Antidifferentiation for a larger problem

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In summary, the conversation focused on the process of antidifferentiating a function in order to find the volume of a solid. The function in question was (2x+1)/(x^2+8x+16), and the attempt at a solution involved trying multiple methods, including partial fractions. The conversation discussed the use of (x+4) and (x+4)^2 as the denominators for the partial fractions in order to obtain the required form. The end result was finding the values of A and B to solve the equation and ultimately understanding the concept of partial fractions better.
  • #1
nealh149
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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
 
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  • #2
nealh149 said:
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
ALright. I don't understand though. Where did the third (x+4) come from. The denominator only factors to (x+4)(x+4)
 
  • #4
Okay even when i try that I get

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

When I plug into eliminate one of the variables (x=-4) then I get 0+0 = -7. I'm probably doing something wrong.
 
  • #5
nealh149 said:
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

nealh149 said:
Okay even when i try that I get

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

When I plug into 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
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
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
nevermind on that last part, i found a mistake. but the previous question is still troubling me
 
  • #9
nealh149 said:
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
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
nealh149 said:
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
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
nealh149 said:
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!
 

1. What is antidifferentiation for a larger problem?

Antidifferentiation for a larger problem is a mathematical process used to find the original function given its derivative. This is especially useful when dealing with more complex functions or problems with multiple variables.

2. How is antidifferentiation different from integration?

Antidifferentiation and integration are often used interchangeably, but there is a subtle difference between the two. Antidifferentiation is the process of finding the original function given its derivative, while integration is the process of finding the area under a curve. In other words, integration is the reverse of differentiation, while antidifferentiation is the reverse of taking a derivative.

3. What are the common techniques used in antidifferentiation for a larger problem?

Some common techniques used in antidifferentiation for a larger problem include the chain rule, the product rule, and the quotient rule. These rules allow us to find the antiderivative of more complex functions by breaking them down into simpler parts.

4. Can antidifferentiation be used to solve real-world problems?

Yes, antidifferentiation is commonly used in physics, engineering, and other fields to solve real-world problems. For example, it can be used to calculate distance traveled, velocity, or acceleration given a function representing the object's motion.

5. Are there any limitations to using antidifferentiation for a larger problem?

While antidifferentiation is a powerful tool, it does have its limitations. It may not always be possible to find the antiderivative of a function, especially for more complex functions. In addition, there may be multiple antiderivatives for a given function, so it is important to check for any potential constants of integration when solving a problem.

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