Question

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

• A. 4/9
• B. 16/9
• C. 8/9
• D. 9/2

Answer 1

The correct option is C 8/9

$y=\sqrt{x}$ ; $y=\sqrt{4-3x}$
$\stackrel{1}{\underset{0}{\int }}\sqrt{x}dx+\stackrel{4/3}{\underset{1}{\int }}\sqrt{4-3x}dx$
$\frac{2}{3}1+\frac{2}{9}(4-3x{)}^{3/2}{|}_1^{4/3}$
$\frac{2}{3}-\frac{2}{9}(0-1)$
$=\frac{2}{3}+\frac{2}{9}=\frac{6+2}{9}=\frac{8}{9}squnits$