Why is this integral problem wrong?

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  • #1
i2c
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Homework Statement


[tex]\int\frac{3}{\sqrt{6x-x^2}}dx[/tex]

Homework Equations



NA

The Attempt at a Solution



http://img138.imageshack.us/img138/3744/001llo.jpg [Broken]


I know I should *should* have completed the square first, however I didn't see it at the time, so I tried this and got it wrong, no big deal, but I'm just wondering where I went wrong. Also, I got the 'calc answer' with my TI-89Titanium.
 
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Answers and Replies

  • #2
Tom Mattson
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I haven't checked all your steps, but are you really sure you got it wrong? You checked by equating the antiderivative from your calculator to your answer. But two antiderivatives don't have to be equal, they just have to be equivalent up to a constant (Hence the +C at the end of every antiderivative). The way to really check your answer is not to see if it's equal to some other antiderivative, but rather to take its derivative and verify that it equals the integrand.
 
  • #3
i2c
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I haven't checked all your steps, but are you really sure you got it wrong? You checked by equating the antiderivative from your calculator to your answer. But two antiderivatives don't have to be equal, they just have to be equivalent up to a constant (Hence the +C at the end of every antiderivative). The way to really check your answer is not to see if it's equal to some other antiderivative, but rather to take its derivative and verify that it equals the integrand.

Yes it is wrong. I copied and pasted both my answer and the calculator answer into Y= and graphed them, and it graphed 2 separate graphs which were not different by just a constant (i.e. a phase shift up or down)

Also, when I plug in my answer into my calculator and take the derivative of it, I get back


[tex]\frac{-sign(x-3)}{\sqrt{6x-x^2}}[/tex]
 
  • #4
699
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Yes it is wrong. I copied and pasted both my answer and the calculator answer into Y= and graphed them, and it graphed 2 separate graphs which were not different by just a constant (i.e. a phase shift up or down)

Also, when I plug in my answer into my calculator and take the derivative of it, I get back


[tex]\frac{-sign(x-3)}{\sqrt{6x-x^2}}[/tex]
I think Tom is basically correct. I think your method is right, but you are off by a factor of 3 in your answer. Also, your two answers are valid over different ranges (0<x<3 and 3<x<6). The antiderivative you calculate back is showing this with the sign function.

By the way, that was a creative way to find 3-x in the substitution. It creates a little issue as you see, but overall your method works. You just need to track down where you lost the factor of 3, and you need to identify where your two solutions are valid so that the derivative of your answer won't have the "sign" function in it.


EDIT: In case you are unclear on how to determine why your two solutions have limited range of validity, carefully check each step in the section where you generate the substitution for 3-x. At every step think about the range of validity, based on what is under the square roots.
 
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  • #5
i2c
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But when I graph my solution and the calculator solution I get 2 completely different graphs, not just a C difference, so where is my problem?
 
  • #6
699
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But when I graph my solution and the calculator solution I get 2 completely different graphs, not just a C difference, so where is my problem?
Your problem is that your solution is off by a factor of 3. Double check your work and you should find your mistake. Also, which solution did you plot? You have two solutions and each one is part of the answer. One answer is valid from 0<x<3 and the other from 3<x<6.
 
  • #7
i2c
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My answers graphed

http://img8.imageshack.us/img8/5345/screenshot000us.jpg [Broken]

'Correct' calculator answer

http://img153.imageshack.us/img153/3094/screenshot001ir.jpg [Broken]
 
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  • #8
699
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My answers graphed

http://img8.imageshack.us/img8/5345/screenshot000us.jpg [Broken]

'Correct' calculator answer

http://img153.imageshack.us/img153/3094/screenshot001ir.jpg [Broken]
Yes. Do you see how to make your solution equal the other one?

I do.

First, you need to identify which of your two solutions is valid over which region of x. You are then free to shift those solutions by a constant such that you match the other method.

As Tom pointed out, the solutions only need to agree to within an additative arbitrary constant.

