Q. 36 - 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 $f\left(x\right)=x^2\sqrt{3-x}$ and the x-axis on [a, b] = [0, 3]. (Compare with Exercise 32.)

Step-by-Step Solution

Answer

The centroid of the region between $f\left(x\right)=x^2\sqrt{3-x}$ and x-axis is $(2,3)$.

Step 1. Given Information.

The function: 

$f\left(x\right)=x^2\sqrt{3-x}$

Step 2. Centroid of region under curves.

The centroid of the region between the curve f(x) and x-axis is:

$(\overline{x},\overline{y})=(\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx},\frac{{\int }_a^{b}f{\left(x\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx})$

Step 3. Find the denominator.

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx={\int }_0^3{x}^2\sqrt{3-x}dx \\ \approx 7.1262 \\ {\int }_a^{b}xf\left(x\right)dx={\int }_0^{3}x.x^2\sqrt{3-x}dx \\ =14.252 \end{gathered}$$

Step 4. Find ∫ a b f ( x ) 2 d x

$$\begin{gathered} {\int }_a^{b}f{\left(x\right)}^{2}dx={\int }_0^3{\left(x^2\right)}^2{\left(\sqrt{3-x}\right)}^{2}dx \\ ={\int }_0^3{x}^4(3-x)dx \\ =24.3 \end{gathered}$$

Step 5. Substitute the known values in the formula.

$$\begin{gathered} (\overline{x},\overline{y})=(\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx},\frac{{\int }_a^{b}f{\left(x\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx}) \\ =(\frac{14.25}{7.13},\frac{24.3}{7.13}) \\ =(1.998, 3.42) \\ =(2,3) \end{gathered}$$