Understanding the fundamental theorem of calculus 
I'm having problems understanding why $$ \frac{d}{dx} \int_{a}^{x} f(t)\,dt = f(x)$$
I'm somewhat weirded out by the fact that there is a $dt$ at the end of $$F(x) = \int_a^x f(t)\,dt$$ too.
We are differentiating with respect to $x$...I understand that $ \frac{d}{dx} \int f(x) \, dx = f(x)$ but why is the $t$ in the definition?
 A: http://en.wikipedia.org/wiki/Free_variables_and_bound_variables
The expression
$$
\sum_{i=1}^3 \cos(i^2 k^3)
$$
means
$$
\cos(1^2k^3) + \cos(2^2k^3)+\cos(3^2k^3),
$$
and that's the same as
$$
\sum_{j=1}^3 \cos(j^2 k^3),
$$
i.e. $i$ and $j$ are "bound variables", whereas $k$ is a "free variable".
The $t$ in
$$
\int_a^x f(t)\,dt
$$
is a bound variable, like $i$ and $j$ above, and $x$ is a free variable, like $k$ above.  The value of the sums above depends on $k$, but not on anything called $i$ or $j$, and similarly the value of the integral above depends on $x$, but not on anything called $t$.
A: You're encountering two standard abuses of notation in calculus:


*

*The integral sign $\int$ is used to denote both "definite" integrals (limits of Riemann sums) and "indefinite" integrals (a.k.a., antiderivatives, with $\int f(x)\, dx$ denoting an arbitrary function $F$ whose derivative is $f$).

*Mathematically, a function is an association of a unique output to each input. Calculus books often write $y = f(x)$ and speak of $f(x)$ as "the function". While this usage is compact and practically convenient, it's potentially misleading (and technically incorrect).
Concretely, suppose $f(x) = x^2$ for all real $x$. The "function" here (the object that can gets differentiated and integrated in calculus) is the squaring operator. The symbol $x$ is merely a dummy variable: In the absence of external restrictions on the meanings of symbols, the equations $f(t) = t^2$, $f(u) = u^2$, and $f(\xi) = \xi^2$ define exactly the same function.

The "indefinite" usage of the integral sign allows us to write, with $C$ denoting a real number,
$$
\int x^2\, dx = \frac{x^3}{3} + C.
$$
Note that "each side is a function of $x$". However, with equivalent meaning we could write: $\int t^2\, dt = \frac{1}{3}t^3 + C$ , $\int u^2\, du = \frac{1}{3}u^3 + C$, or $\int \xi^2\, d\xi = \frac{1}{3}\xi^3 + C$.
The "$dx$" plays a couple of roles, one being to specify the name of the input variable to $f$. Writing $\int x^2\, dx$ means "antidifferentiate the squaring function". Writing (say) $\int x^2\, dt$ is ambiguous; if $x$ denotes a real number, then $\int x^2\, dt = x^2t + C$, but if $x$ denotes a function of $t$ there is no general simplification. Writing $\int x^2$ (with no differential) is therefore a recipe for disaster.

The "definite" usage of the integral sign behaves a little differently. The result of integrating the squaring function over some interval $[a, x]$ can be written
$$
\int_a^x t^2\, dt = \frac{x^3}{3} - \frac{a^3}{3}.
$$
The integral on the left depends on $x$ (and on $a$), but $t$ makes no sense outside the integral, and can be replaced by any letter other than $x$ or $a$ without changing the left-hand side.
If we denote by $F$ the function defined by
$$
F(x) = \int_a^x t^2\, dt = \int_a^x f(t)\, dt,
$$
viewing $a$ as a fixed real number, then indeed $F'(x) = x^2 = f(x)$. Again, we wouldn't expect a $t$ to appear, since nothing in the preceding equation depends on $t$.

Trouble occurs when these notations and conventions about dummy variables get mixed or otherwise abused. The following are tempting to write, but likely to confuse (at best):
$$
\int_a^x f(x)\, dx;\qquad
\frac{d}{dx} \int_a^b f(x)\, dx;\qquad
\int f(x).
$$
A: The t means nothing. It is just a dummy variable that allows you to "see" how you are integrating buy it disappears in the end (when we plug in the bounds of integrstion). In the case of F(x), a is some fixed point but x is actually itself variable and is, in fact, the main (and only true) variable in the expression. Thus, we can differentiate with respect to it. The first integral produces F(x)-F(a), and when we differentiate that wrt x, we kill the subtrahend (since a is fixed and thus f(a) is constant too), and get F'(x) which is f(x) by definition. It is a bit tricky the first time.
A: Fix $a$. Then you can think of the function $F(x) = \int_a^x f(t)\;dt$ as a function which measures the area under $f$ from $a$ to $x$. The amount of area changes as $x$ changes. What FTC says is that the rate of change of this area at $x$, i.e. $\frac{d}{dx}F(x)$, is exactly $f(x)$.
As others have mentioned, the $t$ variable is a dummy variable. We could have instead written something like $F(t) = \int_a^t f(x)\;dx$, but then $F$ is a function of $t$ while our original $f$ is a function of $x$. We want the two to be functions of the same variable. (Much as how if we have $g$ as a function of $x$, then $g'$ is also a function of $x$.)
