em dash; sentence case

quarto/integrals/arc_length.qmd

@@ -557,7 +557,7 @@ Following (faithfully) [Kantorwitz and Neumann](https://www.researchgate.net/pub
 
 @fig-kantorwitz-neumann is clearly of a concave down function. The asymmetry about the critical point will be seen to be a result of the derivative also being concave down. This asymmetry will be characterized in several different ways in the following including showing that the arc length from $(a,0)$ to $(c,f(c))$ is longer than from $(c,f(c))$ to $(b,0)$.
 
-::: {#@fig-kantorwitz-neumann}
+::: {#fig-kantorwitz-neumann}
 
 
 ```{julia}

quarto/integrals/area.qmd

@@ -16,7 +16,7 @@ using Roots
 
 ---
 
-![A jigsaw puzzle needs a certain amount of area to complete. For a traditional rectangular puzzle, this area is comprised of the sum of the areas for each piece. Decomposing a total area into the sum of smaller, known, ones--even if only approximate--is the basis of definite integration.](figures/jigsaw.png)
+![A jigsaw puzzle needs a certain amount of area to complete. For a traditional rectangular puzzle, this area is comprised of the sum of the areas for each piece. Decomposing a total area into the sum of smaller, known, ones---even if only approximate---is the basis of definite integration.](figures/jigsaw.png)
 
 
 The question of area has long fascinated human culture. As children, we learn early on the formulas for the areas of some geometric figures: a square is $b^2$, a rectangle $b\cdot h$, a triangle $1/2 \cdot b \cdot h$ and for a circle, $\pi r^2$. The area of a rectangle is often the intuitive basis for illustrating multiplication. The area of a triangle has been known for ages. Even complicated expressions, such as [Heron's](http://tinyurl.com/mqm9z) formula which relates the area of a triangle with measurements from its perimeter have been around for 2000 years. The formula for the area of a circle is also quite old. Wikipedia dates it as far back as the [Rhind](http://en.wikipedia.org/wiki/Rhind_Mathematical_Papyrus) papyrus for 1700 BC, with the approximation of $256/81$ for $\pi$.
@@ -1067,7 +1067,7 @@ plot!(zero)
 
 We could add the signed area over $[0,1]$ to the above, but instead see a square of area $1$, a triangle with area $1/2$ and a triangle with signed area $-1$. The total is then $1/2$.
 
-This figure--using equal sized axes--may make the above decomposition more clear:
+This figure---using equal sized axes---may make the above decomposition more clear:
 
 ```{julia}
 #| echo: false

quarto/integrals/area_between_curves.qmd

@@ -448,7 +448,7 @@ When doing problems by hand this latter style can often reduce the complications
 Consider two overlapping circles, one with smaller radius. How much area is in the larger circle that is not in the smaller? The question came up on the `Julia` [discourse](https://discourse.julialang.org/t/is-there-package-or-method-to-calculate-certain-area-in-julia-symbolically-with-sympy/99751) discussion board. A solution, modified from an answer of `@rocco_sprmnt21`, follows.
 
