[tex]\bar{x} = 0[/tex]
[tex]\bar{y} =\dfrac{136}{125}[/tex]
Step-by-step explanation:
Let's define our functions [tex]f(x)\:\text{and}\:g(x)[/tex] as follows:
[tex]f(x) = x^2 + 1[/tex]
[tex]g(x) = 6x^2[/tex]
The two functions intersect when [tex]f(x)=g(x)[/tex] and that occurs at [tex]x = \pm\frac{1}{5}[/tex] so they're going to be the limits of integration. To solve for the coordinates of the centroid [tex]\bar{x}\:\text{and}\:\bar{y}[/tex], we need to solve for the area A first:
[tex]\displaystyle A = \int_a^b [f(x) - g(x)]dx[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}[(x^2 + 1) - 6x^2]dx[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}(1 - 5x^2)dx[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\left(x - \frac{5}{3}x^3 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]
[tex]\:\:\:\:\:\:\:= \dfrac{28}{75}[/tex]
The x-coordinate of the centroid [tex]\bar{x}[/tex] is given by
[tex]\displaystyle \bar{x} = \dfrac{1}{A}\int_a^b x[f(x) - g(x)]dx[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:= \frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} (x - 5x^3)dx[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{75}{28}\left(\dfrac{1}{2}x^2 -\dfrac{5}{4}x^4 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]
[tex]\:\:\:\:\:\:\:= 0[/tex]
The y-coordinate of the centroid [tex]\bar{y}[/tex] is given by
[tex]\displaystyle \bar{y} = \frac{1}{A}\int_a^b \frac{1}{2}[f^2(x) - g^2(x)]dx[/tex]
[tex]\displaystyle \:\:\:\:\:\:\:=\frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} \frac{1}{2}(-35x^4 + 2x^2 + 1)dx[/tex]
[tex]\:\:\:\:\:\:\:=\frac{75}{56} \left[-7x^5 + \frac{2}{3}x^3 + x \right]_{-\frac{1}{5}}^{+\frac{1}{5}}[/tex]
[tex]\:\:\:\:\:\:\:=\dfrac{136}{125}[/tex]
