Am I interpreting antiderivatives the right way? 
Here the antiderivative of $f(x)=x^2$ is $F(x) = x^3/3 + C$. If the constant of integration is $C = 0$, then for any $x$, $F(x)$ would give me the area under the curve of $f(x)$ from $0$ till $x$ since $F(0) = 0$. Is this correct? If it is, then for $x<0$ why is $F(x)$ giving a negative value?
Also, what would $F(x) = x^3/3 + 0.5$ represent? The area of $f(x)$ from what seems to be $-1.2$ to $x$?
 A: Yes, this is kind of correct.
$F(x)$ would give you the area between $0$ and $x$, counted positively if $x>0$ since you're going forward from $0$ to $x$, and counted negatively if $x<0$ since you're going backward from $0$ to $x$.
$F(x)=x^3/3+0.5$ gives you the area between $\alpha:=-\sqrt[3]{1.5}$ and $x$, because $F(\alpha)=0$. 
A: I'd like to emphasize the difference between integrals and antiderivatives.
An antiderivative of a function $f : I \to \mathbb{R}$ (where $I$ is any open subset of $\mathbb{R}$) is any function $F : I \to \mathbb{R}$ satisfying $F'(x) = f(x)$ for all $x \in I$. Clearly, is $F$ is an antiderivative of $f$, then so is $F + C$ for any constant $C \in \mathbb{R}$, since the derivative of a constant function is $0$. If $I$ is an open interval, then it follows from the Lagrange mean value theorem that any two antiderivatives of $f$ can only differ by a constant. We often write
$$
\int f(x)dx = F + C
$$
to indicate that all antiderivatives of $f$ can be written as $F + C$ for some constant $C$.
A (definite) integral of a function $f$ over an interval $[a, b]$ has a more technical definition. It can be interpreted as the signed surface area between the $x$-axis, the curve and the endpoints $a$ and $b$. By signed I mean that the area bellow the $x$-axis counts negatively. The definite integral of $f$ from $a$ to $b$ is denoted as
$$
\int_a^b f(x)dx
$$
and the fundamental theorem of calculus says that, under some conditions on $f$,
$$
\int_a^b f(x)dx = F(b) - F(a)
$$
for any antiderivative $F$ of $f$. If you're interested in the positive area between curve, $x$-axis and endpoints, you should calculate
$$
\int_a^b \lvert f(x)\rvert dx.
$$
In your example, since $x^2$ is already a positive function, the area between $x$-axis, curve and the (vertical lines through) the endpoints $a < b$ of an interval is given by
$$
\int_a^b x^2dx = \frac{b^3}{3} - \frac{a^3}{3}
$$
which is always positive, assuming $a < b$.
