align fix; theorem style; condition number

quarto/integrals/arc_length.qmd

@@ -48,14 +48,18 @@ Recall the distance formula gives the distance between two points: $\sqrt{(x_1 -
 
 Consider now two functions $g(t)$ and $f(t)$ and the parameterized graph between $a$ and $b$ given by the points $(g(t), f(t))$ for $a \leq t \leq b$. Assume that both $g$ and $f$ are differentiable on $(a,b)$ and continuous on $[a,b]$ and furthermore that $\sqrt{g'(t)^2 + f'(t)^2}$ is Riemann integrable.
 
+::: {.callout-note icon=false}
+## The arc length of a curve
 
-> **The arc length of a curve**. For $f$ and $g$ as described, the arc length of the parameterized curve is given by
->
-> $L = \int_a^b \sqrt{g'(t)^2 + f'(t)^2} dt.$
->
-> For the special case of the graph of a function $f(x)$ between $a$ and $b$ the formula becomes $L = \int_a^b \sqrt{ 1 + f'(x)^2} dx$ (taking $g(t) = t$).
+For $f$ and $g$ as described, the arc length of the parameterized curve is given by
 
+$$
+L = \int_a^b \sqrt{g'(t)^2 + f'(t)^2} dt.
+$$
 
+For the special case of the graph of a function $f(x)$ between $a$ and $b$ the formula becomes $L = \int_a^b \sqrt{ 1 + f'(x)^2} dx$ (taking $g(t) = t$).
+
+:::
 
 :::{.callout-note}
 ## Note
@@ -126,7 +130,7 @@ $$
 But looking at each term, we can push the denominator into the square root as:
 
 
-
+$$
 \begin{align*}
 d_i &= d_i \cdot \frac{t_i - t_{i-1}}{t_i - t_{i-1}}
 \\
@@ -134,6 +138,7 @@ d_i &= d_i \cdot \frac{t_i - t_{i-1}}{t_i - t_{i-1}}
 \left(\frac{f(t_i)-f(t_{i-1})}{t_i-t_{i-1}}\right)^2} \cdot (t_i - t_{i-1}) \\
 &= \sqrt{ g'(\xi_i)^2 + f'(\psi_i)^2} \cdot (t_i - t_{i-1}).
 \end{align*}
+$$
 
 
 The values $\xi_i$ and $\psi_i$ are guaranteed by the mean value theorem and must be in $[t_{i-1}, t_i]$.
@@ -272,7 +277,7 @@ nothing
 The museum notes have
 
 
-> For his Catenary series (1997–2003), of which Near the Lagoon is    the largest and last work, Johns formed catenaries—a term used to    describe the curve assumed by a cord suspended freely from two    points—by tacking ordinary household string to the canvas or its    supports.
+> For his Catenary series (1997–2003), of which Near the Lagoon is    the largest and last work, Johns formed catenaries—a term used to    describe the curve assumed by a cord suspended freely from two points—by tacking ordinary household string to the canvas or its    supports.
 
 
 
@@ -377,11 +382,12 @@ nothing
 The [nephroid](http://www-history.mcs.st-and.ac.uk/Curves/Nephroid.html) is a curve that can be described parametrically by
 
 
-
+$$
 \begin{align*}
 g(t) &= a(3\cos(t) - \cos(3t)), \\
 f(t) &= a(3\sin(t) - \sin(3t)).
 \end{align*}
+$$
 
 
 Taking $a=1$ we have this graph:
@@ -407,7 +413,7 @@ quadgk(t -> sqrt(𝒈'(t)^2 + 𝒇'(t)^2), 0, 2pi)[1]
 The answer seems like a floating point approximation of $24$, which  suggests that  this integral is tractable. Pursuing this, the integrand simplifies:
 
 
-
+$$
 \begin{align*}
 \sqrt{g'(t)^2 + f'(t)^2}
 &= \sqrt{(-3\sin(t) + 3\sin(3t))^2 + (3\cos(t) - 3\cos(3t))^2} \\
@@ -417,6 +423,7 @@ The answer seems like a floating point approximation of $24$, which  suggests th
 &= 3\sqrt{2}\sqrt{1 - \cos(2t)}\\
 &= 3\sqrt{2}\sqrt{2\sin(t)^2}.
 \end{align*}
+$$
 
