Determine the centroid $of$ the area shown when $a=7in.$

Integral Table :

Area $={\displaystyle \underset{x_1}{\overset{x_2}{}}}\frac{1}{x^2}dx=\frac{1}{x_1}-\frac{1}{x_2}$

$Qy={\displaystyle \underset{x_1}{\overset{x_2}{}}}x\frac{1}{x^2}dx=ln(x_2)-ln(x_1)$

$Qx={\displaystyle \underset{x_1}{\overset{x_2}{}}}\frac{\frac{1}{x^2}}{2}\frac{1}{x^2}dx=\frac{1}{6(x_1{)}^3}-\frac{1}{6(x_2{)}^3}$

These integrals can also $be$ found using MATLAB.

$5ymsy12$;

$f=\frac{1}{x^2}$;

$$ Area $={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(f,x,x1,x2)={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}({\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(1,y,0,f),x,x1,x2)$

$Qy={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x^{*}f,x,x1,x2)={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}({\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(x,y,0,f),x,x1,x2)$

$Qx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}((\frac{f}{2}{)}^{*}f,x,x1,x2)={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}({\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(y,y,0,f),x,x1,x2)$

$\frac{?}{\text{b}\text{a}\text{r}}(Y)$:$,in.$

$\stackrel{}{x}$:$,in.$