I found the intersection point $(-2,4)$ and $(1,1)$
I decided to make everything in terms of $y$. Thus, $y=x^2$, $y = 2-x$.
So I will use the integral bound of -2 to 1. $2-x$ is the higher curve, $x^2$ is the lower curve clearly from graphing
The formula for the $x$ coordinate is $\frac{\int_{-2}^1((2-x)-x^2)x \,dx}{\int_{-2}^1((2-x)-x^2)\,dx}$.
We can also then proceed to find the $y$ coordinate. However my calculations are not yielding the same $x$ value as the solutions which is$(\frac{52}{45},\frac{20}{63})$
Let me redo this... the problem statement is a bit confusing, so I'll re-interpret it. If you were given the problem from a book, please transcribe the problem verbatim so we can best determine what is sought.