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Mathematics-Test

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TN1

posted on Jul, 22 2005 @ 09:51 AM
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Dear Friends,

For those who are interested in mathematics here comes a nice question, not that difficult, although it involves integration.

Prove that the Integral from 0 to pi/2 of the function

F(x) = (e^2x) sinx is equal to (1 + 2e^pi)/ 5

Please give it a try

I am waiting for your replies and methods

TN1




posted on Jul, 22 2005 @ 11:02 AM
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WTF!!!
I have no idea!!
Guess I'm just dumb.



posted on Jul, 22 2005 @ 11:45 AM
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Hi,

No idea where you are going with this, must be just a fun time waster.


I originally by looking at it thought it would involve Euler's formula, but I think I was off track. I believe the correct solution involves integration by parts... Done twice.

It is a long detail, anyone can look up the formula for integration by parts, but start off by using e^2x as your u', and sin(x) as the v. Perform the calculations. You end up with uv - integral of uv'.

Perform the integration by parts on the second term.... and use artihmetic, you will come up with the following equality for the left term in your original equation.

e^2x/3 [2sin(x)] - cox(x)] evaluated at pi/2 and 0.

I come up with the ultimate answer of (2e^pi + 1) / 3.

Tell me if I have gone wrong, or if you are just testing. I am off by a small factor from your answer.

I can also detail the proof, maybe in a web link. It might be better to do it this way, to give someone else a hint on the solution.

TheMesh

[edit on 22-7-2005 by TheMesh]



posted on Jul, 22 2005 @ 12:13 PM
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Please. You're trying to get people to do your homework. Go over here: www.physicsforums.com...

EDIT - Your answer is correct however, so why would want to prove it? YOu already did the steps inbetween to arrive at that answer.... right right right?


EDIT- MESH your are wrong. Check your integration. You should have 1 + 4, ulitiamtely, at the bottom.

[edit on 22-7-2005 by ktprktpr]

[edit on 22-7-2005 by ktprktpr]



posted on Jul, 22 2005 @ 12:21 PM
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Hah, ktprktpr

You are probably right. Maybe it was a take home exam, with multiple choice answers. Since I got it wrong likely, he will be in Calc II again next semester.


In any case, haven't done any calculus in a long time, was a fun exercise. Kind of forgot about that stuff, thank God.

EDIT: Here is the answer.

img14.imgspot.com...

TheMesh

[edit on 22-7-2005 by TheMesh]



posted on Jul, 22 2005 @ 01:17 PM
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IBP...not that complicated...

Wish that was as complicated as calculus gets


If only you had my Complex Variables professor...if only...



posted on Jul, 22 2005 @ 03:48 PM
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Im very confused with these types of mathimatics. Really never was good at it. Teachers told me all the time that I was going to use this in my career. Till this day, I never had. !!!!!



posted on Jul, 22 2005 @ 06:30 PM
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evanfitz

Ya, these guys here saying it's not difficult, WHAT!!!

I guess if you're not solving it. Looked pretty hard to me. Probably only math majors or engineering/physics kinds of people would know WTF to do with it.... I would imagine if it was tossed on a calc exam, maybe 10-20% of the class would get it right at a decent university. Who knows, maybe I'm out of touch, its been 20 years since I took calc II. I sure didn't know what to do with it at first glance, had to dig out a book.

So what was it TN1, a homework problem? And if it's not that difficult, why even post it here, you know we only have the time to mess with the hard stuff, not this trivial algebra stuff!!!


TheMesh


[edit on 22-7-2005 by TheMesh]



posted on Jul, 22 2005 @ 06:49 PM
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Originally posted by evanfitz
Im very confused with these types of mathimatics. Really never was good at it. Teachers told me all the time that I was going to use this in my career. Till this day, I never had. !!!!!



nice message
heh

Down with math !



posted on Jul, 23 2005 @ 07:36 PM
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Well, it's not hard for me considering I've taken many math & physics courses. It's a pretty basic integration for anyone who has taken Calculus II (maybe even I).


TN1

posted on Jul, 25 2005 @ 06:12 AM
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Dear Friends,

Thanks for your replies.

The Function could be integrated if we use the method of integration by parts and also if we consider the following function:

g(x) = (e^2x) cosx, Integrated between the same limits, i.e from 0 to pi/2

If you name the first Integral S and the second Integral C then you end up with the following equation:

5S = 1 + 2 e^pi, and this gives you the correct result.



posted on Aug, 30 2005 @ 06:42 PM
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it's not that difficult.

In France, a 18 years old boy/girl studying science knows how to do :

sin(x)*e^2x is the imaginary part of e^x[2+i]


so you just do the maths : integrale of e^x[2+i] between 0 and pi/2
and then you take the imaginary part of this



posted on Sep, 2 2005 @ 03:07 PM
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it's not that difficult.


[edit] Never mind. I was gonna make a remark. Not gonna go there.


Prolly need to move this post over with the other one. Just a snipers paradise here now.

TheMesh



[edit on 2-9-2005 by TheMesh]



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