Integral Table

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}cf(x)dx=c{\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}f(x)dx$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}cos(x)dx=1$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}sin(x)dx=1$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^2(x)dx=\frac{}{4}$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}sin(x)cos(x)dx=\frac{1}{2}$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}xcos(x)dx=\frac{}{2}-1$ Integral Table

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}cf(x)dx=c{\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}f(x)dx$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}cos(x)dx=1$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}sin(x)dx=1$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}co{s}^2(x)dx=\frac{}{4}$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}sin(x)cos(x)dx=\frac{1}{2}$

${\displaystyle \underset{0}{\overset{\frac{}{2}}{}}}xcos(x)dx=\frac{}{2}-1$

Fill $in$ this table:

$Q_{-}y=\stackrel{}{x}Area={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$xdA,$Q_{-}x\frac{?}{b}=ar(y)$ Area $={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}$ydA

The centroid $of$ the shape $isatx=$

$in.$ and $y=,in.$