many edits

quarto/integrals/area.qmd

@@ -369,8 +369,10 @@ Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c
 
 
 \begin{align*}
-f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1)  + \cdots + f(c_n -c) \cdot (x_n - x_{n-1}) = \\
-& f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
+f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1)  + \cdots\\
+ &\quad + f(c_n -c) \cdot (x_n - x_{n-1})\\
+ &= f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots\\
+ &\quad + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
 \end{align*}
 
 
@@ -636,15 +638,15 @@ With this, we can easily find an approximate answer. We wrote the function to us
 
 
 ```{julia}
-𝒇(x) = exp(x)
-riemann(𝒇, 0, 5, 10)   # S_10
+f(x) = exp(x)
+riemann(f, 0, 5, 10)   # S_10
 ```
 
 Or with more intervals in the partition
 
 
 ```{julia}
-riemann(𝒇, 0, 5, 50_000)
+riemann(f, 0, 5, 50_000)
 ```
 
 (The answer is $e^5 - e^0 = 147.4131591025766\dots$, which shows that even $50,000$ partitions is not enough to guarantee many digits of accuracy.)
@@ -719,7 +721,7 @@ We have to be a bit careful with the Riemann sum, as the left Riemann sum will h
 
 
 ```{julia}
-𝒉(x) = x > 0 ? x * log(x) : 0.0
+h(x) = x > 0 ? x * log(x) : 0.0
 ```
 
 This is actually inefficient, as the check for the size of `x` will slow things down a bit. Since we will call this function 50,000 times, we would like to avoid this, if we can. In this case just using the right sum will work:
@@ -910,7 +912,8 @@ $$
 $$
 
 ```{julia}
-quadgk(x -> x^x, 0, 2)
+u(x) = x^x
+quadgk(u, 0, 2)
 ```
 
 $$

quarto/integrals/area_between_curves.qmd

@@ -79,14 +79,15 @@ For this problem we need to identify $a$ and $b$. These are found numerically th
 
 
 ```{julia}
-a,b = find_zeros(x -> f(x) - g(x), -3, 3)
+h(x) = f(x) - g(x)
+a,b = find_zeros(h, -3, 3)
 ```
 
 The answer then can be found numerically:
 
 
 ```{julia}
-quadgk(x -> f(x) - g(x), a, b)[1]
+first(quadgk(h, a, b))
 ```
 
 ##### Example
@@ -99,18 +100,18 @@ A plot shows the areas:
 
 
 ```{julia}
-𝒇(x) = sin(x)
-𝒈(x) = cos(x)
-plot(𝒇, 0, 2pi)
-plot!(𝒈)
+f(x) = sin(x)
+g(x) = cos(x)
+plot(f, 0, 2pi)
+plot!(g)
 ```
 
 There is a single interval when $f \geq g$ and this can be found algebraically using basic trigonometry, or numerically:
 
 
 ```{julia}
-𝒂,𝒃 = find_zeros(x -> 𝒇(x) - 𝒈(x), 0, 2pi)  # pi/4, 5pi/4
-quadgk(x -> 𝒇(x) - 𝒈(x), 𝒂, 𝒃)[1]
+a, b = find_zeros(x -> f(x) - g(x), 0, 2pi)  # pi/4, 5pi/4
+quadgk(x -> f(x) - g(x), a, b)[1]
 ```
 
 ##### Example
@@ -177,15 +178,15 @@ For concreteness, let $f(x) = 2-x^2$ and $[a,b] = [-1, 1/2]$, as in the figure.
 
 
 ```{julia}
-𝐟(x) = 2 - x^2
-𝐚, 𝐛 = -1, 1/2
-𝐜 = (𝐚 + 𝐛)/2
+f(x) = 2 - x^2
+a, b = -1, 1/2
+𝐜 = (a + b)/2
 
-sac, sab, scb = secant(𝐟, 𝐚, 𝐜), secant(𝐟, 𝐚, 𝐛), secant(𝐟, 𝐜, 𝐛)
+sac, sab, scb = secant(f, a, 𝐜), secant(f, a, b), secant(f, 𝐜, b)
 f1(x) = min(sac(x), scb(x))
 f2(x) = sab(x)
 
-A1 = quadgk(x -> f1(x) - f2(x), 𝐚, 𝐛)[1]
+A1 = quadgk(x -> f1(x) - f2(x), a, b)[1]
 ```
 
 As we needed three secant lines, we used the `secant` function from `CalculusWithJulia` to create functions representing each. Once that was done, we used the `min` function to facilitate integrating over the top bounding curve, alternatively, we could break the integral over $[a,c]$ and $[c,b]$.
@@ -195,7 +196,7 @@ The area of the parabolic segment is more straightforward.
 
