91Ó°ÊÓ

The region between f (x) = ln x and g(x) = 2 − ln x on [a, b] = [1, e^2]

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

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

$$\begin{gathered} {∫}_a^b\left|f\right(x)-g(x\left)\right|dx \\ f\left(x\right)=\ln \left(x\right) \\ g\left(x\right)=2-\ln x \\ intervalisa=1,b=e^2 \\ {∫}_1^{e^2}\left|\ln \right(x\left)\right)-(2-\ln x)|dx \\ ={∫}_1^{e^2}2\ln \left(x\right)-2dx \\ ={∫}_1^{e^2}2\ln \left(x\right)dx-{∫}_1^{e^2}2dx \\ ={\left[2\left[x\ln x-x\right]-2x\right]}_1^{e^2} \\ =2\left(e^2+1\right)-\left(2e^2-2\right) \\ =4 \end{gathered}$$

$$\begin{gathered} {∫}_a^{b}x\left|f\right(x)-g(x\left)\right|dx \\ {∫}_1^{e^2}x\left|\ln \right(x\left)\right)-(2-\ln x)|dx \\ =|{∫}_1^{e^2}2x\ln \left(x\right)-2xdx| \\ =|{∫}_1^{e^2}2x\ln \left(x\right)-2{∫}_1^{e^2}xdx| \\ =|2\left(\ln \left(x\right){∫}_1^{e^2}xdx-{∫}_1^{e^2}\left(\frac{d}{dx}\ln \left(x\right){∫}_1^{e^2}xdx\right)dx\right)-2{∫}_1^{e^2}xdx \\ =|2{\left[\frac{x^2\ln \left(x\right)}{2}-\frac{x^2}{4}\right]}_1^{e^2}-2{\left[\frac{x^2}{2}\right]}_1^{e^2}| \\ Substitutetheintervalsandsimplifyit \\ =|2e^4-\frac{e^4}{2}+\frac{1}{2}-e^4+1| \\ =28.80 \end{gathered}$$

$$\begin{gathered} \frac{1}{2}{∫}_1^{e^2}|\ln {\left(x\right))}^2-{(2-\ln x)}^2|dx \\ =\frac{1}{2}|{∫}_1^{e^2}\ln {\left(x\right))}^2-4+4\ln x-\ln x^2|dx \\ =\frac{1}{2}|{∫}_1^{e^2}-4+4\ln x|dx \\ =\frac{1}{2}{\left[-4x+4(x\ln x-x)\right]}_1^{e^2} \\ Substitutetheintervalsandsimplify \\ =\frac{1}{2}\left(8\right)=4 \end{gathered}$$

substitute all the integral values and find out centroid

$$\begin{gathered} \left(\frac{{∫}_a^{b}x\left|f\right(x)-g(x\left)\right|dx}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx},\frac{{\displaystyle \frac{1}{2}{∫}_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx}}{{∫}_a^b\left|f\right(x)-g(x\left)\right|dx}\right) \\ \left(\frac{28.80}{4},\frac{{\displaystyle 4}}{4}\right) \\ (7.2,1) \\ centroidis(7.2,1) \end{gathered}$$