$$x= \left(\frac{1}{2} y\right)^2 - 3$$
$$Y= x-1 $$
All examples I viewed online seem to have no $x$ in them, should I write the equation in terms of $Y$? Would doing this give me $Y=x-1$ and $y= \left(\frac{1}{2}x\right)^2$. (Nearly $100 \%$ sure that's wrong.)
I can't get to the integration bits without understanding the question first, any tips please?
To find the area bounded by the curves in this case is easier to consider an equivalent system of the two functions
\begin{align} f_1(y)&= \left(\frac{1}{2} y\right)^2 - 3 \\ f_2(y)&=y+1, \end{align}
or, to make it look more familiar,
\begin{align} f_1(x)&= \left(\frac{1}{2} x\right)^2 - 3 \\ f_2(x)&=x+1, \end{align} since $y$ or $x$ is just the name of parameter, which we are free to choose suitable.
Next you need to find the points of intersection, when $f_1(x)=f_2(x)$ which is this simple case are the limits of integration $a=2-2\sqrt5$, $b=2+2\sqrt5$.
The area then is just
\begin{align} S&=\int_a^b f_2(x)-f_1(x) dx. \end{align}