Answer
$4.5$
Work Step by Step
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$
$4.5$
