Question

Find the centroid of the region R enclosed between the curves y = x

2

+2 and y = x + 8.

Answer 1

$\overline{x}=\frac{1}{A}\int x(f(x)-g(x))dx$

$\overline{y}=\frac{1}{A}\int \frac{f(x)+g(x)}{2}(f(x)-g(x))dx$

A is area of region

intersections:

$x^2+2=x+8$

$x^2-x-6=0$

$x=\frac{1\pm \sqrt{25}}{2}$

$x_1=-2,x_2=3$

$A=\displaystyle{\int^3_{-2}}(f(x)-g(x))dx=\displaystyle{\int^3_{-2}}(x+8-x^2-2)dx=$

$=(-x^3/3+x^2/2+6x)|^3_{-2}=-9+9/2+18-8/3-2+12=125/6$

$\overline{y}=\frac{6}{25}\displaystyle{\int^3_{-2}}x(x+8-x^2-2)dx=$

$=\frac{6}{25}(-x^4/4+x^3/3+3x^2)|^3_{-2}=\frac{6}{25}(-81/4+9+27+4+8/3-12)=$

$=\frac{6}{25}\cdot\frac{125}{12}=\frac{5}{2}$

$\overline{x}=\frac{6}{25}\displaystyle{\int^3_{-2}}\frac{x^2+x+10}{2}(x-x^2+6)dx=$

$=\frac{6}{50}(x^4/4+x^3/3+5x^2-x^5/5-x^4/4-10x^3/3+2x^3+3x^2+60x)|^3_{-2}=$

$=\frac{6}{50}(-x^3+8x^2-x^5/5+60x)|^3_{-2}=$

$\frac{6}{50}(-27+72-243/5+180-8-32-32/5+120)=\frac{6}{50}\cdot100=12$