# Let R be the region in the first quadrant bounded above by the graph of y=(6x+4)^(1/2) the line y=2x and the y axis, how do you find the area of region R?

Nov 28, 2017

The area is $= \frac{20}{9} {u}^{2}$

#### Explanation:

The point of intersection of the line $y = 2 x$ with the curve $y = \sqrt{6 x + 4}$ is

$\sqrt{6 x + 4} = 2 x$

$6 x + 4 = 4 {x}^{2}$

$2 {x}^{2} - 3 x - 2 = 0$

$x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm 5}{4}$

$x = 2$ or $x = - \frac{1}{2}$

$\int \sqrt{6 x + 4} = {\left(6 x + 4\right)}^{\frac{3}{2}} / \left(\frac{3}{2} \cdot 6\right) = \frac{1}{9} {\left(6 x + 4\right)}^{\frac{3}{2}} + C$

$\int 2 x \mathrm{dx} = 2 \cdot {x}^{2} / 2 + C$

Therefore,

The area is

$A = {\int}_{0}^{2} \left(\sqrt{6 x + 4} - 2 x\right) \mathrm{dx}$

$= {\left[\frac{1}{9} {\left(6 x + 4\right)}^{\frac{3}{2}} - {x}^{2}\right]}_{0}^{2}$

$= \left(\frac{1}{9} {\left(16\right)}^{\frac{3}{2}} - 4\right) - \left(\frac{1}{9} \cdot {4}^{\frac{3}{2}}\right)$

$= \frac{28}{9} - \frac{8}{9}$

$= \frac{20}{9} {u}^{2}$

graph{(y-sqrt(6x+4))(y-2x)=0 [-3.9, 10.146, -0.91, 6.11]}