Answer
$\frac{196}{3}$
Work Step by Step
$\int^{1}_{-3}(7t^{2}+t+1)dt=7\int^{1}_{-3}t^{2}dt+\int^{1}_{-3}tdt+\int^{1}_{-3}dt $
$=7\times[\frac{t^{3}}{3}]^{1}_{-3}+[\frac{t^{2}}{2}]^{1}_{-3}+[t]^{1}_{-3}$
$=7(\frac{1^{3}}{3}-\frac{(-3)^{3}}{3})+(\frac{1^{2}}{2}-\frac{(-3)^{2}}{2})+(1-(-3))$
$=7(\frac{1}{3}+\frac{27}{3})+(\frac{1}{2}-\frac{9}{2})+(1+3)$
$=\frac{7\times28}{3}-4+4=\frac{196}{3}$
