Answer
$\dfrac {16}{3}=5.333\ldots $
Work Step by Step
$\int ^{3}_{2}\left( t^{2}-1\right) dt=\int ^{3}_{2}\left( t^{2}dt-dt\right) =\left( \dfrac {t^{3}}{3}-t\right) ]^{3}_{2}=\left( \dfrac {3^{3}}{3}-3\right) -\left( \dfrac {2^{3}}{3}-2\right) =\dfrac {16}{3}=5.333\ldots $
