# Mathematical Fallacy

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#21

(Original post by

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....

**El Stevo**)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....

Let F(x) = 2/3(x^2/3)

INT[-1,1] f(x) = INT[-1,0)f(x) + INT(0,1]f(x)

= lim(x-> 0-) F(x) - F(-1) + F(1) - lim(x->0+)F(x)

and lim(x->0) F(x) = 0.

So I = F(1)-F(-1).

I'm sure it's right.

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#22

umm... Int 1/cbrtx dx = Int x^-(1/3) dx =

i agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.

**(3/2)**x^(2/3) + ci agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.

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#23

(Original post by

umm... Int 1/cbrtx dx = Int x^-(1/3) dx =

i agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.

**El Stevo**)umm... Int 1/cbrtx dx = Int x^-(1/3) dx =

**(3/2)**x^(2/3) + ci agree the area of -1 to 0 would be the same as 0 to 1, but you can't work out the area of 0 to 1 as it is a shape with infinite height, and as it never touches the y axis, it has some width. a shape with infinite height and some width has infinite area.

The area is not only at 1 point or 1 line x = 0. So if e is a very small value, you can assumes area is the integral from -1 to -e, and from e to 1.

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#24

i *thought* (my thoughts have been known to be inconsequential blabber before) that that was only the case 'continuous' line with a negligible break in it. as opposed to a line disapearing down and re-emerging from up top?

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#25

This similar to integral of the funtion but has only one boundary. Like this integral:

INT[0, +infinite) of (e^-x). The answer is 1.

They are all improper integrals.

INT[0, +infinite) of (e^-x). The answer is 1.

They are all improper integrals.

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#26

(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.

Generally the integral of x^a on (0,1) does exist improperly for a>-1.

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#27

(Original post by

The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

Generally the integral of x^a on (0,1) does exist improperly for a>-1.

**RichE**)The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

Generally the integral of x^a on (0,1) does exist improperly for a>-1.

Gawd, Riemann, I spit in his general direction! Such a Looooohooooserrrrr!

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#28

**RichE**)

The Fundamental Theorem of Calculus doesn't apply to x^(-1/3) on (0,1) say, though you're right to say that it does exist as an improper (Riemann) integral.

Generally the integral of x^a on (0,1) does exist improperly for a>-1.

I don't really know about The Fundamental Theorem of Calculus

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#29

(Original post by

Yap, maybe...

I don't really know about The Fundamental Theorem of Calculus

**BCHL85**)Yap, maybe...

I don't really know about The Fundamental Theorem of Calculus

Int[a->b] f'(x) dx = f(b)-f(a)

But you can't just apply it to any old function like

f(x) = -1/x and f'(x) = 1/x^2

- for example you usually insist f'(x) is continuous.

There are much simpler examples (rather than ones where the function goes off to infinity) where it doesn't work.

e.g. set

f(x) = 0 for 0=<x<1 and 1 for 1=<x=<2

Then f'(x) =0 except at x=1 (but f(x) hasn't zoomed off to infinity, just had a jump) and we see

0 = Int[0->2] f'(x) dx =/= f(2)-f(0) = 1.

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