Find the area under the curve `y=sqrt(6x+4)`(above the x-axis) from x=0 to x=2

Explanation

Here the equation `y^{2}=6 x+4` is a parabola having its vertex at `x=-2 / 3`

The line parallel to `y`-axis is `x=2` at a 2 units distance

Similarly `y^{2}=6 x+4` has even power of `y` and is symmetrical about `x`-axis

So the required area can be written as

Area of `{OBCD}=int_{0}^{2} y {dx}=int_{0}^{2} sqrt{6 x+4} d x`

By integrating w.r.t `x`

``
=`[frac{(6 x+4)^{frac{3}{2}}}{frac{3}{2}(6)}]_{0}^{2}=frac{1}{9}[(6 x+4)^{frac{3}{2}}]_{0}^{2}`

...