My sketch is right so that's not the issue. I know the formula for finding the area. I just wanted to know if my work is right. Is the area just:
$$A= \int_{a}^{b} g(t)f'(t) dt$$
$$A= \int_{-1}^{1} (t^3-t)(2t) dt$$
?
The point where is intesects itself is at $t=0,-1,1 $ but since $t_{1} \neq t_{2}$ , we can only have $t=-1,1$ for the points of self intersection.
It is very important to have a sketch of the curve (see below: here for values of the parameter between $t=-1.15$ and $t=1.15$). $P=(1,0)$ is the only double point (this can be shown rigorously) corresponding to values $t=-1$ and $t=1$ (value $t=0$ does not play a special role). Thus your formula for $A$ is right, but you have to take it with an absolute value.
Edit : The problem with formula $A= \int_{a}^{b} g(t)f'(t) dt$ is that it gives a signed result but you are unable to predict the meaning of this sign. It is all right if you are interested in the "arithmetical" area, just take the absolute value. But if you want to interpret the sign, prefer the following formula :
$$A=\dfrac{1}{2} \int_{a}^{b} (f(t)g'(t)-g(t)f'(t)) dt$$
This expression of $A$ will have a "plus" sign (resp. "minus" sign) if the loop is done with direct, i.e., trigonometric orientation (resp. non-direct, i.e., clockwise).
(see for example here).