Answer to Question #249727 in Calculus for SHAIKAT

Question #249727

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

2

+2 and y = x + 8.


1
Expert's answer
2021-10-12T10:24:35-0400

"\\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"


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