 
 The second to last line comes from a double angle formula expansion of $\cos(3t - t)$ and the last line from the half angle formula for $\cos$.
@@ -452,13 +459,14 @@ A teacher of small children assigns his students the task of computing the lengt
 Mathematically, suppose a curve is described parametrically by $(g(t), f(t))$ for $a \leq t \leq b$. A new parameterization is provided by $\gamma(t)$. Suppose $\gamma$ is strictly increasing, so that an inverse function exists. (This assumption is implicitly made by the teacher, as it implies the student won't start counting in the wrong direction.) Then the same curve is described by composition through $(g(\gamma(u)), f(\gamma(u)))$, $\gamma^{-1}(a) \leq u \leq \gamma^{-1}(b)$. That the arc length is the same follows from substitution:
 
 
-
+$$
 \begin{align*}
 \int_{\gamma^{-1}(a)}^{\gamma^{-1}(b)} \sqrt{([g(\gamma(t))]')^2 + ([f(\gamma(t))]')^2} dt
 &=\int_{\gamma^{-1}(a)}^{\gamma^{-1}(b)} \sqrt{(g'(\gamma(t) )\gamma'(t))^2 + (f'(\gamma(t) )\gamma'(t))^2 } dt \\
 &=\int_{\gamma^{-1}(a)}^{\gamma^{-1}(b)} \sqrt{g'(\gamma(t))^2 + f'(\gamma(t))^2} \gamma'(t) dt\\
 &=\int_a^b \sqrt{g'(u)^2 + f'(u)^2} du = L
 \end{align*}
+$$
 
 
 (Using $u=\gamma(t)$ for the substitution.)
@@ -483,12 +491,13 @@ For a simple example, we have $g(t) = R\cos(t)$ and $f(t)=R\sin(t)$ parameterizi
 What looks at first glance to be just a slightly more complicated equation is that of an ellipse, with $g(t) = a\cos(t)$ and $f(t) = b\sin(t)$. Taking $a=1$ and $b = a + c$, for $c > 0$ we get the equation for the arc length as a function of $t$ is just
 
 
-
+$$
 \begin{align*}
 s(u) &= \int_0^u \sqrt{(-\sin(t))^2 + b\cos(t)^2} dt\\
      &= \int_0^u \sqrt{\sin(t)^2 + \cos(t)^2 + c\cos(t)^2} dt  \\
 	 &=\int_0^u \sqrt{1 + c\cos(t)^2} dt.
 \end{align*}
+$$
 
 
 But, despite it not looking too daunting, this integral is not tractable through our techniques and has an answer involving elliptic integrals. We can work numerically though. Letting $a=1$ and $b=2$, we have the arc length is given by:
@@ -588,11 +597,12 @@ $$
 So
 
 
-
+$$
 \begin{align*}
 \int_0^1 (tf'(u) + (1-t)f'(v)) dt &< \int_0^1 f'(tu + (1-t)v) dt, \text{or}\\
 \frac{f'(u) + f'(v)}{2} &< \frac{1}{v-u}\int_u^v f'(w) dw,
 \end{align*}
+$$
 
 
 by the substitution $w = tu + (1-t)v$. Using the fundamental theorem of calculus to compute the mean value of the integral of $f'$ over $[u,v]$ gives the following as a consequence of strict concavity of $f'$:
@@ -684,24 +694,26 @@ which holds by the strict concavity of $f'$, as found previously.
 Using the substitution $x = f_i^{-1}(u)$ as needed to see:
 
 
-
+$$
 \begin{align*}
 \int_a^u f(x) dx &= \int_0^{f(u)} u [f_1^{-1}]'(u) du \\
 &> -\int_0^h u [f_2^{-1}]'(u) du \\
 &= \int_h^0 u [f_2^{-1}]'(u) du \\
 &= \int_v^b f(x) dx.
 \end{align*}
+$$
 
 
 For the latter claim, integrating in the $y$ variable gives
 
 
-
+$$
 \begin{align*}
 \int_u^c (f(x)-h) dx &= \int_h^m (c - f_1^{-1}(y)) dy\\
 &> \int_h^m (c - f_2^{-1}(y)) dy\\
 &= \int_c^v (f(x)-h) dx
 \end{align*}
+$$
 
