The $\overline{x}$ coordinate of the center of mass of a plane region is calculated as
$$ \overline{x} = \frac{M_y}{M} = \frac{\int_a^b xf(x) \, \textrm{d}x}{\int_a^b f(x) \, \textrm{d}x} $$
And the $\overline{y}$ coordinate as
$$ \overline{y} = \frac{M_x}{M} = \frac{\frac{1}{2} \int_a^b [f(x)]^2 \, \textrm{d}x}{\int_a^b f(x) \, \textrm{d}x} $$
Is it possible to simplify the division of two integrals so that the above coordinates look something like this where I assume the values outside the integrals remain that way:
$$ \overline{x} = \int_a^b g(x) \, \textrm{d}x \\ \overline{y} = \frac{1}{2} \int_a^b h(x) \, \textrm{d}x $$
If this really depends on the function, let $f(x) = x^2$ for fun.
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$\begingroup$You can differentiate the division of the antiderivative of your functions. That's a mouthful so this is what I mean using $f(x) = x^2$ as you suggested:
$$ \begin{eqnarray} \overline{x} &=& \int_a^b \left[ \frac{d}{dx} \left( \frac{\int xf(x) dx}{\int f(x) dx} \right) \right] dx \\ &=& \int_a^b \left[ \frac{d}{dx} \left( \frac{\int x^3 dx}{\int x^2 dx} \right) \right] dx \\ &=& \int_a^b \left[ \frac{d}{dx} \left( \frac{\frac{x^4}{4} + C_1}{\frac{x^3}{3} + C_2} \right) \right] dx \\ &=& \int_a^b \left[ \frac{d}{dx} \left( \frac{3x}{4} + C_3 \right) \right] dx \\ &=& \int_a^b \frac{3}{4} dx \end{eqnarray} $$
So $g(x) = \frac{3}{4}$ and you can apply the same process for $h(x)$.
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