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gordysc
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1987 B1 Not making sense
« on: Oct 25th, 2005, 12:45pm » |
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Ok, a group of us here are working on B1 from 1987:
Integrate (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2))dx from 2 to 4. Now, the archives answer is to substitute x=6-t in and then add the integrals together to get 1. But, when we substitute x=6-t, we get -(ln(3+t)^1/2)/((ln(9-t)^1/2)+(ln(3+t)^1/2))dt
So how are we suppose to add these two different variables together? It just isn't clicking. Any feedback would be helpful and very much appreciated!
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towr
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Re: 1987 B1 Not making sense
« Reply #1 on: Oct 25th, 2005, 3:01pm » |
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INT 2 to 4 (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
== { sub x = 6-t }
INT 4 to 2 (ln(3+t)^1/2)/((ln(3+t)^1/2)+(ln(9-t)^1/2)) d (6-t)
==
INT 2 to 4 (ln(3+t)^1/2)/((ln(3+t)^1/2)+(ln(9-t)^1/2)) dt
== { sub t = x }
INT 2 to 4 (ln(3+x)^1/2)/((ln(3+x)^1/2)+(ln(9-x)^1/2)) dx
Add first and last, to get twice the value you want
INT 2 to 4 (ln(9-x)^1/2)/((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
+ INT 2 to 4 (ln(3+x)^1/2)/((ln(3+x)^1/2)+(ln(9-x)^1/2)) dt
== { move addition inside integration }
INT 2 to 4 (ln(9-x)^1/2 + (ln(3+x)^1/2) /((ln(9-x)^1/2)+(ln(3+x)^1/2)) dx
== {simplify}
INT 2 to 4 1 dx = 2
which is twice our integral at the start, which is thus 1.
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| « Last Edit: Oct 25th, 2005, 3:10pm by towr » |
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Luke Gordon
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Re: 1987 B1 Not making sense
« Reply #2 on: Oct 25th, 2005, 6:54pm » |
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Ok, this is where we get confused. We state that x=6-t, but then you substitute x back into the equation again....x!=6-x We understand everything else. It's just that x has already been defined as equaling 6-t...
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towr
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Re: 1987 B1 Not making sense
« Reply #3 on: Oct 26th, 2005, 1:07am » |
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It's not the same x. You could have done the substitution x -> 6-x, in one go. But that would be more confusing (and easier to make mistakes with).
The thing is, the x is a bounded variable, it doesn't exist outside the scope of integration. So it can't carry meaning over the equality signs either.
[sum]Nx=1 x = N (N+1)/2
{sub x -> y}
[sum]Ny=1 y = N (N+1)/2
{sub y -> z-20}
[sum]N+20z=21 (z-20) = N (N+1)/2
{sub z -> x}
[sum]N+20x=21 (x-20) = N (N+1)/2
{sub x -> x+20}
[sum]Nx=1 x = N (N+1)/2
The same thing keeps happening; 1+2+3+4+...+N And that's what's important, not the names of the variables involved (which is why you can substitute).
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Luke Gordon
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Re: 1987 B1 Not making sense
« Reply #4 on: Oct 26th, 2005, 11:01am » |
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Ok, that makes some sense. So then, we can say x=x^2 or anything we want to for a bounded variable?
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towr
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Re: 1987 B1 Not making sense
« Reply #5 on: Oct 26th, 2005, 1:51pm » |
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on Oct 26th, 2005, 11:01am, Luke Gordon wrote:| Ok, that makes some sense. So then, we can say x=x^2 or anything we want to for a bounded variable? |
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Yes, as long as both expressions are equivalent after substitution it doesn't matter what the substitution is (or what the names of the variables in it are).
It's really just to make an expression more readable, or easier to manipulate. And naturally you have to be mindful of additional constarints; if x is in the range [-1,1], you can't substitute x with x2, because it can't cover the same range.
I think the substitution should be reversable (for the range the variables are used in)
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| « Last Edit: Oct 26th, 2005, 1:53pm by towr » |
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Icarus
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Re: 1987 B1 Not making sense
« Reply #6 on: Oct 26th, 2005, 2:47pm » |
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The variable in a definite integral is a dummy variable. That means it is only there as a placeholder. The integral itself does not depend in any way on a value for this variable.
So [int]ab f(x) dx = [int]ab f(t) dt = [int]ab f(y) dy = ..., entirely independent of any other meanings given to the variables x, t, y, etc, in the calculation. The value of the integral depends only on the values of a and b, and the definition of the function f.
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towr
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Re: 1987 B1 Not making sense
« Reply #7 on: Oct 26th, 2005, 2:52pm » |
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On thing you do have to be carefull of is that the new variable wasn't already present in the expression being integrated.
[int]ab f(x,y) dx is not [int]ab f(y,y) dy
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