+\frac{B}{(x+4)^2}=\frac{A(x+4)+B}{(x+4)^2}[/tex] which is of the required form.
 
  • #10
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Alright that makes sense. I'm trying to solve that equation now. I can solve for B=-7, but I can't solve for A because I have two variables: x and A.
 
  • #11
cristo
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Alright that makes sense. I'm trying to solve that equation now. I can solve for B=-7, but I can't solve for A because I have two variables: x and A.
Just put, say, x=0.
 
  • #12
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Oh nvm. I got it. I have the problem solved now. I really appreciate the help cristo, you've enhanced my understanding of partial fractions.
 
  • #13
cristo
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Oh nvm. I got it. I have the problem solved now. I really appreciate the help cristo, you've enhanced my understanding of partial fractions.
You're very welcome; glad to be of help!
 

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