Answer
$\frac{21}{5}$
Work Step by Step
$\int^{0}_{-2}(\frac{1}{2}t^4+\frac{1}{4}t^3-t) dt$
Integrating each separate term:
$=[\frac{1}{2\times 5}t^5+\frac{1}{4 \times 4}t^4-\frac{1}{2}t^2]|^0_{-2}$
$=[\frac{1}{10}t^5+\frac{1}{16}t^4-\frac{1}{2}t^2]|^0_{-2}$
$=[\frac{1}{10}(0)^5+\frac{1}{16}(0)^4-\frac{1}{2}(0)^2]-
[\frac{1}{10}(-2)^5+\frac{1}{16}(-2)^4-\frac{1}{2}(-2)^2]$
$=0-
[\frac{1}{10}(-32)+\frac{1}{16}(16)-\frac{1}{2}(4)]$
$=0-
[\frac{-32}{10}+1-2]$
$=\frac{32}{10}+1$
$=\frac{42}{10}$
$=\frac{21}{5}$