Mathematical FallacyWatch

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Thread starter 14 years ago
#1
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.
0
14 years ago
#2
(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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Thread starter 14 years ago
#3
(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.
0
14 years ago
#4
(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
0
14 years ago
#5
has someone been getting letters from ucl 0
14 years ago
#6
(Original post by danmint)
has someone been getting letters from ucl 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.
0
14 years ago
#7
what ucl send these [email protected]
0
14 years ago
#8
cos thats the same fallacy ucl sent me ... out of curiosity what course did you choose?
0
14 years ago
#9
ah, ok. I've applied to 4 year Mathematics.
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Thread starter 14 years ago
#10
(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.
0
14 years ago
#11
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.
0
14 years ago
#12
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
0
14 years ago
#13
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
0
14 years ago
#14
(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.
0
14 years ago
#15
with questions like this, it also helps to draw the graph of what you are trying to integrate.
0
14 years ago
#16
(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.
0
14 years ago
#17
(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...
0
14 years ago
#18
(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.
0
14 years ago
#19
(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)
0
14 years ago
#20
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....
0
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