In Exercises 35–40, use definite integrals to calculate the centroid of the region described. Use graphs to verify that your answers are reasonable

The region between $f\left(x\right)=x^{2}andg\left(x\right)=64-x^2$on [a, b] = [0, 8].

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

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

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

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

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

Substitute all the integral values we got

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