Q. 31 - Calculus

Question

In Exercises 31–34, use a weighted average over n rectangles to approximate the centroid of the region described. 

The region between f(x) = √x and the x-axis on [a, b] = [1, 9], with n = 2. 

Step-by-Step Solution

Answer

$Centre of mass:(-0.37,1.22)$

Step1. Given Information

region,

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

Step2. Calculate

$$\begin{gathered} A={\int }_{-1.31}^{0.54}2-x^2-e^{x}dx \\ A=2x-\frac{x^3}{3}-e^x{{|}_{-1.31}}^{0.54} \\ A=1.45 \end{gathered}$$

Step3. Continue

$$\begin{gathered} \overline{x}=\frac{1}{A}{\int }_a^{b}xf\left(x\right)\rho \left(x\right)dx\Rightarrow \overline{x}=\frac{1}{1.45}{\int }_{-1.31}^{0.54}(2x-x^3-x{e}^x)dx \\ \overline{x}=\frac{1}{1.45}[x^2-\frac{x^4}{4}-(x{e}^x-e^x\left){{|}_{-1.31}}^{0.54}\right]=-0.3711 \\ \overline{y}=\frac{1}{A}{\int }_a^b\left(f^2\right(x)-{\rho }^2(x\left)\right)dx \Rightarrow \frac{1}{1.45}{\int }_{-1.31}^{0.54}\left(\frac{1}{2}\right)[{(2-x)}^2-e^{2x}]dx \\ \overline{y}=0.345{\int }_{-1.31}^{0.54}4-4x^2+x^4-e^{2x}dx \\ \overline{y}=0.345[4x-\frac{4x^3}{3}+\frac{x^5}{5}-\frac{e^{2x}}{2}{{|}_{-1.31}}^{0.54}] \\ \overline{y}=1.22 \\ C.o.M:(-0.37,1.22). \end{gathered}$$