The question I'm answering involves a parametric equation and these equations form a loop (according to the sketch). I was wondering if you were to integrate over an interval that was part of the loop would the value be the difference between the two areas??
The equations I got were:
$x=1-t^2$
$y= 2t-t^3$
And for example what if you integrated between x=1 and x=0, what does that value represent?
Thanks in advance!!
In order to do this the way you have in mind, you have to solve for $y$ in terms of $x$ in the interval of interest, so let's start by figuring out what the interval is. There are two points on the curve where $x=0$, corresponding to $t=1$ and $t=-1$. There is one point where $x=1$, corresponding to $t=0$, so we are interested in $-1\leq t\leq 1$. Note that $x(-t)=x(t)$ and $y(-t)=-y(t)$ so if $(x,y)$ is on the curve, so also is $(x, -y)$. That is, the curve is symmetric about the $x$-axis, so the area in question is twice the area in the first quadrant, and we can restrict our attention to the case $y\geq0$, that is $0\leq t\leq1$.
We have $$\begin{align} y&=2t-t^3\\ &=t+t(1-t^2)\\ &=t+tx\\ &=t(1+x) \end{align}$$
Further, $$\begin{align} t^2&=1-x\\ t&=\pm\sqrt{1-x}\\ t&=\sqrt{1-x} \end{align}$$ since $t\geq0$.
Combining these two results gives $y=(1+x)\sqrt{1-x}$ so the area in question is $$2\int_0^1(1+x)\sqrt{1-x}\,\mathrm{d}x$$ which you should have no trouble computing.
In this case, we were able to solve for $y$ explicitly in terms of $x$. Usually, that would not be possible, and other methods that you haven't learned yet would be needed.