What’s the definition indefinite integral? I am very confused that sometimes an indefinite integral represents the anti derivative, and other times it represents the area under the graph I.e. $\int f(x) dx = \lim_{x\rightarrow \infty} \int_{-x}^x f(x) dx$. I wonder if the representation of anti derivative an abuse of notation? 
 A: This is actually a good question and one that is awfully confusing without a proper background in analysis.
An antiderivative of $f(x)$ on a  set $X$ is a function $F(x)$ such that $F'(x)=f(x)$ for all $x\in X$. Note that antiderivatives are certainly not unique, which is why a $+C$ is added (derivative of constant is zero).
The first fundamental theorem of calculus says that if $f$ is integrable on $[a,x]$ and an antiderivative $F(x)$ exists for all $x\in[a,b]$, then $\int_a^x f=F(x)-F(a)$. We call $F(x)=F(a)+\int_a^x f(y)\,\mathrm{d}y$ an indefinite integral. Obviously, $F'(x)=f(x)$ by the second fundamental theorem if $f$ is continuous at $x$.
Where things get subtle is considering $F(x)$ in general when $f$ is integrable, but not necessarily continuous. It is a perfectly legitimate, continuous function. However, it's not necessarily an antiderivative of $f$ if $f$ is not continuous at $x$. 
For example, consider $$f(x)=\begin{cases}1,&\quad x\in[0,1] \\
2,& \quad x\in(1,2]\end{cases}$$
There is no antiderivative function $F(x)$ on $[0,2]$. One might suspect that 
$$
F(x)=\begin{cases}x,&\quad x\in[0,1] \\
2x,& \quad x\in(1,2]\end{cases}
$$
works. However, while each piece of $F(x)$ serves as an antiderivative on $[0,1]$ and $(1,2]$ respectively, this piecewise function is not an antiderivative on $[0,2]$, even though $F(x)=F(0)+\int_0^x f(y)\,\mathrm{d}y$ is continuous. Another way to see it is since $\int_0^2 f=1+2=3$, which is not equal to $F(2)-F(0)=4-0=4$, we would be contradicting the fundamental theorem to claim $F$ is an antiderivative.
Now consider when $f$ is continuous. Then it's always true that $F(x)=C+\int_a^x f(y)\,\mathrm{d}y$ serves the role of an antiderivative. The reason this notation is common is because this holds even if the antiderivative does not have an elementary form. For example, consider $f(x)=e^{x^2}$, which is continuous on any $[a,b]$. Then it's true that $F(x)=\int_a^x e^{y^2}\,\mathrm{d}y$ satisfies $F'(x)=f(x)$, but the notation is more instructive since an elementary form does not exist.
I hope some of this helps. The above also underscores the fact that all differentiable functions are continuous, and that a function can be differentiable without the derivative being continuous.
