Answer
$-\frac{10}{3}$
Work Step by Step
$\int^1_{-1} (t^2 -2)dt $
$\frac{1}{3}t^3 -2t $ , Integrate
${\frac{1}{3}t^3 -2t}]^1_{-1} $ , Definite Integral Form
$(\frac{1}{3} - \frac{6}{3}) - ( -\frac{1}{3}+\frac{6}{3})$ , Take definite integral
$-\frac{10}{3}$, Simplify