 
 Now, the area under $h$ over $[u,c]$ is greater than that over $[c,v]$ as $(u+v)/2 < c$ or $v-c < c-u$. That means the area under $f$ over $[u,c]$ is greater than that over $[c,v]$.
@@ -724,7 +736,7 @@ or $\phi'(z) < 0$. Moreover, we have by the first assertion that $f'(z) < -f'(\p
 Using the substitution $x = \phi(z)$ gives:
 
 
-
+$$
 \begin{align*}
 \int_v^b \sqrt{1 + f'(x)^2} dx &=
 \int_u^a \sqrt{1 + f'(\phi(z))^2} \phi'(z) dz\\
@@ -733,6 +745,7 @@ Using the substitution $x = \phi(z)$ gives:
 &= \int_a^u \sqrt{\phi'(z)^2 + f'(z)^2} dz\\
 &< \int_a^u \sqrt{1 + f'(z)^2} dz
 \end{align*}
+$$
 
 
 Letting $h=f(u \rightarrow c)$ we get the *inequality*
@@ -782,11 +795,12 @@ $$
 with the case above corresponding to $W = -m(k/m)$. The set of equations then satisfy:
 
 
-
+$$
 \begin{align*}
 x''(t) &= - W(t,x(t), x'(t), y(t), y'(t)) \cdot x'(t)\\
 y''(t) &= -g - W(t,x(t), x'(t), y(t), y'(t)) \cdot y'(t)\\
 \end{align*}
+$$
 
 
 with initial conditions: $x(0) = y(0) = 0$ and $x'(0) = v_0 \cos(\theta), y'(0) = v_0 \sin(\theta)$.
@@ -795,28 +809,30 @@ with initial conditions: $x(0) = y(0) = 0$ and $x'(0) = v_0 \cos(\theta), y'(0)
 Only with certain drag forces, can this set of equations be solved exactly, though it can be approximated numerically for admissible $W$, but if $W$ is strictly positive then it can be shown $x(t)$ is increasing on $[0, x_\infty)$ and so invertible, and $f(u) = y(x^{-1}(u))$ is three times differentiable with both $f$ and $f'$ being strictly concave, as it can be shown that (say $x(v) = u$ so $dv/du = 1/x'(v) > 0$):
 
 
-
+$$
 \begin{align*}
 f''(u) &= -\frac{g}{x'(v)^2} < 0\\
 f'''(u) &= \frac{2gx''(v)}{x'(v)^3} \\
 &= -\frac{2gW}{x'(v)^2}  \cdot \frac{dv}{du} < 0
 \end{align*}
+$$
 
 
 The latter by differentiating, the former a consequence of the following formulas for derivatives of inverse functions
 
 
-
+$$
 \begin{align*}
 [x^{-1}]'(u) &= 1 / x'(v) \\
 [x^{-1}]''(u) &= -x''(v)/(x'(v))^3
 \end{align*}
+$$
 
 
 For then
 
 
-
+$$
 \begin{align*}
 f(u)   &= y(x^{-1}(u)) \\
 f'(u)  &= y'(x^{-1}(u)) \cdot {x^{-1}}'(u) \\
@@ -825,6 +841,7 @@ f''(u) &= y''(x^{-1}(u))\cdot[x^{-1}]'(u)^2 + y'(x^{-1}(u)) \cdot [x^{-1}]''(u)
        &= -g/(x'(v))^2 - W y'/(x'(v))^2 - y'(v) \cdot (- W \cdot x'(v)) / x'(v)^3\\
        &= -g/x'(v)^2.
 \end{align*}
+$$
 
 
 ## Questions

quarto/integrals/area.qmd

@@ -230,7 +230,7 @@ To successfully compute a good approximation for the area, we would need to choo
 For Archimedes' problem - finding the area under $f(x)=x^2$ between $0$ and $1$ -  if we take as a partition $x_i = i/n$ and $c_i = x_i$, then the above sum becomes:
 
 
-
+$$
 \begin{align*}
 S_n &= f(c_1) \cdot (x_1 - x_0) + f(c_2) \cdot (x_2 - x_1) + \cdots + f(c_n) \cdot (x_n - x_{n-1})\\
 &= (x_1)^2 \cdot \frac{1}{n} + (x_2)^2 \cdot \frac{1}{n} + \cdots + (x_n)^2 \cdot \frac{1}{n}\\
@@ -238,6 +238,7 @@ S_n &= f(c_1) \cdot (x_1 - x_0) + f(c_2) \cdot (x_2 - x_1) + \cdots + f(c_n) \cd
 &= \frac{1}{n^3} \cdot (1^2 + 2^2 + \cdots + n^2) \\
 &= \frac{1}{n^3} \cdot \frac{n\cdot(n-1)\cdot(2n+1)}{6}.
 \end{align*}
+$$
 
