Answer
$0$
Work Step by Step
$\int ^{1}_{-1}\left( 4t^{3}-2t\right) dt=\left( \dfrac {4}{3+1}\times t^{3+1}-\dfrac {2}{1+1}\times t^{1+1}\right) ]^{1}_{-1}$
$=\left( t^{4}-t^{2}\right) ]^1_{-1}=\left( 1^{4}-1^{2}\right) -\left( \left( -1\right) ^{4}-\left( -1\right) ^{2}\right) =0$
