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Continuity in Integrals and Antiderivatives

  1. Dec 31, 2014 #1
    I was a bit confused by the definition of integrals (both definite and indefinite) and anti-derivatives. The definition for indefinite integrals is-

    The indefinite integral of a function x with respect to f(x) is another function g(x) whose derivative is f(x).
    i.e. g'(x) = f(x) ⇒ Indefinite Integral of f(x) is g(x)
    In mathematical notation we write- ∫f(x) dx = g(x) + C if and only if g'(x) = f(x) (as d/dx(g(x)+C) = f(x)

    And the definition of definite integral is-

    Let f(x) be continuous on [a,b]. If G(x) is continuous on [a,b] and G'(x)=f(x) for all x char32.png (a,b), then G is called an anti-derivative of f.

    We can construct anti-derivatives by integrating. The function

    F(x)= [itex] \int_a^x f(t) dt [/itex]
    is an anti-derivative for f since it can be shown that F(x) constructed in this way is continuous on [a,b] and F'(x) = f(x) for all x char32.png (a,b).
    My question is that, firstly, in the definition of an indefinite integral, how can we be sure that g(x) would have to be continuous ∀ x ∈ R? And in the definition of a definite integral, how can we be sure that g(x) would have to be continuous for x ∈ (a,b)?

    Just because g'(x) = f(x), is this enough to say that g(x) is a continuous function for ∀ x ∈ R and x ∈ (a,b) respectively in the 2 definitions? g'(x) exists means g(x) is a continuous function. But we don't know whether g'(x) exists ∀ x ∈ R. Similarly, [itex] \int_a^x f(t) dt [/itex] is integral in the interval (a,b). So how can we be certain that its anti-derivative would also be valid for only the interval (a,b)? Is this the definition of the indefinite integral that we can't question as to why would the -derivative be valid for only the interval (a,b)?

     
    Last edited: Dec 31, 2014
  2. jcsd
  3. Dec 31, 2014 #2
    Not sure if this answers the question. But in the hypotheses step. You are given that g is a differentiable function. If a function has a derivative at some point (the limit from the left and right exist and is defined at that point) Then the derivative of that function implies that it's anti derivative is also continuous. Note that the converse is not always true.

    What us also cool about this defintiom. Is that a bounded piece wise function that is continuous also follows the fundamental theorem of calculus.
     
  4. Dec 31, 2014 #3
    It actually makes more sense if you a problem from that section and compute it using the 1st fundamental theorem of calculus. Until I did it this way and analyze d my solution I finally understoodmit.
     
  5. Jan 1, 2015 #4
    Okay. I had a new question though. It is- If f(x) is continuous in (a,b) and g'(x) = f(x), doesn't this imply that g(x) is also continuous in (a,b) automatically?
     
  6. Jan 1, 2015 #5
    Yes, because g(x) even has a continuous derivative everywhere in (a,b).
     
    Last edited: Jan 1, 2015
  7. Jan 1, 2015 #6
    Infact, this can be modified a little bit to - If f(x) is defined in (a,b) and g'(x) = f(x), doesn't this imply that g(x) is also continuous in (a,b) automatically?

    Am I right?
     
  8. Jan 1, 2015 #7
    Go back to the section in your book on what it means for a function to be differentialable. Are you good at following proofs? To understand this concept you have to understand what a differential is, mean value theorem, etc.

    Yes a function has a derivative(implies that it is conti/defined) Then it's anti derivative (g (x)) is also continuous there. It is not true however to say if a function is continuous it has a derivative at that point. A classic example would be y=|x| it fails to have a derivative at x=0.
     
  9. Jan 1, 2015 #8
    Yup. So what I was asking was that for f(x), just being defined in a domain is enough and that f(x) need not be continuous in that domain for g(x) to be continuous, right? [Here f(x) = g'(x) just to remove confusions, if any]
     
  10. Jan 1, 2015 #9
    f(x) is automatically continuous on (a,b) if f'(x) exists everywhere on (a,b).

    You can think of it like this maybe: That ##f'(c)## exists means that there is a linear function that is a good approximation to f(x) around c. And linear functions always are continuous.

    Ah, you weren't asking that. Yes, the derivative g'(x) does not have to be continuous for g(x) to be continuous.
     
    Last edited: Jan 1, 2015
  11. Jan 1, 2015 #10
    Sorry, but I didn't understand what you meant by "good approximation to f(x) around c"
    Are you saying that if ##f'(c)## exists it means that a slope/tangent (i.e. a linear function) at x = c exists because of LHD (Left Hand Derivative)= RHD (Right Hand Derivative) at x = c? And so both of these limits cover many points around x = c and so the points covered around x = c are sufficient enough (good approximation) for f to be called continuous at c?
     
    Last edited: Jan 1, 2015
  12. Jan 1, 2015 #11
    If f(x) is differentiable at c, the linear function ##l(x)=f(c)+f'(c)\cdot (x-c)## is an approximation to f(x), but only useful if ##|(x-c)|## does not get too big.
     
  13. Jan 1, 2015 #12
    I just edited post #10 a moment ago. Please have a look at it again. :)
     
  14. Jan 1, 2015 #13
    That's good intuition, but I will try not to commit any more handwavery. :) The linear approximation given by the function ##l(x)## above does not have to actually cover many points, but the function values will gather very close to the line around ##x=c##. That's the way I think about it.
     
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