Answer
$-76$
Work Step by Step
$\int^9_{1}(\frac{\sqrt u-2u^2}{u})du=\int^9_{1}(\frac{1}{\sqrt u}-2u)du$
$=[2\sqrt u-u^2]^9_{1}$.
Sub in the bounds and subtract the lower bound (1) from the upper bound (9):
$=[2\sqrt {(9)}-(9)^2]-[2\sqrt {(1)}-(1)^2] $
$=6-81-(2-1)$
$=-76 $.
