Question

Please help NO LINKS

Answer 1

$\bar{x} = 0$

$\bar{y} =\dfrac{136}{125}$

Step-by-step explanation:

Let's define our functions $f(x)\:\text{and}\:g(x)$ as follows:

$f(x) = x^2 + 1$

$g(x) = 6x^2$

The two functions intersect when $f(x)=g(x)$ and that occurs at $x = \pm\frac{1}{5}$ so they're going to be the limits of integration. To solve for the coordinates of the centroid $\bar{x}\:\text{and}\:\bar{y}$, we need to solve for the area A first:

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

$\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}[(x^2 + 1) - 6x^2]dx$

$\displaystyle \:\:\:\:\:\:\:=\int_{-\frac{1}{5}}^{+\frac{1}{5}}(1 - 5x^2)dx$

$\displaystyle \:\:\:\:\:\:\:=\left(x - \frac{5}{3}x^3 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}$

$\:\:\:\:\:\:\:= \dfrac{28}{75}$

The x-coordinate of the centroid $\bar{x}$ is given by

$\displaystyle \bar{x} = \dfrac{1}{A}\int_a^b x[f(x) - g(x)]dx$

$\displaystyle \:\:\:\:\:\:\:= \frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} (x - 5x^3)dx$

$\:\:\:\:\:\:\:=\dfrac{75}{28}\left(\dfrac{1}{2}x^2 -\dfrac{5}{4}x^4 \right)_{-\frac{1}{5}}^{+\frac{1}{5}}$

$\:\:\:\:\:\:\:= 0$

The y-coordinate of the centroid $\bar{y}$ is given by

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

$\displaystyle \:\:\:\:\:\:\:=\frac{75}{28}\int_{-\frac{1}{5}}^{+\frac{1}{5}} \frac{1}{2}(-35x^4 + 2x^2 + 1)dx$

$\:\:\:\:\:\:\:=\frac{75}{56} \left[-7x^5 + \frac{2}{3}x^3 + x \right]_{-\frac{1}{5}}^{+\frac{1}{5}}$

$\:\:\:\:\:\:\:=\dfrac{136}{125}$