Q. 14 - Calculus

Question

The centroid of the region between the graph of f(x) = x 2

and the x-axis on [0, 2].

Step-by-Step Solution

Answer

The centroid of the region between the $f\left(x\right)=x^2$ and x-axis in the interval $\left(0, 2\right)$ $\left(\frac{3}{2}, \frac{2}{15}\right)$

Step 1. Given information

Given is the graph of $f\left(x\right)=x^2$ and the x-axis on $\left(0, 2\right)$

Step 2. Calculate the centroid of region

$$\begin{gathered} f\left(x\right)=x^2 \\ \left(\overline{x}, \overline{y}\right)=\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx},\frac{{\displaystyle \frac{1}{2}}{\int }_a^{b}f{\left(x\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx} \\ {\int }_0^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_0^2 \\ {\int }_0^2{x}^{2}dx=\frac{8}{3} \\ {\int }_0^{2}x·x^{2}dx={\int }_0^2{x}^{3}dx \\ ={\left[\frac{x^4}{4}\right]}_0^2 \\ \frac{1}{2}{\int }_0^2{\left(x^2\right)}^{2}dx=\frac{1}{2}{\int }_0^2{x}^{4}dx \\ =\frac{1}{2}{\left[\frac{x^5}{5}\right]}_0^2 \\ =\frac{16}{5} \\ {\int }_0^2{x}^{2}dx=\frac{8}{3}, {\int }_0^{2}x·x^{2}dx=4 \\ and \frac{1}{2}{\int }_0^2{\left(x^2\right)}^{2}dx=\frac{16}{5} \\ \left(\overline{x}, \overline{y}\right)=\frac{{\displaystyle \frac{4}{8}}}{3}, \frac{{\displaystyle \frac{16}{5}}}{{\displaystyle \frac{8}{3}}} \\ =\left(\frac{3}{2}, \frac{2}{15}\right) \end{gathered}$$

Step 3. The solution

The centroid of the region between the $f\left(x\right)=x^2$ and x-axis in the interval $\left(0, 2\right)$ is  $\left(\frac{3}{2}, \frac{2}{15}\right)$