Integral of 5x^2 /sqrt(4x-x^2) dx

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  • #1
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Homework Statement


Integral of 5x^2 /sqrt(4x-x^2) dx. (This can also be seen in the attached TheProblem.png file.)

Homework Equations


Integration with variable substitution.

The Attempt at a Solution


Could someone please tell me what I did wrong in my work (which is attached as MyWork.jpg)?
 

Attachments

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Answers and Replies

  • #2
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Homework Statement


Integral of 5x^2 /sqrt(4x-x^2) dx. (This can also be seen in the attached TheProblem.png file.)

Homework Equations


Integration with variable substitution.

The Attempt at a Solution


Could someone please tell me what I did wrong in my work (which is attached as MyWork.jpg)?

You can't do what you did when you squared the integrand. IOW, in the step where the integrand changes from ## \frac{5x^{3/2}}{\sqrt{4 -x}}## to ##\frac{25x^3}{4 - x} ##.
 
  • #3
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Ouch why would anyone ever make you solve this. You have to complete the square, then do a trig sub, followed by a bunch of manipulation of variables.
 
  • #4
s3a
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You can't do what you did when you squared the integrand. IOW, in the step where the integrand changes from ## \frac{5x^{3/2}}{\sqrt{4 -x}}## to ##\frac{25x^3}{4 - x} ##.
I don't know if this is answerable in a way other than “you just can't” but, why can't I do that?

Ouch why would anyone ever make you solve this. You have to complete the square, then do a trig sub, followed by a bunch of manipulation of variables.
[Dark Knight movie quote]Some men just want to see the world burn.[/Dark Knight movie quote] :P

But, on a serious note, are you saying to make what's under the root be -(x-2)^2 + 4?
 
  • #5
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I don't know if this is answerable in a way other than “you just can't” but, why can't I do that?
Because, in general ##\int f(x)dx \neq \int [f(x)]^2 dx##. When you square the integrand, you get a different integrand.
But, on a serious note, are you saying to make what's under the root be -(x-2)^2 + 4?
That seems to be what iRaid was suggesting.
 
  • #6
s3a
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Because, in general ##\int f(x)dx \neq \int [f(x)]^2 dx##. When you square the integrand, you get a different integrand.
Oh, is it because squaring (or something similar) yields volumes?

That seems to be what iRaid was suggesting.
Alright. :) So, is my latest attachment correct so far?
 

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  • #7
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Oh, is it because squaring (or something similar) yields volumes?
No, nothing to do with that. If you square something, what you get is usually different. You wouldn't expect the integral to remain the same, would you?

Looking at it another way, do you think that d/dx(x2) = d/dx(x4)?
Alright. :) So, is my latest attachment correct so far?
You can check for yourself. If you differentiate your answer, you should get back your original integrand.
 
  • #8
s3a
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Looking at it another way, do you think that d/dx(x2) = d/dx(x4)?
No but, I think I see a better way of understanding it; 3/4 ≠ 3^2 / 4^2 = 9/16 so, integrating each yields a different answer (and, it's different from squaring both sides of an equation).

You can check for yourself. If you differentiate your answer, you should get back your original integrand.
I don't see how I can bring back the square root plus, Wolfram Alpha seems to imply that the two expressions are NOT equal … where have I gone wrong this time? It's probably something small like the squaring of the numerator and denominator from last time but, I just can't see what it is!
 
  • #9
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In your latest work, it looks like you just swapped dx for dθ to get the new integral. If that's what you did, that's where you went wrong. If 2sinθ = x - 2, then dx = 2cosθdθ.

Also, in you latest work, you have 5x3 in the numerator - it should be 5x2.

After completing the square, you should have this integral:
$$ \int \frac{5x^2 dx}{\sqrt{4 - (x - 2)^2}}$$

At this point I would bring the 5 outside the integral, and would do an ordinary substitution with u = x - 2. Then I would do the trig substitution, with 2sin(θ) = u and 2cos(θ)dθ = du.
 
Last edited:
  • #10
s3a
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I now see how to do that problem! Thank you (both) very much!
 

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