TITLE: How is $\int f(x) dx$ different from $\int_0^x f(t)dt$?
QUESTION [6 upvotes]: 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?

REPLY [8 votes]: 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 $f(t) = 1/t$.