# Mathematical Fallacy Watch

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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.

Newton.

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#2

(Original post by

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.

**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 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

0

(Original post by

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

**[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

Newton.

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#4

(Original post by

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.

**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.

2 (-1 - -infinity....) = infinity

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#6

(Original post by

has someone been getting letters from ucl

**danmint**)has someone been getting letters from ucl

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(Original post by

cos thats the same fallacy ucl sent me ... out of curiosity what course did you choose?

**danmint**)cos thats the same fallacy ucl sent me ... out of curiosity what course did you choose?

Newton.

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#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.

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#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

Ben

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#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

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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#14

(Original post by

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**)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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#15

with questions like this, it also helps to draw the graph of what you are trying to integrate.

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#16

(Original post by

RichE hit the nail on the head.

**dvs**)RichE hit the nail on the head.

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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#17

(Original post by

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.

**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.

edit: you may get an 'answer', but its wrong...

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#18

(Original post by

no it doesn't.... asymptote at 0 again...

**El Stevo**)no it doesn't.... asymptote at 0 again...

with the function 1/x^2 , the asymptote is x = 0 as well. 0 is the singularity of these function.

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#19

(Original post by

no it doesn't.... asymptote at 0 again...

edit: you may get an 'answer', but its wrong...

**El Stevo**)no it doesn't.... asymptote at 0 again...

edit: you may get an 'answer', but its wrong...

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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#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....

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

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