Answer
$-4$
Work Step by Step
$\int^1_{-1} (\sqrt[3] t -2)dt $
$\frac{3}{4}t^{\frac{4}{3}} -2t$, Integrate
${\frac{3}{4}t^{\frac{4}{3}} -2t}]^1_{-1}$, Definite Integral Form
$((\frac{3}{4}- \frac{8}{4}) - (\frac{3}{4} + \frac{8}{4} ) $, Take Definite Integral
$-4$, simplify
