Answer
$\displaystyle\int\limits_{-1}^2(3u-2)(u+1)du=\dfrac{9}{2}$
Work Step by Step
$\displaystyle\int\limits_{-1}^2(3u-2)(u+1)du$
Evaluate the product inside the integral:
$\displaystyle\int\limits_{-1}^2(3u^{2}+3u-2u-2)du=\int\limits_{-1}^2(3u^{2}+u-2)du=...$
Integrate each term separately and apply the second part of the fundamental theorem of calculus:
$...=\displaystyle3\int\limits_{-1}^2u^{2}du+\int\limits_{-1}^2udu-2\int\limits_{-1}^2du=3\Big(\dfrac{1}{3}\Big)u^{3}+\dfrac{1}{2}u^{2}-2u\Big|_{-1}^2=...$
$...=u^{3}+\dfrac{1}{2}u^{2}-2u\Big|_{-1}^2=...$
$...=\Big[(2)^{3}+\dfrac{1}{2}(2)^{2}-2(2)\Big]-\Big[(-1)^{3}+\dfrac{1}{2}(-1)^{2}-2(-1)\Big]=...$
$...=8+\dfrac{1}{2}(4)-4-\Big[-1+\dfrac{1}{2}+2\Big]=8+2-4+1-\dfrac{1}{2}-2=$
$...=\dfrac{9}{2}$
