Question

Find the coordinates of the centroid of the region between the graph of $f\left(x\right)=x^2$and the x-axis on [0, 3].

Short answer

The points are$\left(\frac{9}{4},\frac{9}{2}\right)$.

Step-by-step solution

Step 1. Given information.

Consider the given information,

$f9x)=x^2\mathrm{and}\left[0,3\right]$

Step 2. Find the centroid coordinate.

Use the given formulas to find the point.

$\begin{array}{rcl}\overline{x}& =& \frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx}\\ & =& \frac{{\int }_0^{3}x·x^{2}dx}{{\int }_0^3{x}^{2}dx}\\ & =& \frac{{\int }_0^3{x}^{3}dx}{{\int }_0^3{x}^{2}dx}\\ & =& \frac{{\left[\frac{x^4}{4}\right]}_0^3}{{\left[\frac{x^3}{3}\right]}_0^3}\\ & =& \frac{\frac{81}{4}}{9}\\ & =& \frac{9}{4}\end{array}$

And,

$\begin{array}{l}\overline{y}=\frac{1}{2}\frac{{\int }_a^b{\left(f\left(x\right)\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx}\\ =\frac{1}{2}\frac{{\int }_0^3{\left(x^2\right)}^{2}dx}{{\int }_0^3{x}^{2}dx}\\ =\frac{1}{2}\frac{{\left[\frac{x^5}{5}\right]}_0^3}{{\left[\frac{x^3}{3}\right]}_0^3}\\ =\frac{1}{2}\frac{\frac{3^5}{3}}{\frac{3^3}{3}}\\ =\frac{9}{2}\end{array}$

The centroid coordinate is$\left(\frac{9}{4},\frac{9}{2}\right)$.