Q. 34 - Calculus

Question

In Exercises 31–34, use a weighted average over n rectangles to approximate the centroid of the region described. (Hint: It may help to draw a picture.)

The region between $f\left(x\right)=x^2 and g\left(x\right)=64-x^2$ on $[a,b]=[0,8]$ with $n=4$.

Step-by-Step Solution

Answer

The region between the function $f\left(x\right)=x^2 and g\left(x\right)=64-x^2$ on [0, 8] is $(6.3,20.8125)$.

Step 1. Given Information.

The function:

$$\begin{gathered} f\left(x\right)=x^2 and \\ g\left(x\right)=64-x^2 \end{gathered}$$

Step 2. Centroid of region between two curves.

Centroid of the region between two curves is given by the formula,

$(\overline{x},\overline{y})=(\frac{{\int }_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx}{{\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx},\frac{{\displaystyle \frac{1}{2}}{\int }_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx}{{\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx})$

Step 3. Find the denominator.

$$\begin{gathered} {\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx={\int }_0^2\left|f\left(0\right)-g\left(0\right)\right|(0.5).dx+{\int }_2^4\left|f\left(2\right)-g\left(2\right)\right|(0.5).dx+{\int }_4^6\left|f\left(4\right)-g\left(4\right)\right|(0.5).dx+{\int }_6^8\left|f\left(6\right)-g\left(6\right)\right|(0.5).dx \\ =[{\int }_0^{2}64.dx+{\int }_2^{4}56.dx+{\int }_4^{6}32.dx+{\int }_6^{8}8.dx]0.5 \\ =\left[64\right(2)+56(2)+32(2)+8(2\left)\right]0.5 \\ =160 \\ {\int }_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx={\int }_0^2\left|f\left(0\right)-g\left(0\right)\right|(0.5).dx+{\int }_2^4\left|f\left(2\right)-g\left(2\right)\right|(0.5).dx+{\int }_4^6\left|f\left(4\right)-g\left(4\right)\right|(0.5).dx+{\int }_6^8\left|f\left(6\right)-g\left(6\right)\right|(0.5).dx \\ =64{\left[\frac{x^2}{2}\right]}_0^2+56{\left[\frac{x^2}{2}\right]}_2^4+32{\left[\frac{x^2}{2}\right]}_4^6+16{\left[\frac{x^2}{2}\right]}_6^8 \\ =128+336+320+224 \\ =1008 \end{gathered}$$

Step 4. Find ∫ a b f ( x ) 2 - g ( x ) 2 d x .

$$\begin{gathered} {\int }_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx=\underset{n=0}{\overset{8}{0.5\sum }}{\int }_0^2\left|{\left(x^2\right)}^2-{(64-x^2)}^2\right|dx \\ =0.5\sum _{n=0}^8{\int }_0^2\left|(x^4-4096+x^4)\right|dx \\ =0.5\sum _{n=0}^8{\int }_0^2\left|(2x^4-4096)\right|dx \\ =0.5\sum _{n=0}^8{\left[\left|\frac{2}{5}{x}^5-4096x\right|\right]}_0^2 =0.5\times \frac{2}{5}[8160+7200+1140+16800] \\ =0.2(33,300) \\ =6660 \end{gathered}$$

Step 5. Substitute the known values in the formula.

$$\begin{gathered} (\overline{x},\overline{y})=(\frac{{\int }_a^{b}x\left|f\left(x\right)-g\left(x\right)\right|dx}{{\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx},\frac{{\displaystyle \frac{1}{2}{\int }_a^b\left|f{\left(x\right)}^2-g{\left(x\right)}^2\right|dx}}{{\int }_a^b\left|f\left(x\right)-g\left(x\right)\right|dx}) \\ =(\frac{1008}{160},\frac{{\displaystyle \frac{1}{2}}\times 6660}{160}) \\ =(6.3,20.8125) \end{gathered}$$