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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?


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

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  • $\begingroup$ $\displaystyle \int f(x)$ 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

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 $f(t) = 1/t$.

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    $\begingroup$ If anyone would like to add to this answer to improve it, I welcome it. $\endgroup$ – Antonio Vargas Sep 27 '15 at 15:21
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    $\begingroup$ 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
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    $\begingroup$ @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

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