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 $f\left(x\right)=\sqrt{x}$and the x-axis on[a, b] = [1, 9]. (Compare with Exercise 31.)

Short answer

The centroid of the region between and x-axis is $(3,60)$.

Step-by-step solution

Step 1. Given Information.

The function:

$f\left(x\right)=\sqrt{x}$

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.

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx={\int }_1^9\sqrt{x}dx \\ ={\left[\frac{1}{\sqrt{x}}\right]}_1^9 \\ =\frac{1}{\sqrt{9}}-\frac{1}{1} \\ =\frac{1}{3}-1 \\ =-\frac{2}{3} \\ {\int }_a^{b}xf\left(x\right)dx={\int }_1^{9}x\sqrt{x}dx \\ ={\int }_1^9{x}^{\frac{3}{2}}dx \\ =\frac{3}{2}{\left[\sqrt{x}\right]}_1^9 \\ =\frac{3}{2}[3-1] \\ =3 \end{gathered}$$

Step 4. Find ∫abf(x)2dx

$$\begin{gathered} {\int }_a^{b}f{\left(x\right)}^{2}dx={\int }_1^9{\left(\sqrt{x}\right)}^{2}dx \\ ={\int }_1^{9}xdx \\ ={\left[\frac{x^2}{2}\right]}_1^9 \\ =40 \end{gathered}$$

Step 5. Substitute the known values in the formula.

$$\begin{gathered} (\overline{x},\overline{y})=(\frac{{\int }_a^{b}xf\left(x\right)dx}{{\int }_a^{b}f\left(x\right)dx},\frac{{\int }_a^{b}f{\left(x\right)}^{2}dx}{{\int }_a^{b}f\left(x\right)dx}) \\ =(\frac{3}{{\displaystyle \frac{2}{3}}},\frac{40}{{\displaystyle \frac{2}{3}}}) \\ =(\frac{9}{2},\frac{120}{2}) \\ =(3,60) \end{gathered}$$