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Integral of 0

  1. Aug 10, 2005 #1
    first the integral of 0 should be a constant rather any constant,which makes the integral of 0 to be 1=2=3=4..........
    does this equate all numbers or am i gettin this mixed up.
    well that is not want i intend sending,find the derivative of 3^3^3^3^3^3....................^x,and then tell me what the integral of 0 really is.I rest my case
  2. jcsd
  3. Aug 10, 2005 #2
    You're not making sense. The antiderivative of 0 is C, that is because information is lost when deriving. The derivative of any constant is 0. Where does "1=2=3=4" come from?

    What's the difference between "a constant" and "any constant"?

    Also, the definate integral of 0 on any interval is 0.
    Last edited by a moderator: Aug 10, 2005
  4. Aug 10, 2005 #3
    If you have an integral of zero, and you evaluate it, you will get 0+c. C is "a constant" But you cant just pick and choose a value for c. In order to find the value of c, you need the value at a point on the function. This will determine the PARTICULAR value of c you have to use. So its not literrally "any constant", it depends on what value the curve has. 1=2=3=4 does not happen, I am not sure how you came to that conclusion, but I think it was via an error in an assumption along the way.
  5. Aug 10, 2005 #4


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    Maybe it's time mathematicians start putting up massive billboards:

  6. Aug 10, 2005 #5


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    I think I saw one of those on I95!
  7. Aug 10, 2005 #6


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    you guys do not seem to realize that the word "integral" does NOT mean antiderivative. the integral of zero, over any interval at all, is definitely just zero.
  8. Aug 10, 2005 #7
    Yes! Exactly! And with this we are free to muse about the integral on its own terms -- it is its own thing. Which allosws us to ask the question: what is zero growing into as we add zero to it?

    Such a question is a geometric-like way of thinking of the integral. And, as such, no one needs to think about antiderivatives to realize that such a thing is "definitely just zero."
  9. Aug 10, 2005 #8
    What ever happened to the constant of integration?
  10. Aug 10, 2005 #9
    I cant read your text because its not long enough rach, sorry. I always thought evaluating the integral was to do the antiderivative to it. Could you explain the difference please.
    Last edited: Aug 10, 2005
  11. Aug 11, 2005 #10


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    The integral evaluated from a to b is 0, but the improper integral, i.e. the anti-derivative, is C.
  12. Aug 11, 2005 #11
    Specifically, the FTC tells us that [itex]\int_a^b 0 dx= F(b)-F(a)=C-C=0[/itex]. An antiderivative is just a function, but an integral is a number.
  13. Aug 11, 2005 #12
    You don't even need the fundamental theorem of calculus. You can start from the Riemann-Steltjes definition of the integral and prove that any finite sum of zeros is zero.
  14. Aug 11, 2005 #13


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    Of course, but it's nice to see how the two "views" concord with each other. I.e. how the FTC "makes it work". :smile:
  15. Aug 11, 2005 #14
    Hey, I was just trying to make things easier to understand. If you're looking for the most general result, then might as well point out that a zero function (real or complex) has Lebesgue integral zero over any measurable set in any measure space.
  16. Aug 11, 2005 #15
    Ah, yes of course. Thanks rachmaninoff. From the wording of the origional question, I thought it was in refrence to doing the antiderivative.
  17. Aug 12, 2005 #16
    so what does the derivative of the function 3^3^3^3^3^3^3^3^...........................^x gives us
  18. Aug 12, 2005 #17
    [tex]\frac{d}{dx} 3^x = 3^x\ln 3[/tex]

    [tex]\frac{d}{dx} 3^{3^x} = 3^x \ln 3 * 3^{3^x} \ln 3[/tex]

    You can do the rest. However, I don't see how this has anything to do with the integral of zero or how 1=2=3=4.
    Last edited by a moderator: Aug 13, 2005
  19. Aug 12, 2005 #18
    Let the function [itex]T(a) = 3^a[/itex]. The derivative of this function with respect to a is [itex]3^a \ln(3)[/itex]. Thus you have the function T(T(...(T(x))...)) where T is composed n times with itself. Let [itex]3_i(x)[/itex] be the power tower of 3 to order i where the ith position is replaced with the variable x, and [itex]3_i[/itex] be the power tower of order i. The derivative of the given function is then [tex]\frac{d}{dx}T(T(...(T(x))...)) = T'(T(...(T(x))...))*T'(...(T(x))...)*...*T'(x)[/tex]
    [tex] = \prod_{i=1}^n \ln(3_{i-1})*3_i(x)[/tex]
    Nasty looking thing. :yuck:
    Last edited: Aug 12, 2005
  20. Aug 13, 2005 #19
    i donot understand,ice breaker i do not buy your idea
  21. Aug 13, 2005 #20


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    Do you mean (3^3^3^3...^3)^x or 3^(3^(^3(....3^(x)))))))... . If you mean the first case then this is just(3^3^3^3^3...^3)^x = e^(ln((3^3^3^3^3....)^x))= e^(x*ln(3^3^3^3^3...^3)) and the derivative is therefore ln(3^3^3^3^3...)e^(ln((3^3^3^3^3....)^x)) = ln(3^3^3^3^3...)(3^3^3^3...^3)^x. In the second case it is like hypermorphism suggested.
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