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Newton
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The integral of 1/(x^2) between -1 and 1. Although we know that the function is always of a positive value how do we obtain the negative area...

Newton.
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[email protected]
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#2
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(Original post by Newton)
The integral of 1/(x^2) between -1 and 1. Although we know that the function is always of a posiitive value how do we obtain the negative area...

Newton.
you can't do this because your integrating thru x=0 and f(x)=1/x^2 so you're dividing by 0
you know its symmetric about x=0
so you can let it equal
2(int x^-2) between 1 and 0 =2[-1/x] between 1 and 0
= 2 (-1 - -infinity....)
infinite area
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Newton
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(Original post by [email protected])
you can't do this because your integrating thru x=0 and f(x)=1/x^2 so you're dividing by 0
you know its symmetric about x=0
so you can let it equal
2(int x^-2) between 1 and 0 =2[-1/x] between 1 and 0
= 2 (-1 - -infinity....)
infinite area
You are just stating the obvious and accounting for the integral between 1 and 0 only, according to your explanation the area between 1 and 0=-1. Explain.

Newton.
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[email protected]
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(Original post by Newton)
You are just stating the obvious and accounting for the integral between 1 and 0 only, according to your explanation the area between 1 and 0=-1. Explain.

Newton.
why?
2 (-1 - -infinity....) = infinity
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danmint
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#5
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has someone been getting letters from ucl :rolleyes:
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MRLX69
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(Original post by danmint)
has someone been getting letters from ucl :rolleyes:
Why have you chosen this thread to post that irrelavent post? (Newton had his interview today)....... I have an offer, I'm sure he'll get one as well.
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idiopathic
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#7
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what ucl send these [email protected]
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danmint
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#8
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cos thats the same fallacy ucl sent me ... out of curiosity what course did you choose?
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MRLX69
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#9
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ah, ok. I've applied to 4 year Mathematics.
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Newton
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#10
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(Original post by danmint)
cos thats the same fallacy ucl sent me ... out of curiosity what course did you choose?
I applied for Mathematics and Physics.

Newton.
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RichE
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The only reason you get -2 is because you're applying the Fundamental Theorem of Calculus to a function to which the theorem doesn't apply. The Theorem applies to continuous functions which 1/x^2 certainly isn't.
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Ben.S.
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#12
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If you split it into two integrals, -1 to 0 and 0 to 1, then it's more obvious why it isn't valid.

Ben
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BCHL85
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Hmm, let's consider
INT of (1/x^2) = -1/x + C.
F(x) = -1/x
I = INT[-1, 1] (1/x^2) = INT[-1,0) (1/x^2) + INT(0, 1] (1/x^2)
I = [lim(x->0-)F(x) - F(-1)] + [F(1) - lim(x->0+)F(x)]
But lim (x->0-)(-1/x) = +infinite
lim(x->0+)(-1/x) = -infinite
So I = F(1) - F(-1) + (+infinite) = +infinite
That's why you obtained negative value for area cuz you forgot +infinite in this value
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dvs
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(Original post by RichE)
The only reason you get -2 is because you're applying the Fundamental Theorem of Calculus to a function to which the theorem doesn't apply. The Theorem applies to continuous functions which 1/x^2 certainly isn't.
RichE hit the nail on the head.
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El Stevo
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with questions like this, it also helps to draw the graph of what you are trying to integrate.
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BCHL85
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#16
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(Original post by dvs)
RichE hit the nail on the head.
Actually, some function can applied it.
Look at this INT[-1,1] of (1/cbrt(x)). Eventhough 1/cbrt(x) is not continuous function, but it still has INT from -1 to 1.
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El Stevo
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(Original post by BCHL85)
Actually, some function can applied it.
Look at this INT[-1,1] of (1/cbrt(x)). Eventhough 1/cbrt(x) is not continuous function, but it still has INT from -1 to 1.
no it doesn't.... asymptote at 0 again...

edit: you may get an 'answer', but its wrong...
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BCHL85
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(Original post by El Stevo)
no it doesn't.... asymptote at 0 again...
Oh... you've just changed.
with the function 1/x^2 , the asymptote is x = 0 as well. 0 is the singularity of these function.
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BCHL85
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#19
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(Original post by El Stevo)
no it doesn't.... asymptote at 0 again...

edit: you may get an 'answer', but its wrong...
I'm sure it's right. Because it's a converge integral, but in the case of 1/x^2, it's diverse integral, so that you won't have the answer.
As the definition of integral, at the singular point b of the function (here is 0)
INT[a, b) of f(x) = F(a) - lim(x->b)F(x)
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El Stevo
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hold on a bit... as x approaches 0 from the negative side, x^-(1/3) starts getting really negative and approaches - 'infinity' as x approaches 0 from the positive side, x^-(1/3) starts getting really huge and approaches +'infinity'

integrals including singularities can be done using limits, but integrals with asymptotes can't....
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