# Mathematical Fallacy

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15 years ago
#21
(Original post by 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....
Hmm, INT (1/cbrt(x) ) = 2/3.(x^2/3) + C
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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15 years ago
#22
umm... Int 1/cbrtx dx = Int x^-(1/3) dx = (3/2)x^(2/3) + c

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.
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15 years ago
#23
(Original post by El Stevo)
umm... Int 1/cbrtx dx = Int x^-(1/3) dx = (3/2)x^(2/3) + c

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. , sorry about my mistake
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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15 years ago
#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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15 years ago
#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.
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15 years ago
#26
(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.
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.
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15 years ago
#27
(Original post by 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!   0
15 years ago
#28
(Original post by 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.
Yap, maybe... I don't really know about The Fundamental Theorem of Calculus 0
15 years ago
#29
(Original post by BCHL85)
Yap, maybe... I don't really know about The Fundamental Theorem of Calculus The Fundamental Theorem of Calculus just essentially says that integration is the opposite of differentiation; namely it says

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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