Antiderivative and definite integral If $f$ is a continuous, real-valued function on interval $[a,b]$, then the fundamental theorem of calculus tells us that
$$\int_a^x f(t)dt=F(x)$$
where $F(x)$ is antiderivative, i.e. $F(x)'=f(x)$.
If so, why I can't find the equality $\int f(x)dx=\int_a^x f(t)dt=F(x)$ anywhere? It expresses the relationship between definite and indefinite integral in such a straightfoward way (assuming this equality is true). So it it true and can I use $\int$ and $\int_a^x$ interchangeably? 
 A: Actually, the Fundamental Theorem of Calculus states that if $f:[a,b]\to \mathbb R$ is a continuous real-valued function, and $F$ is its antiderivative, then
$$\int_a^x f(t)dt= F(x)-F(a)$$
This is precisely why, more accurately,
$$\int_a^x f(t)dt= F(x)-F(a)=\int f(x)dx +C$$
for some $C \in \mathbb R$.
A: You are probably referring to this part of the fundemental theorem of calculus
\begin{equation}
\frac{d}{dx} \int_a^x f(t)dt =f(x)
\end{equation}
And your using the fact that
\begin{equation}
f(x) = \frac{d}{dx} \int f(x) dx
\end{equation}
which yields
\begin{equation}
\frac{d}{dx} \int_a^x f(t)dt = \frac{d}{dx}\int f(x) dx
\end{equation}
While this is true, remember that just because two functions have the same derivative, doesn't mean that they're the same function. Here is an example
\begin{align}
f(x) &= \int_3^x \sin t dt = -\cos x + \cos 3\\
g(x) &= \int \sin x dx = -\cos x
\end{align}
As you can cleary see both functions have the same derivative, but the functions are not the same. Your statement will be true if $a$ is a root of the indefinite integral. Example
\begin{align}
\int_0^x 2t dt= x^2 - 0 = x^2 = \int 2x dx
\end{align}
So it really depends on the function you are using. But the statement is NOT always true.
A: For a F(x) fixed, it's only ONE primitive function of f.
the other functions are of the form F(x) + arbitrary constante.
The form $\int_a^x f(u)du$ is the primitive which is 0 at x=a.
So $\int f(u) du = F(x)+ C$ where $C$ is an arbitrary number.
( the derivative of C is 0)
