Question

find the ordinate of the centroid of the area bounded by the parabola y=x^2 and the line y=2x+3

Answer 1

f(x)= x² , g(x)=2x + 3

A= $\int_a^b$ (g(x) - f(x))dx

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

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

Find a and b;

x²= 2x+3

x² -2x-3=0

x₁=-1, x₂=3

A = $\int_a^b$ (2x-3-x²)dx = x²-3x-$\frac{x^3}{3}$ $\vert_{-1}^3$

= 9+9−9−(1−3−$\frac{1}{3}$ )=11-$\frac{1}{3}$ =$\frac{34}{3}$

$\overline{x}$ = $\frac{3}{34}$ $\int_{-1}^3$ x(2x +3 -x²)dx = $\frac{3}{34}$ $\int_{-1}^3$ (2x²+3x-x³)dx =

$\frac{3}{34}$ ($\frac{2x^3}{3}$ +$\frac{3x^2}{2}$ -$\frac{x^4}{4}$$\vert_{-1}^{3}$ ) = $\frac{3}{34}$ (18+$\frac{27}{2}$- $\frac{81}{4}$ -($-\frac{2}{3} + \frac{3}{2} -\frac{1}{4}$ ) ) =$\frac{16}{17}$

$\overline{y}$ = $\frac{3}{34}$ $\int_{-1}^3$ $\frac{1}{2}$ ((2x +3)² -x⁴)dx = $\frac{3}{68}$ $\int_{-1}^3$ (4x²+12x +9-x⁴)dx =

$\frac{3}{68}$ ($\frac{4x^3}{3}$ -6x² -9x-$\frac{x^5}{5}$$\vert_{-1}^{3}$ ) = $\frac{3}{68}$ (36 + 54 + 27$-\frac{243}{5}-$ ($-\frac{4}{3} +6-9 -\frac{1}{5}$ ) ) =$\frac{16}{5}$

The coordinate of centroid is ($\frac{16}{17} , \frac{16}{5}$ )