6.4. Real-World Applications of Integration - Calculus

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 $f \left(x\right) = \sin x and the line y =1/2 on [a, b] = [0, \pi ].$

Step-by-Step Solution

Answer

the centroid of the region is $\left(\frac{\mathrm{\pi }}{2},0.91\right)$

Step 1: Given Information

Given two functions 

 $$\begin{gathered} f\left(x\right)= \sin x \\ y=\frac{1}{2} \\ Interval [0,\mathrm{\pi }] \end{gathered}$$

Step 2: 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,

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

Step 3: Integrate

Find the value of integration.

$\begin{array}{rcl}m& =& {\int }_a^b\left|f\right(x)-g(x\left)\right|dx\\ & =& {\int }_0^{\mathrm{\pi }}|\sin x-\frac{1}{2}|dx\\ & =& {\left(-\cos x-\frac{x}{2}\right)}_0^{\mathrm{\pi }}\\ & =& \left(-\mathrm{cos\pi }-\frac{\mathrm{\pi }}{2}\right)\\ & =& -\left(-\cos 0-\frac{0}{2}\right)\\ & =& 2-\frac{\mathrm{\pi }}{2}\\ {\int }_a^b\left|f\right(x)-g(x\left)\right|dx& =& 2-\frac{\mathrm{\pi }}{2}\end{array}$

Step 4 : Integration

Find the value 

$\begin{array}{rcl}x& =& {\int }_a^{b}x\left|f\right(x)-g(x\left)\right|dx\\ & =& {\int }_0^{\mathrm{\pi }}x|\sin x- \frac{1}{2})|dx\\ & =& |{\int }_0^{\pi }x\sin \left(x\right)-\frac{x}{2}dx|\\ & =& {\int }_0^{\pi }x\sin \left(x\right)dx-{\int }_0^{\pi }\frac{x}{2}dx\\ & =& {\left[-x\cos \left(x\right)-\int \hspace{0.17em}-\cos \left(x\right)dx-\frac{x^2}{4}\right]}_0^{\pi }\\ & =& {\left[-x\cos \left(x\right)-\left(-\sin \left(x\right)\right)-\frac{x^2}{4}\right]}_0^{\pi }\\ & & compute the boundaries\\ & =& \left(-\mathrm{\pi cos}\left(\mathrm{\pi }\right)-\left(-\sin \left(\mathrm{\pi }\right)\right)-\frac{{\mathrm{\pi }}^2}{4}\right)-\left(-0\cos \left(0\right)-\left(-\sin \left(0\right)\right)-\frac{0^2}{4}\right)\\ & =& \pi -\frac{{\pi }^2}{4}\end{array}$

Step 5: Integrate

Find the value of y by using integration.

$\begin{array}{rcl}y& =& \frac{1}{2}{\int }_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx\\ & =& \frac{1}{2}|\hspace{0.17em}{\int }_0^{\pi }\hspace{0.17em}\left(si{n}^{2}x-{\left(\frac{1}{2}\right)}^2\right)dx|\\ & =& \frac{1}{2}\left({\int }_0^{\pi }{\sin }^2\left(x\right)dx-{\int }_0^{\pi }{\left(\frac{1}{2}\right)}^{2}dx\right)\\ & =& \frac{1}{2}\left({\int }_0^{\pi }\frac{1-\cos \left(2x\right)}{2}dx-{\int }_0^{\pi }{\left(\frac{1}{2}\right)}^{2}dx\right)\\ & =& \frac{1}{2}\left(\frac{1}{2}\left({\int }_0^{\pi }1dx-{\int }_0^{\pi }\cos \left(2x\right)dx\right)-{\int }_0^{\pi }{\left(\frac{1}{2}\right)}^{2}dx\right)\\ & =& \frac{1}{2}{\left(\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)-\frac{x}{4}\right)}_0^{\mathrm{\pi }}\\ & & \\ & & compute the boundaries\\ & =& \left[\frac{1}{2}\left(\frac{1}{2}\left(\mathrm{\pi }-\frac{1}{2}\sin \left(2\mathrm{\pi }\right)\right)-\frac{\mathrm{\pi }}{4}\right)\right]-\left[\frac{1}{2}\left(\frac{1}{2}\left(0-\frac{1}{2}\sin \left(0\right)\right)-\frac{0}{4}\right)\right]\\ & =& \frac{\pi }{8}\\ & & \end{array}$

Step 6: Substitute all integral values

Substitute the values in the formula to find the cor

$\begin{array}{rcl}& & \\ (\overline{x},\overline{y})& =& \left(\frac{{\int }_a^{b}x\left|f\right(x)-g(x\left)\right|dx}{{\int }_a^b\left|f\right(x)-g(x\left)\right|dx},\frac{{\displaystyle \frac{1}{2}{\int }_a^b|f{\left(x\right)}^2-g{\left(x\right)}^2|dx}}{{\int }_a^b\left|f\right(x)-g(x\left)\right|dx}\right)\\ & =& \left(\frac{ \mathrm{\pi }-{\displaystyle \frac{{\mathrm{\pi }}^2}{4}}}{ 2-{\displaystyle \frac{\mathrm{\pi }}{2}}},\frac{{\displaystyle \frac{\mathrm{\pi }}{8}}}{ 2-\frac{\mathrm{\pi }}{2}}\right)\\ & =& \left(\frac{\mathrm{\pi }}{2},0.91\right)\end{array}$

Step 7: Graph both equations