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Integration problem

  • Thread starter UNIQNESS
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  • #1
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http://img.photobucket.com/albums/v295/uniqness/math.jpg [Broken]

I'm having a hard time trying to figure out how the last two steps were taken to obtain the answer. I actually have an exam in couple hours and I would really appreciate if someone could explain carefully as soon as possible. Thanks in advance.
 
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Answers and Replies

  • #2
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In the first line, the x integration is being done first. The whole square root bit is a function of z and not x, so you just treat it like a constant. So, when you do the x integral, all you get is

[tex]x\sqrt{17 + 4z^2}[/tex]

Next you have to put in the x limits: 0-1. Clearly, putting in x = 0 will give you nothing and putting in x = 1 gives you

[tex]\sqrt{17 + 4z^2}[/tex]

This leaves you with the z integral at the end of the first line.

As for the second line, I'm a bit stuck right now :/.
 
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  • #3
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lol thanks for the reply but the second line is what i'm having a problem with..
 
  • #4
arildno
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The logarithmic expression is nothing else than the cosh or sinh inverse 8don't remember which)
In order to crack the integral, remember that [tex]Cosh^{2}y=1+Sinh^{2}y[/tex]
Hence, make the change of varibles: [tex]z=a*Sinh(y)[/tex], where "a" is a constant to be fitted in the nicest manner possible.
 
  • #5
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I still don't understand how the first line became the second line.. Also, how would i use the expression Cosh^2y = 1 + Sinh^2y there? I'd really appreciate if you could tell me step by step.. I just need to know how the first line became the second line.. This is not a homework problem as you can tell the answer is already there.. I'm just studying for a test.
 
  • #6
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I know it's very early but is there anyone who can help me?
 
  • #7
arildno
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OK, let's rewrite the integral in the first line as:
[tex]\int_{0}^{1}\sqrt{17+4z^{2}}dz=\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{\sqrt{17}})^{2}}dz[/tex]
Thus, the change of variables [tex]\frac{2z}{\sqrt{17}}=Sinh(y)[/tex] is rather natural..
 
  • #8
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Thank you for the help.. I guess I just don't remember the properties of sinh & cosh.. I thought there was an easier way to do it because I never used cosh or sinh to solve an integral.. I guess I will get this problem wrong if it comes out on the test.. Thanks though.
 
  • #9
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my math exam is over so i don't need to know how to solve it any more but can anybody solve this thing step by step? when i usually post a problem, many people respond, but it seems like not many people know how to solve this..
 
  • #10
arildno
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Well, you now have [tex]z=\frac{\sqrt{17}}{2}Sinh(y)[/tex], which implies that:
[tex]dz=\frac{\sqrt{17}}{2}Cosh(y)dy[/tex]
Along with the new limits [tex]0\leq{y}\leq{Sinh}^{-1}(\frac{2}{\sqrt{17}})[/tex]
Therefore, we get:
[tex]\sqrt{17}\int_{0}^{1}\sqrt{1+(\frac{2z}{17})^{2}}dz=\frac{17}{2}\int_{0}^{{Sinh}^{-1}(\frac{2}{\sqrt{17}})}Cosh^{2}ydy[/tex]

All right?
 
  • #11
HallsofIvy
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C'mon- in the very first post, Nylex told you how to go from the first line to the second- how the integration was done- and you replied "lol thanks for the reply but the second line is what i'm having a problem with..".
When it was pointed out that the second line was just a matter of evaluating functions you said "I still don't understand how the first line became the second line..", so we're back to what was said in the first post!
 

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