Answer
$5.25$
Work Step by Step
$\int^1_{0}y(y^2+1)^5dy$
$=\frac{1}{2}\times\frac{1}{6}(y^2+1)^6$
$=\frac{1}{12}(y^2+1)^6$
$=[\frac{1}{12}(y^2+1)^6] ^1_{0}$.
Sub in the bounds and subtract the lower bound (0) from the upper bound (1):
$=[\frac{1}{12}((1)^2+1)^6]-[\frac{1}{12}((0)^2+1)^6]$
$=(\frac{1}{12}\times64)-\frac{1}{12}$
$=5.25$.