Your method is perfectly good, but it turns out that your approach requires breaking the domain into two regions 0<x<3 and 3<x<6. You can see this if you carefully check each step in your 3-x substitution.

EDIT: By the way, it looks like you still have not corrected your factor of three error. Are you fully and carefully reading my posts?
 
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  • #9
i2c
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Yes I read your posts, but where did I go wrong? Where did I lose that factor of 3?
 
  • #10
699
6
Yes I read your posts, but where did I go wrong? Where did I lose that factor of 3?
Looks to me that the error is in the step that should read as follows.

[tex]3\int (3^2-u^2)^{-1/2} du=3 \; {\rm arcsin}(u/3)+C[/tex]
 
  • #11
i2c
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Oh right I threw a 1/3 in there by mistake because I was thinking of it like arctan(x), how you do 1/a in front, ok so now my final answer is as follows

[tex]3arcsin(\frac{\sqrt{6x-x^2}}{3})[/tex] However that is not correct, I can see if you split it between [0,3] and [3,6] and phase shifted each one up or down respectively by a constant, you would obtain a function equal to

[tex]3arcsin(\frac{\sqrt{3-x}}{3})[/tex]

But I don't see where it would split at 3, I don't see how you'd get 2 different domains
 
  • #12
699
6
my final answer is as follows
[tex]3 {\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})[/tex]

However that is not correct, I can see if you split it between [0,3] and [3,6] and phase shifted each one up or down respectively by a constant, you would obtain a function equal to

[tex]-3 {\rm arcsin} (\frac{{3-x}}{3})[/tex]

But I don't see where it would split at 3, I don't see how you'd get 2 different domains
OK, you are not following my explanation, although that is probably my fault for not being clear. You are correct to say that [tex]3{\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})[/tex] is not the correct answer, but this is not what I'm saying.

What I'm saying is that you determined two possible solutions [tex]3 {\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})[/tex] and [tex]-3 {\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})[/tex] , but you were not clear about what domains each of these solutions is valid over. One solution is valid in the domain 0<x<3 and the other in the domain 3<x<6. I'm not going to tell you which is which, because you can figure this out if you step through your derivation carefully. The domain limits need to be imposed in the section where you derive the substitution for 3-x. (EDIT: actually, you can also figure this out by comparing to the other solution, but that's cheating :smile:)

So, first you need to identify which of your solutions is valid in the domain 0<x<3, and which is valid in the domain 3<x<6. Then you are free to add a constant to one of the solutions so that the total solution is continuous at the boundary x=3. Once you do this, you will have a solution which differs from the solution, obtained by the other method, by a constant shift only.
 
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  • #13
699
6
I thought I might as well write out the full solution, in case what I'm saying is not clear. You have certainly done enough work and shown enough creative thought to have the final solution. Hopefully, I have not made a silly mistake, but double check the answer.

I believe your method will yield the following solution, if you try to match it up with the other solution.

[tex]3 {\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})-{{3\pi}\over{2}}[/tex] for 0<x<3

and

[tex]-3 {\rm arcsin}(\frac{\sqrt{6x-x^2}}{3})+{{3\pi}\over{2}}[/tex] for 3<x<6

Assuming I haven't made any errors, if you plot this solution out, you will see that it matches the other solution.

[tex]-3 {\rm arcsin} (\frac{{3-x}}{3})[/tex]

Note that each of your solutions (valid in different domains) needs a different constant added to it in order to maintain continuity at the boundary, and to match the other solution.

Now, if you really want to have some fun, prove this trig. identity! :smile:

Not too hard if you know the relation [tex] {\rm arcsin} a={\rm arccos}\sqrt{1-a^2} [/tex]
 
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  • #14
i2c
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Yes that makes perfect sense, I just didn't know what the constants were for each, but now it looks perfect.

But I still don't know how you figure out which domain it's valid over, I'll re-do the problem and play with it to see if I can find that out for myself. (other than just graphing and cheating)
 

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