Question $2$

Answer saved

$I_1={\displaystyle \underset{-}{\overset{2}{}}}(t^2-2)(t+3)dt$

$I_2={\displaystyle \underset{-5}{\overset{2}{}}}(2t+1)(t-3)dt$

$I_3={\displaystyle \underset{0}{\overset{4}{}}}(t^2-2)(2t-4)dt$

$I_4={\displaystyle \underset{0}{\overset{+}{}}}t*cos(-\frac{}{3}t)(t+3)dt$

$I_1=,$;$I_2=,$;$I_3=,$;$I_4=$