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Integral of x^n ?

  1. Nov 11, 2011 #1
    can someone show me how to prove that ∫x^n is (x^n+1)/(n+1)?
    i know that derivative of this function gives us x^n....but is there some theoratical proof also?
    like there is a theoratical proof for derivative of x^n

    i have a second problem too.....if the integral of a function is 0 then does it necessarily mean that the function is 0?
    i read a note written in my class notes that it is not necessarily 0 ...but i can not remember the reason why it is not 0?
    i mean if the integral of a function is 0 then it makes perfect sense to say that the function is 0......if this is wrong then please give an example explaining how?
  2. jcsd
  3. Nov 11, 2011 #2


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    For n≠-1 there's no direct proof. As for the definite integral being 0, take any uneven power of x and integrate it between any symmetric interval wrt 0.
  4. Nov 11, 2011 #3


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    If a pretty theoretical proof that
    ∫x^n dx=(x^n+1)/(n+1)
    What kind of proof were you looking for?
    More constructive?
    Using sums?
    Using newton quotient?

    The integral of a continuous function is zero everywhere is and only if the function is zero. A function that is not continuous can have zero integral every where, and a continuous function can have zero integral some places without being zero.
  5. Nov 11, 2011 #4
    i was looking for a more theoratical proof ...like using maybe the binomial theorem or something

    what about a point function that is not 0. its integral is also 0 even if the function is not o itself.
    how can a continuous function have integral 0 without being equal to 0? are you indicating a function like 'cosine function' from 0 to pi/2?
    also how can a discontinuous function integral 0 everywhere?
    explain your answers please
  6. Nov 11, 2011 #5


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    You mean you want to expand (x+a)^n with the binomial theorem? Go ahead, but I do not see how that is more theoretical. What kind of integral is that? Is it an antiderivative or limit of a sum? Do you know the fundamental theorem of calculus? What about using induction?
    ∫x^n dx= (x^n+1)/(n+1)
    consider ∫x^(n+1) dx
    integrate by parts to reduce it to the given case

    The integral of a nonzero continuous function can be zero some places. Actually it is better to consider differences as integrals have arbitrary constants. Yes trig functions are good examples.

    The integral of a nonzero discontinuous function can be zero everywhere. These functions are called null functions and they are in some sense almost zero. A common example is
    f(x)=1 if 1/x is a positive integer
    f(x)=0 otherwise
  7. Nov 12, 2011 #6
    [tex]\int {x}^{n} \mbox{d}x=A{x}^{b}+C[/tex]

    [tex]\int {x}^{n} \mbox{d}x=A{x}^{b}+C=\frac{1}{n+1} {x}^{n+1}+C[/tex]
    [tex]\int {x}^{n} \mbox{d}x=A{x}^{b}+C=\frac{{x}^{n+1}}{n+1} +C[/tex]

    And the second, only the indefinite integral of 0 dx or 0 dt or 0 du etc can be 0. Maybe your notes are talking about the definite integral?

    But actually, only the indefinite integral of 0 d something can be 0 but the indefinite integral of 0 d something can actually be any constant.
  8. Nov 12, 2011 #7


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    The definition of "indefinite integral" or "anti-derivative" is that [itex]\int f(x)dx= F(x)[/itex] if and only if F'(x)= f(x). That is, the fact that the derivative of [itex]x^{n+1}/(n+1)[/itex] is [itex]x^n[/itex] is a perfectly good "theoretical" proof that it is the anti-derivative.
    If you want, instead, to show that the "limit of the Riemann sums" gives the same result, you could simply follow the standard proof, given in any Calculus text, that the limit is the integral, using [itex]x^n[/itex] as the function.
    Last edited by a moderator: Nov 14, 2011
  9. Nov 13, 2011 #8


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    Actually there is but it will take some work. I posted some notes on the learning materials page on integration, read that and you will have an idea how you can generalise for the case where [itex]n[/itex] is a natural number. You will need to know how to sum a series though but this can be found on wikipedia.

    The usual way is via the fundamental theorem of calculus.
  10. Nov 13, 2011 #9


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    ^Again khurram usman has offered not definition of "more theoretical" nor described the type of proof desired in this case. It is not even clear if that integral is to be interpreted as a limit of a sum or an antiderivative. There is no reason to that any of the obvious proofs are any better than the others in general.
  11. Nov 15, 2011 #10


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    Perfectly good and this is the way most of us learnt, but would you not agree that it is excellent thinking by the student to want to find another perfectly good proof? At the end they might be shown to be equivalent. As a habit it will surely benefit a sense of mathematical structure? (Personally I find myself wanting to not just get a result or proof but to get it in the way I want to get it, perhaps from within just one particular set of ideas or assumptions, I want to be sure these are sufficient or the theory can be developed with them.)

    And you could argue that the integral is a more elementary thing than differentiation, it is the idea of area. Archimedes came before Newton. ∫x dx is the area of a triangle. Done quite early at school. Then ∫x2 dx can be done by parts as shown above, then you can get ∫x3 dx in the same way and all ∫xn dx. So actually you could do some integral calculus without tears quite early at school.
    Last edited: Nov 15, 2011
  12. Nov 15, 2011 #11

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    An alternative to the perfectly valid (and simple) solution provided by differentiation is to use induction and integration by parts.

    For n=0, [itex]\int x^ndx = \int 1\,dx = x[/itex]. This gives us the base case. Assume [itex]\int x^ndx = \frac{x^{n+1}}{n+1}[/itex] for some n≥0. What is [itex]\int x^{n+1}dx[/itex]? If the hypothesis is true, this should be [itex]\frac{x^{n+2}}{n+2}[/itex].

    Writing [itex]\int x^{n+1}dx = \int xx^n dx[/itex] and integrating by parts yields [itex]\int x^{n+1}dx = \frac 1 {n+1} \left(x^{n+2} - \int x^{n+1}\,dx\right)[/itex]. Solving for [itex]\int x^{n+1}dx[/itex] yields [itex]\int x^{n+1}dx = \frac{x^{n+2}}{n+2}[/itex]. By induction, this holds for all non-negative integers n.

    Note that this proof only holds for the natural numbers. Generalizing to negative integers, rationals, and reals is an exercise left to the reader.

    Or you could just use the perfectly valid (and simple) solution provided by differentiation.
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