calculusintegration
The way I have seen the fundamental theorem of calculus stated is:
$$\int_a^bf(x)\ dx=F(b)-F(a)$$
Where $F(x)$ is any antiderivative of $f(x)$. Does the "any antiderivative" refer to the constant $C$ that is always added to the integral, or are there functions that have several completely different antiderivatives?
The definite integral $\int_a^b f(x) dx$ can be expressed, via antiderivatives and the fundamental theorem as \begin{equation} \int_a^b f(x) dx = G(b) - G(a). \end{equation} Here you have specified from where to where you want to integrate, making the integral definite. On the other hand, the indefinite integral is the collection of all these integrals, for all possible limit points. So once you make a choice about the integration limits, everything is perfectly well-defined and single-valued, but if you do not do that, you can only determine the (indefinite) integral up to a constant: $$ \int f(x) dx = G + C $$ Since no matter which number $C$ you plug into this equation, the derivative of the right-hand side will always be just $f(x)$. In this way, the indefinite integral is really a collection of functions: $$ \int f(x)dx = \{ G + C ~|~ C \in \mathbb{R}\}. $$ But for the matter of convenience, we leave away the brackets etc and write the integral as if it was a function with an additive, unspecified constant. So I think your misunderstanding lies within the distinction of indefinite (no limits) and definite (with limits) integrals.
Think of it like this:
A definite integral has the form $$ \int_a^b f(x) dx = G(b) - G(a), $$ where $a, b$ are fixed real numbers. The result will be a real number.
Now you allow one of them, say $b$, to be variable. This then gives you an integral FUNCTION of the form $$ I_a (x) = \int_a^x f(t) dt = G(x) - G(a). $$ Two words on notation: I changed the integration variable from $x$ to $t$ to avoid confusion as $x$ now denotes the upper limit of the integral. Furthermore, since the integral function depends on the lower limit $a$ (different $a$'s will give different functions), I stressed this dependence by writing $I_a$ for the function.
Now if you let $a$ remain unspecified, you will get a set of functions which is the indefinite integral. So it is a set as written above, and once we specify the lower bound $a$ and hence the constant $C = -G(a)$ we get one particular function $I_a$. If we then, finally, ask for the value of this function in a point $b$, we get the definite integral from $a$ to $b$ which is a number.
So: a definite integral is a number, an integral function is a function and an indefinite integral is a collection (or set) of functions.
So in the two sources that you provide, the authors each use a different definition for "indefinite integral". The first one uses the definition of step 3 above whereas the second one chose to stop at step 2.
Let me give you an example: Consider the function $$ f(x) = x^2 + 1. $$ An antiderivative is given by $$ G(x) = \frac{1}{3} x^{3} + x $$ but also for any $C \in \mathbb{R}$ by $$ G_{C}(x) = \frac{1}{3} x^{3} + x + C = G(x) + C. $$ Now to evaluate the integral with known bounds is relatively straightforward, you just have to plug the values of one of the antiderivatives into the formula (it doesn't matter which one since you subtract! So the extra constant $C$ will cancel in the evaluation). You could, for example, calculate $$ \int_{3}^{6} f(t) dt = G(6) - G(3) = G(6) - 12 = 75 - 12 = 63. $$ Now let us fix the lower bound, i.e. we let $a = 3$, and vary the upper bound, i.e. $b = x$. Then we will get a function $$ I_{3}(x) := \int_{3}^{x} f(t) dt = G(x) - G(3) = G(x) - 12 = \frac{1}{3} x^{3} + x - 12. $$ As you can see when looking at the right-hand side of the equation, this is a perfectly normal function and you can perform all kinds of calculations with it. This function can be thought of as a special case of the function $G_{C}$ defined above where $C = - 12$. If we now choose to let our integrals start at a different lower bound, say $a = 1$, we will get a related, but different function: $$ I_{1}(x) := \int_{1}^{x} f(t) dt = G(x) - G(1) = G(x) - (\frac{1}{3} + 1) = \frac{1}{3} x^{3} + x - \frac{4}{3}. $$ This corresponds to $G_{\frac{4}{3}}$ in the above notation. So you see that in this way you really get a collection of functions, via the process described in steps 1-3 above, but you can stop this process at any of the steps and it will be clear how to get to the previous/next step.
Hope that helps.
Andre
What you have written is true, but it is not the fundamental theorem of calculus. You've just rearranged the definition $$ F(x) = \int_c^x f(t) dt $$ using integral properties.
If we define $F(x) = \int_c^x f(t) dt$, then one result worthy of the name "fundamental theorem of calculus" says $$ F'(c) = f(c) $$ whenever $f$ is continuous at $c$. Note that we need $f$ to be continuous at $c$ or else the result can fail to be true.
Why is this true? You've made the key insight. $$ F'(c) = \lim_{h \to 0} \frac{F(c + h) - F(c)}{h} = \lim_{h \to 0} \frac{\int_c^{c + h} f(t) dt}{h} $$ To finish the proof, you just need to explain why that last limit is equal to $f(c)$. This is where you will have to use continuity at $c$.
The other result that goes by the name "fundamental theorem of calculus" says that if we have a function $F$ such that $F' = f$ and $f$ is integrable on $[a,b]$, then $$ \int_a^b f(t) dt = F(b) - F(a) $$ The classic proof uses a completely different trick. We use the mean value theorem with $F$ and $f$ and the integrability of $f$ to set up a telescoping sum.
People will often give a simpler proof which is more properly a corollary of the first fundamental theorem. If we assume $f$ is continuous on $[a,b]$, then we can argue as in RafGaming's answer to prove the same result. But note that the mean value theorem proof, though more complicated, proves the result without making any assumptions about the continuity of $f$.
Suppose $f$ is integrable on $[a,b]$ and $c \in (a,b)$
Schematically, in the first theorem, we have $$ \left[ F(x) = \int_c^x f(t) dt \quad \& \quad \lim_{x \to c} f(x) = f(c) \right] \implies F'(c) = f(c) $$ Note, again, that we need continuity of $f$ at $c$. The statement $F(x) = \int_c^x f(t) dt \implies F'(c) = f(c)$ is not true.
In the second theorem, we have $$ F' = f \quad \implies \int_a^b f(t) dt = F(b) - F(a) $$ Though similar looking, these results are not the same. In fact, they are not even exactly converses. They're two distinct results that both formalize the intuition that "differentiation undoes integration and vice versa."
Best Answer
Yes, your first hunch is right. "Any" antiderivative refers to the fact that an antiderivate is really a family of functions which differ only by a constant.