Find approximately, $by$ using a calculator $or$ other method $to$ estimate the area, ${\displaystyle \underset{0}{\overset{1}{}}}{u}^{2}arcsin(u^3)du=$

Then find the answer exactly, $by$ applying the Fundamental Theorem $of$ Calculus: $ifis$ the antiderivative $of$ having $no$ added constant, then ${\displaystyle \underset{0}{\overset{1}{}}}{u}^{2}arcsin(u^3)du=F(1)-F(0),$

where

$F(1)$

and

$F(0)$