## Calculus: Early Transcendentals 8th Edition

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