6
$\begingroup$

This question already has an answer here:

Something niggling at me from way back. Is our definition of an antiderivative $\int f(x)dx = F(x)$ (such that $F'(x) = f(x)$) different in any way from the definite integral with variable limits, i.e. the function $f(x) = \int_0^x f(t)dt$ ?

It seems that I can think of them both as operators which take in a function and give back a function. Doesn't the Fundamental Theorem of Calculus then give us that $\frac{d}{dx}\int f(x)dx = f(x)$?

This came up because Wolfram says, of the FTC

$$\int_a^b = f(x)dx = F(a) - F(b)$$

that, "This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral."

I suppose my question is: why do we need these two distinct concepts, the algebraic and the geometric? Why can't we get by with only definite integrals?

$\endgroup$

marked as duplicate by Mankind, graydad, Strants, colormegone, Harish Chandra Rajpoot Sep 28 '15 at 0:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ $\displaystyle \int f(x)$ isn't proper notation. If you have a particular meaning of the symbol $\int$ in mind, you have to be precise. $\endgroup$ – GFauxPas Sep 27 '15 at 15:14
  • $\begingroup$ @GFauxPas: By $\int f(x)dx$ I mean "the function $F(x)$ such that $F'(x) = f(x)$". $\endgroup$ – Eli Rose Sep 27 '15 at 15:18
  • $\begingroup$ @EliRose : You write $\int f(x)$ instead of $\int f(x) dx$ in the question. Hence the confusion I suppose.. $\endgroup$ – user99914 Sep 27 '15 at 15:19
  • $\begingroup$ @JohnMa: You're right; edited. $\endgroup$ – Eli Rose Sep 27 '15 at 15:30
8
$\begingroup$

The function defined by

$$ F(x) = \int_0^x f(t)\,dt $$

is one antiderivative of $f$. There are infinitely many antiderivatives of $f$, and they are collectively represented by the symbol

$$ \int f(t)\,dt, $$

sometimes called the indefinite integral. This is why you always add "$+C$" to the end when evaluating the indefinite integral; each choice of $C$ gives a different antiderivative.

It should also be noted that sometimes $\int_0^x f(t)\,dt$ doesn't exist, for instance when $f(t) = 1/t$.

$\endgroup$
  • 1
    $\begingroup$ If anyone would like to add to this answer to improve it, I welcome it. $\endgroup$ – Antonio Vargas Sep 27 '15 at 15:21
  • 1
    $\begingroup$ Interesting ... so the first operator actually yields an equivalence class of functions, while the second gives a single function. Could we just define an equivalent class of operators $\int_C$ by $\int_C f(x) = \int_0^x f(t) dt + C$? $\endgroup$ – Eli Rose Sep 27 '15 at 15:35
  • $\begingroup$ With regards to your second point, you're saying that $\int_0^x f(t) dt$ may not be expressible in a closed form involving $x$, right? But certainly the function $f(x) = \int_0^x f(t)dt$ is well-defined and has a value for any value of $x$. $\endgroup$ – Eli Rose Sep 27 '15 at 15:37
  • 1
    $\begingroup$ @EliRose, Re your first comment: Basically yes, as long as $\int_0^x f(t)\,dt$ exists. Re your second comment: No, $\int_0^x f(t)\,dt$ simply doesn't exist in the particular case with $f(t) = 1/t$. It is not well-defined for any nonzero value of $x$ due to the nature of the singularity of $1/t$ at $t=0$. $\endgroup$ – Antonio Vargas Sep 27 '15 at 15:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.