Question

Find the coordinates of the centroid of the region between$f\left(x\right)=\cos x$and the x-axis on $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.

Short answer

The coordinate point is$\left(0,\frac{\pi }{4}\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given information.

$f\left(x\right)=\cos x\mathrm{and}\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Step 2. Find the centroid coordinate.

Use the given formulas to find the point.

$\begin{array}{l}\overline{x}=\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx}\\ =\frac{{\int }_{-\pi /2}^{\pi /2}x\cos xdx}{{\int }_{-\pi /2}^{\pi /2}\cos xdx}\\ =\frac{{\left[x\sin x-\left(-\cos \left(x\right)\right)\right]}_{-\pi /2}^{\pi /2}}{{\left[\sin \left(x\right)\right]}_{-\pi /2}^{\pi /2}}\\ =\frac{0}{2}\\ =0\end{array}$

And,

$\begin{array}{l}\overline{y}=\frac{{\int }_a^b{\left(f\left(x\right)\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx}\\ =\frac{{\int }_{-\pi /2}^{\pi /2}{\cos }^{2}xdx}{{\int }_{-\pi /2}^{\pi /2}\cos xdx}\\ =\frac{{\left[\frac{1}{2}\left(x+\frac{1}{2}\sin \left(2x\right)\right)\right]}_{-\pi /2}^{\pi /2}}{{\left[\sin \left(x\right)\right]}_{-\pi /2}^{\pi /2}}\\ =\frac{\frac{\pi }{2}}{2}\\ =\frac{\pi }{4}\end{array}$