Q. 37 - 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 $f\left(x\right)=x^3$ and the line y = 8 on [a, b] = [0, 2]. (Compare with Exercise 33.)

Step-by-Step Solution

Answer

The centroid of the region between $f\left(x\right)=x^3$ and y=8 is $(0,0.2)$.

Step 1. Given Information.

The function: 

$$\begin{gathered} f\left(x\right)=x^3 \\ y=8 \\ on [0,2] \end{gathered}$$

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.

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx={\int }_a^b(x^3-8)dx \\ ={[\frac{x^4}{4}-8x]}_0^2 \\ =12 \\ {\int }_a^{b}xf\left(x\right)dx={\int }_a^{b}x(x^3-8)dx \\ ={[\frac{x^5}{5}-4x^2]}_0^2 \\ =9.6 \end{gathered}$$

Step 4. Find ∫ a b f ( x ) 2 d x

$$\begin{gathered} {\int }_a^{b}f{\left(x\right)}^{2}dx={\int }_a^b\left(x^6\right)-8)dx \\ =2.29 \end{gathered}$$

Step 5. Substitute the known values in the formula.

$$\begin{gathered} (\overline{x},\overline{y})=(\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx},\frac{{\int }_a^{b}f{\left(x\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx}) \\ =(\frac{9.6}{12},\frac{2.29}{12}) \\ =(0.08,0.19) \\ =(0,0.2) \end{gathered}$$