The area of the region bounded by the curves $\mathrm{y}=\sqrt{x}$ and $y=\sqrt{4-3x}$ and $\mathrm{y}=0$ is:

$\mathrm{y}=\sqrt{x}$ ; $y=\sqrt{4-3x}$

$\overset{1}{\underset{0}{\int}}\sqrt{x}dx+\overset{4/3}{\underset{1}{\int}}\sqrt{4-3x}dx$

$\dfrac{2}{3}1+\dfrac{2}{9}(4-3x)^{3/2}|_{1}^{4/3}$

$\dfrac{2}{3}-\dfrac{2}{9}(0-1)$

$=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{6+2}{9}=\dfrac{8}{9}sq\:units$