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Basic question, integral at a specific point

  1. Jul 31, 2012 #1
    1. The problem statement, all variables and given/known data

    This isn't a specific problem ;)

    2. Relevant equations
    3. The attempt at a solution

    My thought process, skip to the last paragraph if you want the question:

    I was wondering how you compute an integral at a specific x value, lets say c. Based on the fact that the integral is a function, my very first thought was that I should treat it like any other function and just "plug c into" the integral like -

    If [itex]F(x)[/itex] is the integral and [itex]f(x)[/itex] is the function you're integrating,

    [tex]\int{f(c)}=F(c)[/tex]

    Then I thought that you can't treat the integral like any other function because [itex]F(c)[/itex] actually represents the sum of the values of [itex]f(x)[/itex] from x=0 to x=c, and therefore the proper way to do this would be to evaluate a definite integral from 0 to c. I don't know if there's a flaw in my thinking so I've included it.

    -------

    Question: If I want to compute an integral at x=c, should I calculate the definite integral from x=0 to x=c?
    Example:

    [tex]\int_{0}^{c} cos(x) dx = \left[ sin(x) \right]_{0}^{c} = sin(c) -sin(0) = sin(c)[/tex]

    I have this follow-up question, but first I want to see if this is right.

    Thanks! (And if there's a better way to use LaTeX, please tell me, I'm clueless :D)
    Similarly- sorry if the way I'm expressing the math is wrong, if so please correct me.
     
  2. jcsd
  3. Jul 31, 2012 #2

    chiro

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    Hey Flumpster and welcome to the forums.

    The way to think about the indefinite-integral relates to having some initial condition at some particular value of x and then using this to get the final description of the function F(x).

    If your initial condition corresponds to x=0 then the description you have given is correct: if it's some other initial condition at say x = a onwards and f(x) is correct, then yout limits will start from x=a to some x=c.

    In short it depends on the context of your function, the initial condition, and also the context of the problem you are solving, and how the initial value plays into the definition of the integral.
     
  4. Jul 31, 2012 #3
    Thanks :)
    That's interesting chiro - in what kind of problems would you use a non-zero initial condition?

    This sort of leads me to my second question:

    What I want to evaluate the integral of cos(x) at x=0. I'd have to set up a definite integral

    [tex]\int_{a}^{b} cos(x) dx[/tex]

    where both [tex]a[/tex] and [tex]b[/tex] = 0. Can both limits equal each other/ 0?
     
  5. Jul 31, 2012 #4

    chiro

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    Yes they can and the integral will always equal 0 even if a and b don't equal 0: as long as they are equal and the integral is defined at a and b, it will always be zero.

    The way to think about this is that the rectangle has zero width if you want to use the Riemann-Integral analogy for integration: in the Riemann integration explanation, the rectangles widths approach zero but never get there, but when your limits are the same, the width is zero (not just approaching zero) which means you get 0*whatever = 0.

    This is the intuitive explanation but there are other ways of showing it that involve theorems of integration, the theory of the real numbers and so on, but all this stuff is a lot harder to explain than the idea that our rectangle width (or the length of our interval) is just zero.
     
  6. Jul 31, 2012 #5

    HallsofIvy

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    The "indefinite integral" or "anti-derivative", [itex]\int f(x)dx[/itex], always has an undetermined constant, so is a family of functions rather than a single function. The "definite integral", [itex]\int_a^b f(x)dx[/itex] is a number, not a function, and cannot be evaluated at a specific x.

    It is common to "choose" one of the infinite number of functions defined by the anti-derivative by fixing the lower bound: [itex]\int_a^x f(t)dt[/itex]. The choice of a determines the constant in the indefinite integral so this is really the same as choosing a value for that constant.
     
  7. Jul 31, 2012 #6
    HallsofIvy, could you explain a bit about the undetermined constant? I haven't been taught this rigorously. Why is there always a constant?

    Does what chiro talked about imply

    [tex]\int_{b}^{c} f(x) = \int_{0}^{c} f(x) - \int_{0}^{b} f(x)[/tex] ?

    Sorry for asking question after question, thinking about this makes me, well, think about it more. :)
     
  8. Jul 31, 2012 #7
    Just to expand on this.

    The definite integral is a function of the integration limits, let's call it F(a, b). It is closely related to the family of the functions produced by the indefinite integral. Specifically, we can choose one of them, say G(x), and then F(a, b) = G(b) - G(a). F(a, b) is independent of the choice of G(x) because any other function from the family, say E(x), is just G(x) + C, so E(b) - E(a) = G(b) + C - G(a) - C = F(a, b).

    By denoting "b" as "x" to emphasize that it is a variable and fixing "a" at a particular value we obtain F(x) = F(a, x).
     
  9. Aug 1, 2012 #8
    I appreciate your replies - and the bits I understand have been really helpful - but I haven't learned this rigorously enough to understand other parts.

    I obviously need to find a good text on the subject because there are simply too many issues here that I haven't learned (I'm in a high-school program which doesn't really enter the theoretical / pure side of calculus, and I want to get a rigorous introduction to it before next year). I didn't realize my question would involve so many issues.

    Anyway, thanks for you replies!
     
  10. Aug 1, 2012 #9

    HallsofIvy

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    Surely, one of the first things you learned in Calculus is that the derivative of a constant is 0? And because of that, if the derivative of F(x) is f(x), then the derivative of F(x) plus any constant is f(x)+ 0= f(x). So the anti-derivative of f(x) is any of those functions: F(x)+ C where C is any number.
    If, on a test, you were asked to find [itex]\int 3x^2 dx[/itex], would your answer be [itex]x^3[/itex] or [itex]x^3+ C[/itex]? If you did not answer the latter, you should be marked wrong.

    Yes, in fact that is true with any number replacing 0:
    [tex]\int_b^c f(x)dx= \int_b^a f(x)dx+ \int_a^c f(x)dx= \int_a^c f(x)dx- \int_a^b f(x)dx[/tex]
    (I notice you did not write the "dx" in your integrals. That is bad notation. Always write the "dx".)

    OOh, that's a bad habit! It will never let you rest!:tongue:
     
  11. Aug 1, 2012 #10
    This I did learn.

    I was never explained this, nor was I *formally* explained the definition of the integral as an anti-derivative (so you can see how I don't understand this, for example:

    )

    I know I have huge gaps in my knowledge. Basically, I don't know how it works in the states, but where I am you're either learning math at the first, second or third level (third being most advanced math). I really hated math so I was in the first level. Now (with a new teacher) I love it, and have learned all the stuff on the first level and most on the second level. But for two reasons I don't have optimal access to calculus material: 1) my school is a special-needs school and doesn't really cater for third-level students (who learn calculus), and 2) even though I've done the equivalent of 3 or 4 years of work in one year, the school so far refuses to bump me up a level (there are signs this will change next year).

    Anywayyy, I know how to do Riemann sums, but don't know basic definitions, I can differentiate polynomials and irrational functions, but can't do epsilon-delta proofs, can evaluate definite integrals but don't know basic integration theorems.
     
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