The parabola and line intersect when x²=2x+3, so x²-2x-3=0=(x-3)(x+1).
They intersect at (-1,1) and (3,9). This gives us the limits for x for the integrals.
Area A of the region=∫(2x+3-x²)dx=(x²+3x-x³/3)[-1,3]=(9+9-9-(1-3+1/3))=32/3 sq units.
To find x̄, we need ∫x(2x+3-x²)dx=∫(2x²+3x-x³)dx =
(2x³/3+3x²/2-x⁴/4)[-1,3]=
18+27/2-81/4-(-2/3+3/2-1/4)=32/3.
Then we divide by A=32/3, so x̄=1.
To find ȳ, ½∫((2x+3)²-x⁴)dx=½∫(4x²+12x+9-x⁴)dx=
½(4x³/3+6x²+9x-x⁵/5)[-1,3]=171/5-(-31/15)=544/15.
Then we divide by A: ȳ=(544/15)/(32/3)=17/5.
So the centroid is (1,17/5).