Answer
$-\frac{37}{6}$
Work Step by Step
Evaluate the integral: $\int^{1}_{0}(u+2)(u-3) du$
Recall the 2nd part of the Fundamental Theorem of Calculus:
$∫^b_af(x)dx=F(b)−F(a)$
Find $F(x)$:
$F(x)=\int(u+2)(u-3) du$
$F(x)=\int(u^2 - u -6)dx$
$F(x)=\frac{u^3}{3} -\frac{u^2}{2} - 6u$
Now evaluate $F(b)−F(a)$:
$F(1)−F(0)$
$(\frac{(1)^3}{3} -\frac{(1)^2}{2} - 6(1)) - (\frac{(0)^3}{3} -\frac{(0)^2}{2} - 6(0))$
$\frac{1}{3} - \frac{1}{2} -6 -0$
$\frac{2}{6}- \frac{3}{6} - \frac{36}{6}$
$-\frac{37}{6}$
