align fix; theorem style; condition number

quarto/integrals/center_of_mass.qmd

@@ -158,23 +158,29 @@ The figure shows the approximating rectangles and circles representing their mas
 Generalizing from this figure shows the center of mass for such an approximation will be:
 
 
-
+$$
 \begin{align*}
 &\frac{\rho f(c_1) (x_1 - x_0) \cdot x_1 + \rho f(c_2) (x_2 - x_1) \cdot x_1 + \cdots + \rho f(c_n) (x_n- x_{n-1}) \cdot x_{n-1}}{\rho f(c_1) (x_1 - x_0)  + \rho f(c_2) (x_2 - x_1)  + \cdots + \rho f(c_n) (x_n- x_{n-1})} \\
 &=\\
 &\quad\frac{f(c_1) (x_1 - x_0) \cdot x_1 + f(c_2) (x_2 - x_1) \cdot x_1 + \cdots + f(c_n) (x_n- x_{n-1}) \cdot x_{n-1}}{f(c_1) (x_1 - x_0)  + f(c_2) (x_2 - x_1)  + \cdots + f(c_n) (x_n- x_{n-1})}.
 \end{align*}
+$$
 
 
 But the top part is an approximation to the integral $\int_a^b x f(x) dx$ and the bottom part the integral $\int_a^b f(x) dx$. The ratio of these defines the center of mass.
 
+::: {.callout-note icon=false}
+## Center of Mass
 
-> **Center of Mass**: The center of mass (in the $x$ direction) of a region in the $x-y$ plane described by the area under a (positive) function $f(x)$ between $a$ and $b$ is given by
->
-> $\text{Center of mass} = \text{cm}_x = \frac{\int_a^b xf(x) dx}{\int_a^b f(x) dx}.$
->
-> For regions described by a more complicated set of equations, the center of mass is found from the same formula where $f(x)$ is the total height in the $x$ direction for a given $x$.
+The center of mass (in the $x$ direction) of a region in the $x-y$ plane described by the area under a (positive) function $f(x)$ between $a$ and $b$ is given by
 
+$$
+\text{Center of mass} =
+\text{cm}_x = \frac{\int_a^b xf(x) dx}{\int_a^b f(x) dx}.
+$$
+
+For regions described by a more complicated set of equations, the center of mass is found from the same formula where $f(x)$ is the total height in the $x$ direction for a given $x$.
+:::
 
 
 For the triangular shape, we have by the fact that $f(x) = 1 - \lvert x \rvert$ is an even function that $xf(x)$ will be odd, so the integral around $-1,1$ will be $0$. So the center of mass formula applied to this problem agrees with our expectation.