use quarto, not Pluto to render pages

CwJ/integrals/center_of_mass.jmd

@@ -192,13 +192,9 @@ cm = top / bottom
 
 Our guess from the diagram proves correct.
 
-```julia; echo=false
-note("""
+!!! note
+	It proves convenient to use the `->` notation for an anonymous function above, as our function `h` is not what is being integrated all the time, but some simple modification. If this isn't palatable, a new function could be defined and passed along to `quadgk`.
 
-It proves convenient to use the `->` notation for an anonymous function above, as our function `h` is not what is being integrated all the time, but some simple modification. If this isn't palatable, a new function could be defined and passed along to `quadgk`.
-
-""")
-```
 
 ##### Example
 
@@ -220,13 +216,9 @@ For fun, we compare this to the median, which is the value $M$ so that the total
 
 Solving $1/2 = 1 - e^{-M}$ gives $M=\log(2) = 0.69...$, The median is to the left of the mean in this example.
 
-```julia; echo=false
-note("""
+!!! note
+	In this example, we used an infinite region, so the idea of "balancing" may be a bit unrealistic, nonetheless, this intuitive interpretation is still a good one to keep this in mind. The point of comparing to the median is that the balancing point is to the right of where the area splits in half. Basically, the center of mass follows in the direction of the area far to the right of the median, as this area is skewed in that direction.
 
-In this example, we used an infinite region, so the idea of "balancing" may be a bit unrealistic, nonetheless, this intuitive interpretation is still a good one to keep this in mind. The point of comparing to the median is that the balancing point is to the right of where the area splits in half. Basically, the center of mass follows in the direction of the area far to the right of the median, as this area is skewed in that direction.
-
-""")
-```
 
 ##### Example
 
@@ -308,13 +300,9 @@ integrate(x*fx, (x, 0, b)) / integrate(fx, (x, 0, b))
 
 But really, we should have just noted that simply by switching the labels $a$ and $b$ in the diagram we could have discovered this formula.
 
-```julia; echo=false
-note(L"""
+!!! note
+	The [centroid](http://en.wikipedia.org/wiki/Centroid) of a region in the plane is just $(\text{cm}_x, \text{cm}_y)$. This last fact says the centroid of the right triangle  is just $(b/3, a/3)$. The centroid can be found by other geometric means. The link shows the plumb line method. For triangles, the centroid is also the intersection point of the medians, the lines that connect a vertex with its opposite midpoint.
 
-The [centroid](http://en.wikipedia.org/wiki/Centroid) of a region in the plane is just $(\text{cm}_x, \text{cm}_y)$. This last fact says the centroid of the right triangle  is just $(b/3, a/3)$. The centroid can be found by other geometric means. The link shows the plumb line method. For triangles, the centroid is also the intersection point of the medians, the lines that connect a vertex with its opposite midpoint.
-
-""")
-```
 
 ##### Example