 
 The latter uses a well-known formula for the sum of squares of the first $n$ natural numbers.
@@ -301,13 +302,18 @@ The general statement allows for any partition such that the largest gap goes to
 Riemann sums weren't named after Riemann because he was the first to approximate areas using rectangles. Indeed, others had been using even more efficient ways to compute areas for  centuries prior to Riemann's work. Rather, Riemann put the definition of the area under the curve on a firm theoretical footing with the following theorem which gives a concrete notion of what functions are integrable:
 
 
-> **Riemann Integral**: A function $f$ is Riemann integrable over the interval $[a,b]$ and its integral will have value $V$ provided for every $\epsilon > 0$ there exists a $\delta > 0$ such that for any partition $a =x_0 < x_1 < \cdots < x_n=b$ with $\lvert x_i - x_{i-1} \rvert < \delta$ and for any choice of points $x_{i-1} \leq c_i \leq x_{i}$ this is satisfied:
->
-> $$
-> \lvert \sum_{i=1}^n f(c_i)(x_{i} - x_{i-1}) - V \rvert < \epsilon.
-> $$
->
-> When the integral exists, it is written $V = \int_a^b f(x) dx$.
+::: {.callout-note icon=false}
+## Riemann Integral
+
+A function $f$ is Riemann integrable over the interval $[a,b]$ and its integral will have value $V$ provided for every $\epsilon > 0$ there exists a $\delta > 0$ such that for any partition $a =x_0 < x_1 < \cdots < x_n=b$ with $\lvert x_i - x_{i-1} \rvert < \delta$ and for any choice of points $x_{i-1} \leq c_i \leq x_{i}$ this is satisfied:
+
+$$
+\lvert \sum_{i=1}^n f(c_i)(x_{i} - x_{i-1}) - V \rvert < \epsilon.
+$$
+
+When the integral exists, it is written $V = \int_a^b f(x) dx$.
+
+:::
 
 
 
@@ -370,13 +376,14 @@ The area is invariant under shifts left or right.
 
 Any partition $a =x_0 < x_1 < \cdots < x_n=b$ is related to a partition of $[a-c, b-c]$ through $a-c < x_0-c < x_1-c < \cdots < x_n - c = b-c$. Let $d_i=c_i-c$ denote this partition, then we have:
 
-
+$$
 \begin{align*}
 f(c_1 -c) \cdot (x_1 - x_0) &+ f(c_2 -c) \cdot (x_2 - x_1)  + \cdots\\
  &\quad + f(c_n -c) \cdot (x_n - x_{n-1})\\
  &= f(d_1) \cdot(x_1-c - (x_0-c)) + f(d_2) \cdot(x_2-c - (x_1-c)) + \cdots\\
  &\quad + f(d_n) \cdot(x_n-c - (x_{n-1}-c)).
 \end{align*}
+$$
 
 
 The left side will have a limit of $\int_a^b f(x-c) dx$ the right would have a "limit" of $\int_{a-c}^{b-c}f(x)dx$.
@@ -471,7 +478,7 @@ Using the definition, we can compute a few definite integrals:
 This is just the area of a trapezoid with heights $a$ and $b$ and side   length $b-a$, or $1/2 \cdot (b + a) \cdot (b - a)$. The right sum   would be:
 
 
-
+$$
 \begin{align*}
 S &= x_1 \cdot (x_1 - x_0) + x_2 \cdot (x_2 - x_1) + \cdots x_n \cdot (x_n - x_{n-1}) \\
 &= (a + 1\frac{b-a}{n}) \cdot \frac{b-a}{n} + (a + 2\frac{b-a}{n}) \cdot \frac{b-a}{n} + \cdots  (a + n\frac{b-a}{n}) \cdot \frac{b-a}{n}\\
@@ -480,6 +487,7 @@ S &= x_1 \cdot (x_1 - x_0) + x_2 \cdot (x_2 - x_1) + \cdots x_n \cdot (x_n - x_{
 & \rightarrow a \cdot(b-a) + \frac{(b-a)^2}{2} \\
 &= \frac{b^2}{2} - \frac{a^2}{2}.
 \end{align*}
+$$
 
