I know that a similar question exists at here but I need further explanation so with your excuse, I am asking:

Let $F(x)$ be antiderivative of $f(x)$. Now $F(c_1) - F(c_2)$ means "The area under $f(x)$ between $c_1$ and $c_2$". Since that's the case, I thought that: "Well, if that's the case, then $F(c_1)$ may be corresponding to "the area under $f(x)$ from minus infinity to $c_1$". Because when I consider it this way, $F(c_1) - F(c_2)$ becomes equal to: "The area under $f(x)$ between $c_1$ and $c_2$". But apparently this is not true. Because when I take a function like, say $y = 5$, then its antiderivative is $5x + C$ and this function at a given point has a finite value (if $C$ is constant), unlike the "area under the function" interpretation, which is infinite (area from minus infinity to a given point). So my question is: Do "value of an antiderivative at a given point" have a casual meaning like "value of an integral" has?

P.S: I would really appreciate if you could suggest me good further readings on this.

  • $\begingroup$ I'd imagine it simply means the value of the function at the point. $\endgroup$ – AvZ Mar 1 '15 at 10:35
  • 1
    $\begingroup$ Under any sensible definition of the integral, one must let $$\int_a^a f(t)dt=0$$ whenever $f$ is defined at $a$. After all, $F(a)-F(a)=0$. On the other hand, the claim that $F(c_1)$ is $$\int_{-\infty}^{c_1} f(t)dt$$ is true under the constraint that $\lim_{t\to -\infty}F(t)=0$. For a (counter)example, consider $$\int_{-\infty}^c \frac{1}{1+t^2}dt=\arctan c+\frac \pi 2$$ $\endgroup$ – Pedro Tamaroff Mar 1 '15 at 10:36
  • $\begingroup$ @AvZ Yes, but thing is, "The relation of F(x) with f(x)". In an integral with bounds, one can tell something about this relation but in an antiderivative, I am not sure what to tell. $\endgroup$ – Utku Mar 1 '15 at 10:37
  • $\begingroup$ @PedroTamaroff Thanks. Yes, I too discovered that defining F(c1) that way does not hold. But my question is: "Does F(c1) has a casual meaning like an integral with bounds has"? $\endgroup$ – Utku Mar 1 '15 at 10:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.