91Ó°ÊÓ

Two functions

$$\begin{gathered} f\left(x\right)=x^{2}andg\left(x\right)=64−x^2 \\ Interval[0,8] \end{gathered}$$

Centroid formula

Let f and g be integral functions on [a, b]. The centroid (x¯, y¯) of the region between the graphs of f (x) and g(x) on the interval [a, b] is the point

$\left(\frac{{∫}_a^{b}x\left|f\right(x)-g(x\left)\right|dx}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx},\frac{{\displaystyle \frac{1}{2}{∫}_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx}}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx}\right)$

$$\begin{gathered} {∫}_a^b\left|f\right(x)-g(x\left)\right|dx \\ |{∫}_0^8{x}^2-\left(64-x^2\right)|dx \\ =|{∫}_0^{8}2x^2-64|dx \\ =|{∫}_0^{8}2x^{2}dx-{∫}_0^{8}64dx| \\ =|{\left[\frac{2x^3}{3}=64x\right]}_0^8 \\ =|\frac{1024}{3}-512| \\ =170.67 \end{gathered}$$

$$\begin{gathered} {∫}_a^{b}x\left|f\right(x)-g(x\left)\right|dx \\ \left|{∫}_0^{8}x\left(x^2-\left(64-x^2\right)\right)dx\right| \\ ={∫}_0^{8}x\left(2x^2-64\right)dx \\ ={∫}_0^8\left(2x^3-64x\right)dx \\ ={\left[\frac{2x^4}{4}-\frac{64x^2}{2}\right]}_0^8 \\ =\frac{2{\left(8\right)}^4}{4}-\frac{64{\left(8\right)}^2}{2} \\ =0 \end{gathered}$$

$$\begin{gathered} \frac{1}{2}{∫}_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx \\ \frac{1}{2}\left|{∫}_0^8\left({\left(x^2\right)}^2-{\left(64-x^2\right)}^2\right)dx\right| \\ =\frac{1}{2}|{∫}_0^8{\left(x^2\right)}^{2}dx-{∫}_0^8{\left(64-x^2\right)}^{2}dx| \\ =|\frac{1}{2}{∫}_0^8{\left(x^2\right)}^{2}dx-{∫}_0^{8}4096-128x^2+x^{4}dx| \\ =|\frac{1}{2}{\left[\frac{x^5}{5}-4096x+\frac{128x^3}{3}-\frac{x^5}{5}\right]}_0^8 \\ =|\frac{1}{2}\left[\frac{32768}{5}-32768+\frac{65536}{3}-\frac{32768}{5}\right] \\ =5461.33 \end{gathered}$$

Substitute all the integral values we got

$$\begin{gathered} \left(\frac{{∫}_a^{b}x\left|f\right(x)-g(x\left)\right|dx}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx},\frac{{\displaystyle \frac{1}{2}{∫}_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx}}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx}\right) \\ \left(\frac{0}{170.67},\frac{{\displaystyle 5461.33x}}{170.67}\right) \\ (0,32) \end{gathered}$$