TLDR
Why is the center of mass formula for a lamina equal to $A_1$ but not $A_2$? (Specifically the $0.5$ in the $x$-coordinate integral).
$$ A_1=\left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$
$$ A_2=\left(\frac{\int_a^b0.5xf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx}\right)$$
Longer Explanation and Question
The center of mass of $n$ point masses in the $xy$-plane is point $P$ where $P$ is defined as:
$$ P = \left(\frac{\sum_{i=1}^n m_ix_i}{\sum_{i=1}^n m_i}, \frac{\sum_{i=1}^n m_iy_i}{\sum_{i=1}^n m_i} \right) .$$
When applying this to functions we could find the center of mass for an area under a curve bounded by two endpoints. For a given function $f(x)$ bounded by points $a$ and $b$, we can chop up the function into smaller and smaller rectangles and average out their center of masses. Point $Q$ represents the center of mass within this interval and it is defined as:
$$ Q = \left(\frac{\int_a^bxf(x)dx}{\int_a^bf(x)dx}, \frac{\int_a^b 0.5 f(x)^2dx}{\int_a^bf(x)dx} \right) . $$
When taking the center of mass for the $y$-coordinate, it makes sense that we would be summing up the halfway height of every rectangle. However, we do not sum up the halfway width of every rectangle. The center of mass of a rectangle lies directly in the middle of the $dx$ slice we use to integrate the region. However, the formula doesn't include this, and it is wrong.
I cannot find an explanation as to why it is wrong. As the widths of our rectangles approach zero, I guess the center of mass can be approximated as being $x$ instead of $\frac{x}{2}$. We could just as easily factor out the $\frac{1}{2}$ value.
Best Answer
In the $x$-dimension, the midpoint of your infinitesimal is not $x / 2$. Given a slice between $x_1$ and $x_2$ with width $\Delta x = x_2 - x_1$, the midpoint is
$$ \frac{x_1 + x_2}{2} $$ $$ = \frac{x_1 + x_1 + \Delta x}{2} $$ $$ = x_1 + \frac{\Delta x}{2} . $$
In the limit as $x_2 \to x_1$, we have $\Delta x \to 0$, and
$$ x_1 + \frac{\Delta x}{2} \to x_1 .$$