So finding the inverse of the functions gives us $$f(x)=\sqrt{x},-\sqrt{x}$$ $$g(x)=x-2$$ Having two different outcomes for the $f(x)$ threw me off at first but, after looking at the graph with all the functions on it, I thought of this $$\int_0^1\sqrt{x}dx-\int_0^1-\sqrt{x}dx$$ I do this because $f(x)=-\sqrt{x}$ goes into the $3$rd quadrant and $g(x)=x-2$ intersects with $-\sqrt{x}$at $(1,-1)$
This gave me $\frac{4}{3}$ (I just looked at he graph again and doing it this way this misses the region from $x=1$ to $x=2$ below the x-axis).
Then I went on to do $$\int_1^4\sqrt{x}-x+2dx$$ which gave me $-\frac{13}{3}$ and this where I knew I messed up because it should be a positive area. Also if someone could edit-in a graph with the functions above,(I don't know how to) it would be greatly appreciated.
The curves intersect at $y = -1$ and at $y = 2$
The direct way to do this would be to say ' $\int_{-1}^2 (y+2)-y^2 \ dy = \frac 92$
I would not mess with inverses. But if you did want to use inverses, we need to split the integral in two. In the first integral the bottom of the region is defined by the curve $y = -\sqrt x$ in the other $y = x-2$. $y = \sqrt x$ is the top of the region in both integrals.
$\int_{0}^1 2\sqrt x\ dx + \int_1^4 \sqrt x - (x-2) \ dx$
You should get the same result after changing the index of differentiation.