What would you say is the integral of F(x) from a point A to A? (The same point). Before immediately saying zero, think about the definition of integral. Is 1/infinity the same as 0?
Thanks,
Lewkarg
2006-07-09 05:30:24 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Another way too look at it: In statistics, the area under a curve can represent the probability of an event, assuming, of course, that the total area is 1. Imagine f(x)=1 from x=0 to x=1. The probability of .5 to 1 can be found from integrating f(x) from .5 to 1. You will get 1/2, which is of course the probability. What about the probability of a point like .6 it is the integral of f(x) from .6 to .6. Think about it, there is infinite number of points from 0-1.
2006-07-09 05:46:07 · update #1
Not to be argumentative- it is I who asked the question- but I had a few thoughts. By saying the probability of choosing a single number on a continuous number distribution is 0 is saying u have NO chance of choosing that number. I can't accept this fact; I have to have some chance of choosing it. I understand it is, for all practical purposes it is extremely 0, but not actually 0.
2006-07-09 06:26:40 · update #2
Since there are several definitions of the integral, we will take it to be the area under a curve made by a function F.
If F does not go in a loop (for example a straight line), then the integral is indeed 0. This would normally be the case for functions on the Cartesian coordinate system based on x and y.
However, if the function loops around (like a circle), then the integral will not be 0, but the area of the enclosed space. In the Cartesian coordinate system, This can be done by taking the area under the top half of the curve, then the area the bottom half, then subtracting the areas. This can also be done (sometimes much easier) with other coordinate systems.
Finally, take this pattern:
1/2=0.5
1/3=0.3333
1/4=0.25
.
.
.
1/10=0.1
.
.
.
1/100=0.01
.
.
.
As the denominator gets larger, the value of the fraction becomes smaller, closer and closer to 0. So as we approach infinity (we cannot actually equal infinity) the number becomes very small. Basically, since infinity is a very large number that is undefined, 1/infinity =0
2006-07-09 05:55:43 · answer #1 · answered by dennis_d_wurm 4 · 1⤊ 1⤋
The integral of F(x) from point A to point B where A is not equal to B is the area under the curve bounded by A and B.
If F(x) were a flat line you can see that the area is simply the absolute product of the difference between A and B and the absolute height. A similar thing is true no matter what is the shape of F(x). If A equals B, then the difference is zero and the area would also be zero. So the integral of F(x) from A to A would be zero.
2006-07-09 12:42:37 · answer #2 · answered by Anonymous · 0⤊ 0⤋
As others have said before, it's zero. The probability of obtaining any one real number in a continuous distribution would be zero simply because any real number has an infinite numbers after the decimal point (for integers and numbers like a half, there'd be an infinite number of 0's after the decimal point). If you had, say, a uniform distribution from 0 to 1, the probably of obtaining exactly 0.5 (which is 0.50000....) would be 0.1for the 5, 0.1 for the first 0, 0.1 for the second 0, and so on. And 0.1^x as x -> infinity would be 0.
2006-07-09 13:08:13 · answer #3 · answered by Kyrix 6 · 0⤊ 0⤋
That integral will be zero. Remember the fundamental theorem of calculus:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
Particularly look at the last equation under the heading "Formal Statements" for the expression that should clear your doubts about the integral being zero.
Also, 1/infinity is zero for all practical purposes.
2006-07-09 12:37:59 · answer #4 · answered by anonymous 7 · 0⤊ 0⤋
are yu dealin with alph numberals ?if so that would definately be diferant than 0 but note the circle symbol from point a to a =an orbital circle but since F(x) is ur begining symbol it would be F(x) so yu get the whole effects,but in alpha numarics 6(24)=F(x) and if yu was to go though all the alpha numarals then you would return to the same yu would be on the 6th omaga=across and 24th alpha=down this is the a-z trograph multiples of 26 across and down but f to what power?cuz there are several F omagas
2006-07-09 13:05:24 · answer #5 · answered by freikeygee@sbcglobal.net 2 · 0⤊ 0⤋
Depends on what type of integral ...
If you take a contour integral in complex plane the answer can be different than 0.
2006-07-09 12:38:39 · answer #6 · answered by gjmb1960 7 · 0⤊ 0⤋
Yes, it is Zero. you are there where you were. No displacement. And 1/infinity is also nearer to Zero and all practical cases it is considered to be Zero
2006-07-22 04:08:12 · answer #7 · answered by sharanan 2 · 0⤊ 0⤋
STOP thinking in such roundabout ways. The answer to both the questions is Always zero.
2006-07-09 12:35:17 · answer #8 · answered by Anonymous · 0⤊ 0⤋
zero
2006-07-22 12:16:03 · answer #9 · answered by Anonymous · 0⤊ 0⤋