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koban [17]
3 years ago
9

Please help NO LINKS

Mathematics
1 answer:
Alona [7]3 years ago
6 0

\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}

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