## Calculus 8th Edition

$4.5$
Evaluate the integral: $\int^2_{-1}(3u-2)(u+1)du$ Recall the 2nd part of the Fundamental Theorem of Calculus: $∫^b_af(x)dx=F(b)−F(a)$ Find $F(x)$: $F(x)=\int(3u-2)(u+1)du$ $F(x) = \int3u^2+u-3$ $F(x) = u^3 + \frac{u^2}{2} - 2u$ Now evaluate$F(b)−F(a)$: $F(2)−F(-1)$ $(2^3 +\frac{2^2}{2} -2(2)) - (-1^3 + \frac{-1^2}{2} -2(-1))$ $(8 + 2 - 4 ) -(-1 + \frac{1}{2} +2)$ $6 + 1 - 2.5$ $4.5$