 
 ```{julia}
-A2 = quadgk(x -> 𝐟(x) - f2(x), 𝐚, 𝐛)[1]
+A2 = quadgk(x -> f(x) - f2(x), a, b)[1]
 ```
 
 Finally, if Archimedes was right, this relationship should bring about $0$ (or something within round-off error):

quarto/integrals/ftc.qmd

@@ -202,7 +202,7 @@ $$
 \int_{-\pi/2}^{\pi/2} \cos(x) dx = F(\pi/2) - F(-\pi/2) = 1 - (-1) = 2.
 $$
 
-### An alternate notation for $F(b) - F(a)$
+### An alternate notation
 
 
 The expression $F(b) - F(a)$ is often written in this more compact form:
@@ -617,7 +617,7 @@ We need to figure out when this is $0$. For that, we use some numeric math.
 
 ```{julia}
 F(x) = exp(-x^2) * quadgk(t -> exp(t^2), 0, x)[1]
-Fp(x) = -2x*F(x) + 1
+Fp(x) = -2x * F(x) + 1
 cps = find_zeros(Fp, -4, 4)
 ```
 

quarto/integrals/improper_integrals.qmd

@@ -34,8 +34,9 @@ function make_sqrt_x_graph(n)
     a = 1/2^n
     xs = range(1/2^8, stop=b, length=250)
     x1s = range(a, stop=b, length=50)
+    @syms x
     f(x) = 1/sqrt(x)
-    val = N(integrate(f, 1/2^n, b))
+    val = N(integrate(f(x), (x, 1/2^n, b)))
     title = "area under f over [1/$(2^n), $b] is $(rpad(round(val, digits=2), 4))"
 
     plt = plot(f, range(a, stop=b, length=251), xlim=(0,b), ylim=(0, 15), legend=false, size=fig_size, title=title)

quarto/integrals/integration_by_parts.qmd

@@ -440,10 +440,10 @@ Apply the formula to a parameterized circle to ensure, the signed area is proper
 
 ```{julia}
 #| hold: true
-@syms 𝒓 t
-𝒙 = 𝒓 * cos(t)
-𝒚 = 𝒓 * sin(t)
--integrate(𝒚 * diff(𝒙, t), (t, 0, 2PI))
+@syms r t
+x = r * cos(t)
+y = r * sin(t)
+-integrate(y * diff(x, t), (t, 0, 2PI))
 ```
 
 We see the expected answer for the area of a circle.
@@ -461,9 +461,9 @@ Working symbolically, we have one arch given by the following described in a *cl
 ```{julia}
 #| hold: true
 @syms t
-𝒙 = t - sin(t)
-𝒚 = 1 - cos(t)
-integrate(𝒚 * diff(𝒙, t), (t, 0, 2PI))
+x = t - sin(t)
+y = 1 - cos(t)
+integrate(y * diff(x, t), (t, 0, 2PI))
 ```
 
 ([Galileo](https://mathshistory.st-andrews.ac.uk/Curves/Cycloid/) was thwarted in finding this answer exactly and resorted to constructing one from metal to *estimate* the value.)

quarto/integrals/partial_fractions.qmd

@@ -261,15 +261,15 @@ We again just let `SymPy` do the work. A partial fraction decomposition is given
 
 
 ```{julia}
-𝒒 =  (x^2 - 2x - 3)
-apart(1/𝒒)
+q =  (x^2 - 2x - 3)
+apart(1/q)
 ```
 
 We see what should yield two logarithmic terms:
 
 
 ```{julia}
-integrate(1/𝒒, x)
+integrate(1/q, x)
 ```
 
 :::{.callout-note}