Answer
$(a) \int_{0}^{2}(-x+2) d x=2$
(b) $\int_{-1}^{1} 2 d x=4$
$(\mathrm{c}) \int_{1}^{3}(1+x) d x=6$
Work Step by Step
Using the rules of integration:
a)
$\begin{aligned} \int_{0}^{2}(-x+2) d x &=2 \int_{0}^{2} d x-\int_{0}^{2} x d x \\ &=\left.\right|_{0} ^{2}-\left.\frac{x^{2}}{2}\right|_{0} ^{2}2x \\ &=(-0+2)2-\frac{2^{2}-0}{2} \\ &=-2+4 \\ &=2 \end{aligned}$
b)
$\int_{-1}^{1} 2 d x=2 \int_{-1}^{1} d x$
$=\left.2 x\right|_{-1} ^{1}$
$=(1-(-1))2$
$=4$
$\begin{aligned} \int_{1}^{3}(1+x) d x &=\int_{1}^{3} d x+\int_{1}^{3} x d x \\ &=\left.\frac{x^{2}}{2}\right|_{1} ^{3}+\left.x\right|_{1} ^{3} \\ &=\frac{-1+9}{2}+(-1+3) \\ &=6 \end{aligned}$
