Answer
$\frac{(7-3y^{2})\times\sqrt (7-3y^{2})}{3}$ + C
Work Step by Step
3$\int$y$\sqrt (7-3y^{2})$dy
using the substitution t=7-$3y^{2}$ transform the integral
3 $\times$ $\int$ -$\frac{1}{6}$ $\times$ $\sqrt t$ dt
use the property of integral $\int$a$\times$f(x)dx=a$\times$$\int$f(x)dx
3$\times$ ($-\frac{1}{6}$)$\times$$\sqrt t$ dt
-$\frac{1}{2}$$\times$$\int$$t^{\frac{1}{2}}$ dt
-$\frac{1}{2}$$\times$ $\frac{2t\sqrt t}{3}$ =$\gt$ back t=7-$3y^{2}$
-$\frac{1}{2}$ $\times$ $\frac{{2(7-3y^{2})\sqrt (7-3^{2})}}{3}$
$\frac{(7-3y^{2})\times\sqrt (7-3y^{2})}{3}$ + C
