Question

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 $f\left(x\right)=lnxandg\left(x\right)=2-lnxon[a,b]=[1,e^2]$

Short answer

The coordinates of centroid is (7.2, 1)

Step-by-step solution

Given Information 

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

Step 2:  use the formula to find centroid 

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

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

Find the integral 

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

Integrate 

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

Integrate 

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

find centroid 

substitute all the integral values and find out centroid

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

Graph both the curves