Question

Find the centroid of the area bounded by the parabola y=x² and the line 2x+3.

Answer 1

Let

$f(x) = x^2\ ,g(x) = 2x+3$

Formulas for centroid of area:

$A = \int_a^b(g(x)-f(x))dx\\ \overline x = \cfrac{1}{A} \int_a^b x(g(x)-f(x))dx\\ \overline y = \cfrac{1}{A} \int_a^b \cfrac{1}{2}((g(x))^2 - (f(x))^2)dx$

Find the $a$ and $b$:

$x^2 = 2x+3\\ x^2 -2x-3 = 0\\ x_1 = -1;x_2 = 3$

$A = \int_{-1}^3(2x+3-x^2)dx = x^2+3x-\cfrac{x^3}{3}|_{-1}^3 =\\ =9+9-9-(1-3-\cfrac{1}{3})=11+\cfrac{1}{3} = \cfrac{34}{3}$

$\overline x = \cfrac{3}{34}\int_{-1}^3 x(2x+3-x^2)dx = \cfrac{3}{34}\int_{-1}^3(2x^2+3x-x^3)dx =\\ =\cfrac{3}{34}(\cfrac{2x^3}{3}+\cfrac{3x^2}{2}-\cfrac{x^4}{4}|_{-1}^3)=\cfrac{3}{34}(18+\cfrac{27}{2}-\cfrac{81}{4}-(-\cfrac{2}{3}+\cfrac{3}{2}-\cfrac{1}{4})) = \cfrac{16}{17}$

$\overline y= \cfrac{3}{34}\int_{-1}^3\cfrac{1}{2}((2x+3)^2-x^4)dx = \cfrac{3}{68}\int_{-1}^3(4x^2+12x+9-x^4)dx = \\ = \cfrac{3}{68}(\cfrac{4x^3}{3}+6x^2+9x-\cfrac{x^5}{5})|_{-1}^3 = \cfrac{3}{68}(36+54+27-\cfrac{243}{5}-(-\cfrac{4}{3}+6-9+\cfrac{1}{5})) = \\ =\cfrac{16}{5}$

So centroid point's coordinate is $(\cfrac{16}{17};\cfrac{16}{5})$