 
 > $$
@@ -502,7 +510,7 @@ This is similar to the Archimedes case with $a=0$ and $b=1$ shown above.
 Cauchy showed this using a *geometric series* for the partition, not the arithmetic series $x_i = a + i (b-a)/n$. The series defined by $1 + \alpha = (b/a)^{1/n}$, then $x_i = a \cdot (1 + \alpha)^i$. Here the bases $x_{i+1} - x_i$ simplify to $x_i \cdot \alpha$ and $f(x_i) = (a\cdot(1+\alpha)^i)^k = a^k (1+\alpha)^{ik}$, or $f(x_i)(x_{i+1}-x_i) = a^{k+1}\alpha[(1+\alpha)^{k+1}]^i$, so, using $u=(1+\alpha)^{k+1}=(b/a)^{(k+1)/n}$, $f(x_i) \cdot(x_{i+1} - x_i) = a^{k+1}\alpha u^i$. This gives
 
 
-
+$$
 \begin{align*}
 S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
 &= a^{k+1} \cdot \alpha \cdot (u^0 + u^1 + \cdot u^{n-1}) \\
@@ -510,6 +518,7 @@ S &= a^{k+1}\alpha u^0 + a^{k+1}\alpha u^1 + \cdots + a^{k+1}\alpha u^{n-1}\\
 &= (b^{k+1} - a^{k+1}) \cdot \frac{\alpha}{(1+\alpha)^{k+1} - 1} \\
 &\rightarrow \frac{b^{k+1} - a^{k+1}}{k+1}.
 \end{align*}
+$$
 
 
 > $$
@@ -541,9 +550,12 @@ Certainly other integrals could be computed with various tricks, but we won't pu
 ### Some other consequences
 
 
-  * The definition is defined in terms of any partition with its norm bounded by $\delta$. If you know a function $f$ is Riemann integrable, then it is enough to consider just a regular partition $x_i = a + i \cdot (b-a)/n$ when forming the sums, as was done above. It is just that showing a limit for just this particular type of partition would not be sufficient to prove Riemann integrability.
-  * The choice of $c_i$ is arbitrary to allow for maximum flexibility. The Darboux integrals use the maximum and minimum over the subinterval. It is sufficient to prove integrability to show that the limit exists with just these choices.
-  * Most importantly,
+* The definition is defined in terms of any partition with its norm bounded by $\delta$. If you know a function $f$ is Riemann integrable, then it is enough to consider just a regular partition $x_i = a + i \cdot (b-a)/n$ when forming the sums, as was done above. It is just that showing a limit for just this particular type of partition would not be sufficient to prove Riemann integrability.
+
+* The choice of $c_i$ is arbitrary to allow for maximum flexibility. The Darboux integrals use the maximum and minimum over the subinterval. It is sufficient to prove integrability to show that the limit exists with just these choices.
+
+
+* Most importantly,
 
 
 > A continuous function on $[a,b]$ is Riemann integrable on $[a,b]$.
@@ -553,13 +565,13 @@ Certainly other integrals could be computed with various tricks, but we won't pu
 The main idea behind this is that the difference between the maximum and minimum values over a partition gets small. That is if $[x_{i-1}, x_i]$ is like $1/n$ is length, then the difference between the maximum of $f$ over this interval, $M$, and the minimum, $m$ over this interval will go to zero as $n$ gets big. That $m$ and $M$ exists is due to the extreme value theorem, that this difference goes to $0$ is a consequence of continuity. What is needed is that this value goes to $0$ at the same rate – no matter what interval is being discussed – is a consequence of a notion of uniform continuity, a concept discussed in advanced calculus, but which holds for continuous functions on closed intervals. Armed with this, the Riemann sum for a general partition can be bounded by this difference times $b-a$, which will go to zero. So the upper and lower Riemann sums will converge to the same value.
 
 
-  * A "jump", or discontinuity of the first kind, is a value $c$ in $[a,b]$ where $\lim_{x \rightarrow c+} f(x)$ and $\lim_{x \rightarrow c-}f(x)$ both exist, but are not equal. It is true that a function that is not continuous on $I=[a,b]$, but only has discontinuities of the first kind on $I$ will be Riemann integrable on $I$.
+* A "jump", or discontinuity of the first kind, is a value $c$ in $[a,b]$ where $\lim_{x \rightarrow c+} f(x)$ and $\lim_{x \rightarrow c-}f(x)$ both exist, but are not equal. It is true that a function that is not continuous on $I=[a,b]$, but only has discontinuities of the first kind on $I$ will be Riemann integrable on $I$.
 
