Is $\int f(x) dx$ and $\int_{0}^{x} f(x) dx$ the same thing? If I plug in a value into the result of an indefinite integral, which part of the area do I get?
 A: Suppose that $F(x)$ is a primitive of $f(x)$, that is:
$$\frac{d}{dx}F(x) = f(x).$$
Then:
$$\int f(x) dx = F(x) + c, ~\forall c \in \mathbb{R},$$
while
$$\int_{0}^{x} f(x) dx = F(x) - F(0).$$
They are the same if you consider $c = -F(0).$
A: In general you have
$$\frac{d}{dx}F(x)=f(x)$$
$$\implies$$
$$\int_a^xf(t)dt=\int f(x)dx=F(x)+c\iff F(a)=-c$$
A: No it's not, in general:
$$\int f(x)\ \text{d}x = F(x) + c$$
Where $F(x)$ is the primitive of $f(x)$, provided it exist.
Whilst
$$\int_0^x f(x')\ \text{d}x' = F(x) - F(0)$$
And you cannot know if $F(0) = 0$ a priori.
Take $F(0) = -c$ and you're done, eventually.
A: You need to be concern with the variables (not that it's a big deal if you know what you're doing but still....)
Then from the definition of the anti-derivative and Fundamental theorem of Calculus-part II, it follows respectively the answer given below.
A: $\color{green}{\int_0^xf(x)\,\mathrm{d}x} = \color{orange}{F(x)-F(0)}$, which is a $\color{green}{\bf{definite\;integral}}$, or a $\color{orange}{\bf{value}}$.
$\color{green}{\int f(x)\,\mathrm{d}x} = \color{orange}{F(x)+c} = \color{blue}{\{F(x)\,|\,\frac{\mathrm{d}}{\mathrm{d}x}F(x) = f(x)\}} = \color{purple}{[F(x)]}$ which is an $\color{green}{\bf{indefinite\;integral}}$ or an $\color{orange}{\bf{antiderivative}}$ or a $\color{blue}{\bf{set}}$, or a $\color{purple}{\bf{class}}$.
