## Calculus: Early Transcendentals 8th Edition

$80$
We expand the sum: $\displaystyle \sum_{i=4}^{8}(3i-2)$ $=[3(4)-2]+[3(5)-2]+[3(6)-2]+[3(7)-2]+[3(8)-2]$ $=(12-2)+(15-2)+(18-2)+(21-2)+(24-2)$ $=10+13+16+19+22$ $=80$