Answer
$\frac{52}{9}$
Work Step by Step
$\int^2_{0}y^2\sqrt {1+y^3}dy = \int^2_{0}y^2 (1+y^3)^\frac{1}{2}dy$
$= [\frac{2}{3}\times\frac{1}{3y^2}\times y^2 (1+y^3)^\frac{3}{2}] ^2_{0}$
$= [\frac{2}{9}(1+y^3)^\frac{3}{2}] ^2_{0}$.
Sub in the bounds and subtract the lower bound (0) from the upper bound (2):
$= [\frac{2}{9}(1+(2)^3)^\frac{3}{2}] - [\frac{2}{9}(1+(0)^3)^\frac{3}{2}] $
$= (\frac{2}{9}(9)^\frac{3}{2}) - \frac{2}{9} $
$= 6 - \frac{2}{9} $
$=\frac{52}{9}$.
