Answer
$18$
Work Step by Step
Since, we have $ x=2y^2$
$\implies y=\dfrac{\sqrt x}{\sqrt 2}$ and $\dfrac{\sqrt x}{\sqrt 2}=3 \implies 3-\dfrac{\sqrt x}{\sqrt 2}$
We know that $\int x^{n} dx=\dfrac{x^{n+1}}{n+1}+C$
where $C$ is a constant of proportionality.
$\int_0^{18} (3-\dfrac{\sqrt x}{\sqrt 2})dx=[3x-\dfrac{1}{\sqrt 2}(\dfrac{x^{3/2}}{1/(3/2)})]_0^{18}$
or, $=3(18-0)-\dfrac{2}{3\sqrt 2}( 18^{3/2}-0)$
or $=18$
