How do I find the area bounded by a $y=\frac{4}{3}x^2+\frac{12}{3}x-3$ and $y=\sqrt{x}$

Question

I am having difficulties with this problem:$y=\frac{4}{3}x^2+\frac{12}{3}x-3$ and $y=\sqrt{x}$.
Graphng two of the functions I get the following:
Graphing both functions shows that they intersect at $x=0.771$.
Now, the thing thats bugging me about this problem is that both graph's don't seem to have a bounded area. For this particular situation, $y=\sqrt{x}$ is only valid for values that are positive only.
I took a rough guess and did the integral like this: $$\int_{0}^{0.771}(\sqrt{x})-(\frac{4}{3}x^2+\frac{12}{3}x-3)$$ $$\int_{0}^{0.771}(\sqrt{x})-\frac{4}{3}x^2-\frac{12}{3}x+3)$$ $$\int\sqrt{x}-\frac{4}{3}\int(x^2)-\frac{12}{3}\int(x)+\int(3)$$ $$\frac{x^{\frac{3}{2}}}{3/2}-\frac{4x^3}{9}-\frac{12x^2}{6}+3x$$ Evaluating the integral: $$[\frac{(0.771)^{\frac{3}{2}}}{3/2})-\frac{4(0.771)^3}{9}-\frac{12(0.771)^2}{6}+3(0.771)]^{0.771}-[0]^{0}$$ $$\therefore \int_{0}^{0.771}(\sqrt{x})-(\frac{4}{3}x^2+\frac{12}{3}x-3)=0.9204 \text{square units}$$
I don't know if my approach is correct. I need help.