 Without losing too-much generality, we can consider the smaller circle to have radius $a$, the larger circle to have radius $b$ and centered at $(0,c)$.
-We assume some overlap -- $a \ge c-b$, but not too much -- $c-b \ge 0$ or $0 \le c-b \le a$.
+We assume some overlap---$a \ge c-b$, but not too much---$c-b \ge 0$ or $0 \le c-b \le a$.
 
 ```{julia}
 @syms x::real y::real a::positive b::positive c::positive

quarto/integrals/center_of_mass.qmd

@@ -1,4 +1,4 @@
-# Center of Mass
+# Center of mass
 
 
 {{< include ../_common_code.qmd >}}

quarto/integrals/improper_integrals.qmd

@@ -1,4 +1,4 @@
-# Improper Integrals
+# Improper integrals
 
 
 {{< include ../_common_code.qmd >}}

quarto/integrals/integration_by_parts.qmd

@@ -1,4 +1,4 @@
-# Integration By Parts
+# Integration by parts
 
 
 {{< include ../_common_code.qmd >}}
@@ -116,7 +116,7 @@ $$
 
 $B$ is similar with the roles of $u$ and $v$ reversed.
 
-----
+---
 
 Informally, the integration by parts formula is sometimes seen as $\int udv = uv - \int v du$, as well can be somewhat confusingly written as:
 
@@ -382,7 +382,7 @@ Recall, just using *either* $x_i$ or $x_{i-1}$ for $c_i$ gives an error that is
 This [proof](http://www.math.ucsd.edu/~ebender/20B/77_Trap.pdf) for the error estimate is involved, but is reproduced here, as it nicely integrates many of the theoretical concepts of integration discussed so far.
 
 
-First, for convenience, we consider the interval $x_i$ to $x_i+h$. The actual answer over this is just $\int_{x_i}^{x_i+h}f(x) dx$. By a $u$-substitution with $u=x-x_i$ this becomes $\int_0^h f(t + x_i) dt$. For analyzing this we integrate once by parts using $u=f(t+x_i)$ and $dv=dt$. But instead of letting $v=t$, we choose to add--as is our prerogative--a constant of integration $A$, so $v=t+A$:
+First, for convenience, we consider the interval $x_i$ to $x_i+h$. The actual answer over this is just $\int_{x_i}^{x_i+h}f(x) dx$. By a $u$-substitution with $u=x-x_i$ this becomes $\int_0^h f(t + x_i) dt$. For analyzing this we integrate once by parts using $u=f(t+x_i)$ and $dv=dt$. But instead of letting $v=t$, we choose to add---as is our prerogative---a constant of integration $A$, so $v=t+A$:
 
 
 $$

quarto/integrals/partial_fractions.qmd

@@ -1,4 +1,4 @@
-# Partial Fractions
+# Partial fractions
 
 
 {{< include ../_common_code.qmd >}}
@@ -14,7 +14,7 @@ using SymPy
 Integration is facilitated when an antiderivative for $f$ can be found, as then definite integrals can be evaluated through the fundamental theorem of calculus.
 
 
-However, despite differentiation being an algorithmic procedure, integration is not. There are "tricks" to try, such as substitution and integration by parts. These work in some cases--but not all!
+However, despite differentiation being an algorithmic procedure, integration is not. There are "tricks" to try, such as substitution and integration by parts. These work in some cases---but not all!
 
 However, there are classes of functions for which algorithms exist. For example, the `SymPy` `integrate` function mostly implements an algorithm that decides if an elementary function has an antiderivative. The [elementary](http://en.wikipedia.org/wiki/Elementary_function) functions include exponentials, their inverses (logarithms), trigonometric functions, their inverses, and powers, including $n$th roots. Not every elementary function will have an antiderivative comprised of (finite) combinations of elementary functions. The typical example is $e^{x^2}$, which has no simple antiderivative, despite its ubiquitousness.
 

quarto/integrals/surface_area.qmd

@@ -1,4 +1,4 @@
-# Surface Area
+# Surface area
 
 {{< include ../_common_code.qmd >}}
 

quarto/integrals/twelve-qs.qmd

@@ -14,7 +14,7 @@ using LaTeXStrings
 gr();
 ```
 
-----
+---
 
 In the March 2003 issue of the College Mathematics Journal, Leon M Hall posed 12 questions related to the following figure:
 
@@ -80,7 +80,7 @@ zs = solve(f(x) ~ nl, x)
 q = only(filter(!=(a), zs))
 ```
 
-----
+---
 
 The first question is simply:
 
@@ -115,7 +115,7 @@ In the remaining examples we don't show the code by default.
 
 :::
 
-----
+---
 
 > 1b. The length of the line segment $PQ$
 
@@ -133,7 +133,7 @@ lseg = sqrt((f(a) - f(q))^2 + (a - q)^2);
 ```
 
 
-----
+---
 
 > 2a. The horizontal distance between $P$ and $Q$
 
@@ -151,7 +151,7 @@ plot!([q₀, a₀], [f(a₀), f(a₀)], linewidth=5)
 
 hd = a - q;
 ```
-----
+---
 
 > 2b. The area of the parabolic segment
 
@@ -172,7 +172,7 @@ plot!(xs, ys, fill=(:green, 0.25, 0))
 A = simplify(integrate(nl - f(x), (x, q, a)));
 ```
 
-----
+---
 
 > 2c. The volume of the rotated solid formed by revolving the parabolic segment around the vertical line $k$ units to the right of $P$ or to the left of $Q$ where $k > 0$.
 
