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  #1  
Old 26-09-2014, 12:50 PM
Nortilus (Josh)
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Looking for Maths Genius for Uni help

Hi,
I really need some help with some math I'm having trouble with. PM me if you can and I'll send you an email address or facebook friends request. Getting desperate.
Thanks
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Old 26-09-2014, 01:12 PM
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Post it here - you'll be surprised.
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Old 26-09-2014, 01:16 PM
Nortilus (Josh)
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ok

ʃ{e^2x/5+e^2x}dx

so the (intergral of e to the power of 2x over 5 + e to the power of 2x) dx
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Old 26-09-2014, 01:21 PM
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GRRRR integration and log... not my area sorry!
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Old 26-09-2014, 01:30 PM
Nortilus (Josh)
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this is spose to be math B equivilent and know one knows it.
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Old 26-09-2014, 01:32 PM
julianh72 (Julian)
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Quote:
Originally Posted by Nortilus View Post
ok

ʃ{e^2x/5+e^2x}dx

so the (intergral of e to the power of 2x over 5 + e to the power of 2x) dx
There are two possible interpretations of your problem, depending on where the implied parentheses go:

Do you mean:

ʃ{(e^(2x))/5+e^(2x)}dx

or:

ʃ{e^(2x/5)+e^(2x)}dx

See attached - hope this helps!
Attached Files
File Type: pdf Integrals.pdf (6.5 KB, 40 views)
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Old 26-09-2014, 01:44 PM
Nortilus (Josh)
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written like this (see attached)
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  #8  
Old 26-09-2014, 01:45 PM
Nortilus (Josh)
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sorry that its sideways. straight from phone
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  #9  
Old 26-09-2014, 02:07 PM
julianh72 (Julian)
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Quote:
Originally Posted by Nortilus View Post
written like this (see attached)
There are THREE (or more!) possible interpretations of your problem, depending on where the implied parentheses go:

ʃ{(e^(2x))/(5+e^(2x))}dx = ln(e(2x) + 5) / 2

See attached!
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Click for full-size image (Integral.JPG)
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  #10  
Old 26-09-2014, 03:53 PM
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mithrandir (Andrew)
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Quote:
Originally Posted by julianh72 View Post
There are THREE (or more!) possible interpretations of your problem, depending on where the implied parentheses go:

ʃ{(e^(2x))/(5+e^(2x))}dx = ln(e(2x) + 5) / 2

See attached!
Julian you left out a "^" on the RHS.
Going by Josh's handwritten version that's what he meant. If you go to the LaTeX test site http://arachnoid.com/latex/ and paste in:

\int{\frac{e^{2x}}{{5+e^{2x}}}dx} \rightarrow \frac{\ln(5+e^{2x})}{2}

you'll get Julian's image - or something equivalent (commutative law).
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  #11  
Old 26-09-2014, 05:04 PM
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GeoffW1 (Geoff)
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Good stuff Julian. Takes me back to those late nights studying

Cheers
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