I want to understand the connection between the primitive function or antiderivative and the definite integral.
My problem with this is the independent variable called t in the formula for the first part of the Fundamental Theorem of Calculus.
Here's a composite of the answers I've already seen for this question. Because of t I don't understand it:
The primitive is a function $F(x)$ such that
$F′(x) = f(x)$
The first part of the FTC, which is the thing that connects differentiation/antidifferentiation to the definite integral, is this:
$$F(x) = \int_a^x f(t) dt $$
or
“$F(x)$ is a primitive function of $f(x)$. The lower bounds a is fixed. The upper bounds $x$ is variable.”
You can write $\int f(x) dx$
as
$$\int_a^x f(t) dt + C$$
A complete analysis of this problem is given in Richard Courant’s calculus book (linked below) page 109+.
Some of the sites I’ve seen:
http://ia700700.us.archive.org/34/items/DifferentialIntegralCalculusVolI/Courant-DifferentialIntegralCalculusVolI.pdf
http://www.ugrad.math.ubc.ca/coursedoc/math101/notes/integration/ftc.html
web.utk.edu/~wneilson/mathbook.pdf page 153-154
www.physicsforums.com/showthread.php?t=212449&highlight=x+dillema
www.jirka.org/diffyqs/htmlver/diffyqsse3.html
math.stackexchange.com/questions/105937/what-does-integration-do
www.math.hmc.edu/calculus/tutorials/fundamental_thm/
www.intuitive-calculus.com/fundamental-theorem-of-calculus.html
Best Answer
I'll try to be a succint as possible but answering your doubts as far as I can, but you made too many questions. Try to summarize as far as you can, adressing your main concerns that will clear the secondary ones.
The indefinite integral, in my opinion, should be called "primitive" to avoid confusions, as many people call it. The idea is that we learn how to find derivatives, and then are told: "Well, but what about the inverse problem: If we have a known function, what function should we differentiate to get it?" And here comes the idea of primitive of a function, or rather, primitives. The primitive of a given function, which we denote by
$$\int{ f(x) }dx = F(x)$$
is a function such that $F'(x) = f(x) \text{ ; } (1)$.
The notation was used by Leibniz to denote a function that satisfied $(1)$, using the arbitrary constant $C$ to denote that the function wasn't unique - rather, there was a family of primitives of a given function $f$, since the derivative of a constant is null. This is simple notation, but it has nothing to do with $\int_a^b f$ in the sense that $\int_a^b f$ is a number representing the limit of an integral sum of $f$ over $I = (a,b)$, and $\displaystyle \int f(x) dx +C $ is representing a function. As you say, $$\int{ f(x) }dx + C = F(x)$$ is not an integral sum, but a symbolysm for the function that satisfies $(1)$.
Here you're getting confused. The FTC states:
Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by
$$F(x) = \int_a^x f(t) dt $$
Then $$F'(x) = f(x)$$
for all $x$ in $[a, b]$.
This, in short, says that $F$ is indeed another primitive of $f$.
A consequence of this is the so called second FTC and a corollary. Since two primitives are only different by a constant, we should have
$$F(x) - \int_a^x f(t) dt = C$$
But then putting, $x = a$ gives
$$F(a) = C$$
which states that
$$F(x) - \int_a^x f(t) dt = F(a)$$
or
$$F(x) - F(a) = \int_a^x f(t) dt$$
Plugging in $b$ as the upper bound gives the famous corollary:
$$F(b) - F(a) = \int_a^b f(t) dt$$
which states that if $F$ is a primitive of $f$, the previous equality holds.
The second FTC says:
Let $f$ be a function defined on a closed interval $[a, b]$ that admits a primitive $F$ on $[a, b]$, i.e.:
$$f(x) = F'(x)$$
If $f$ is integrable on $[a, b]$ then
$$\int_a^b f(x) dx = F(b) - F(a)$$
(And this didn't depend on the continuity of $f$! You can try and plot the integrals dependeing on the upper bound of discountinous functions to see how the integral always "behaves" much better than the integrand.)
The connection between the primitives and the deinite integral is thus: If we know that a function admits a primitive in an interval, we can easily calculate it's definite integral by means of the primitive.