@@ -185,7 +185,7 @@ A = simplify(integrate(nl - f(x), (x, q, a)));
 V = simplify(integrate(2PI*(nl-f(x))*(a - x + k),(x, q, a)));
 ```
 
-----
+---
 
 > 3. The $y$ coordinate of the centroid of the parabolic segment
 
@@ -214,7 +214,7 @@ yₘ = integrate( (1//2) * (nl^2 - f(x)^2), (x, q, a)) / A
 yₘ = simplify(yₘ);
 ```
 
-----
+---
 
 > 4. The length of the arc of the parabola between $P$ and $Q$
 
@@ -233,7 +233,7 @@ p
 L = integrate(sqrt(1 + fp(x)^2), (x, q, a));
 ```
 
-----
+---
 
 > 5. The $y$ coordinate of the midpoint of the line segment $PQ$
 
@@ -254,7 +254,7 @@ p
 mp = nl(x => (a + q)/2);
 ```
 
-----
+---
 
 > 6. The area of the trapezoid bound by the normal line, the $x$-axis, and the vertical lines through $P$ and $Q$.
 
@@ -273,7 +273,7 @@ p
 trap = 1//2 * (f(q) + f(a)) * (a - q);
 ```
 
-----
+---
 
 > 7. The area bounded by the parabola and the $x$ axis and the vertical lines through $P$ and $Q$
 
@@ -295,7 +295,7 @@ p
 pa = integrate(x^2, (x, q, a));
 ```
 
-----
+---
 
 > 8. The area of the surface formed by revolving the arc of the parabola between $P$ and $Q$ around the vertical line through $P$
 
@@ -321,7 +321,7 @@ vv(x) = f(a - uu(x))
 SA = 2PI * integrate(uu(x) * sqrt(diff(uu(x),x)^2 + diff(vv(x),x)^2), (x, q, a));
 ```
 
-----
+---
 
 > 9. The height of the parabolic segment (i.e. the distance between the normal line and the tangent line to the parabola that is parallel to the normal line)
 
@@ -350,7 +350,7 @@ segment_height = sqrt((b-b′)^2 + (f(b) - nl(x=>b′))^2);
 ```
 
 
-----
+---
 
 > 10. The volume of the solid formed by revolving the parabolic segment around the $x$-axis
 
@@ -371,7 +371,7 @@ end
 Vₓ = integrate(pi * (nl^2 - f(x)^2), (x, q, a));
 ```
 
-----
+---
 
 > 11. The area of the triangle bound by the normal line, the vertical line through $Q$ and the $x$-axis
 
@@ -392,7 +392,7 @@ plot!([p₀,q₀,q₀,p₀], [0,f(q₀),0,0];
 triangle = 1/2 * f(q) * (a - f(a)/(-1/fp(a)) - q);
 ```
 
-----
+---
 
 > 12. The area of the quadrilateral bound by the normal line, the tangent line, the vertical line through $Q$ and the $x$-axis
 
@@ -417,7 +417,7 @@ x₁,x₂,x₃,x₄ = (a,q,q,tl₀)
 y₁, y₂, y₃, y₄ = (f(a), f(q), 0, 0)
 quadrilateral = (x₁ - x₂)*(y₁ - y₃)/2 - (x₁ - x₃)*(y₁ - y₂)/2 + (x₁ - x₃)*(y₁ - y₄)/2 - (x₁ - x₄)*(y₁ - y₃)/2;
 ```
-----
+---
 
 The answers appear here in sorted order, some given as approximate floating point values: