find the centroid of the region bounded by curves y=2x+3, y=x^2

Question

applications of centroids

Answer 1

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).