 
 For example, the function $f(x) = 1$ for $x$ in $[0,1]$ and $0$ otherwise will be integrable, as it is continuous at all but two points, $0$ and $1$, where it jumps.
 
 
-  * Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used as an example.
+* Some functions can have infinitely many points of discontinuity and still be integrable. The example of $f(x) = 1/q$ when $x=p/q$ is rational, and $0$ otherwise is often used as an example.
 
 
 ## Numeric integration
@@ -799,7 +811,7 @@ While such bounds are disappointing, often, when looking for specific values, th
 The Riemann sum above is actually extremely inefficient. To see how much, we can derive an estimate for the error in approximating the value using an arithmetic progression as the partition. Let's assume that our function $f(x)$ is increasing, so that the right sum gives an upper estimate and the left sum a lower estimate, so the error in the estimate will be between these two values:
 
 
-
+$$
 \begin{align*}
 \text{error} &\leq
 \left[
@@ -809,6 +821,7 @@ f(x_1) \cdot (x_{1} - x_0)  + f(x_2) \cdot  (x_{2} - x_1) + \cdots + f(x_{n-1})(
 &= \frac{b-a}{n} \cdot (\left[f(x_1) + f(x_2) + \cdots + f(x_n)\right] - \left[f(x_0) + \cdots + f(x_{n-1})\right]) \\
 &= \frac{b-a}{n} \cdot (f(b) - f(a)).
 \end{align*}
+$$
 
 
 We see the error goes to $0$ at a rate of $1/n$ with the constant depending on $b-a$ and the function $f$. In general, a similar bound holds when $f$ is not monotonic.
@@ -853,11 +866,12 @@ This formula will actually be exact for any 3rd degree polynomial. In fact an en
 The formulas for an approximation to the integral $\int_{-1}^1 f(x) dx$ discussed so far can be written as:
 
 
-
+$$
 \begin{align*}
 S &= f(x_1) \Delta_1 + f(x_2) \Delta_2 + \cdots + f(x_n) \Delta_n\\
   &= w_1 f(x_1) + w_2 f(x_2) + \cdots + w_n f(x_n).
 \end{align*}
+$$
 
 
 The $w$s are "weights" and the $x$s are nodes. A [Gaussian](http://en.wikipedia.org/wiki/Gaussian_quadrature) *quadrature rule* is a set of weights and nodes for $i=1, \dots n$ for which the sum is *exact* for any $f$ which is a polynomial of degree $2n-1$ or less. Such choices then also approximate well the integrals of functions which are not polynomials of degree $2n-1$, provided $f$ can be well approximated by a polynomial over $[-1,1]$. (Which is the case for the "nice" functions we encounter.) Some examples are given in the questions.

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.

quarto/integrals/ftc.qmd

@@ -100,7 +100,7 @@ where we define $g(i) = f(a + ih)h$. In the above, $n$ relates to $b$, but we co
 Again, we fix a large $n$ and let $h=(b-a)/n$. And suppose $x = a + Mh$ for some $M$. Then writing out the approximations to both the definite integral and the derivative we have
 
 
-
+$$
 \begin{align*}
 F'(x) = & \frac{d}{dx} \int_a^x f(u) du \\
 & \approx \frac{F(x) - F(x-h)}{h} \\
@@ -113,17 +113,19 @@ F'(x) = & \frac{d}{dx} \int_a^x f(u) du \\
 \left(f(a + 1h) + f(a + 2h) + \cdots + f(a + (M-1)h) \right) \\
 &= f(a + Mh).
 \end{align*}
+$$
 
 
 If $g(i) = f(a + ih)$, then the above becomes
 
 
-
+$$
 \begin{align*}
 F'(x) & \approx D(S(g))(M) \\
 &= f(a + Mh)\\
 &= f(x).
 \end{align*}
+$$
 
 
 That is $F'(x) \approx f(x)$.
@@ -138,13 +140,14 @@ $$
 
 With these heuristics, we now have:
 
+::: {.callout-note icon=false}
+## The fundamental theorem of calculus
 
-> **The fundamental theorem of calculus**
->
-> Part 1: Let $f$ be a continuous function on a closed interval $[a,b]$ and define $F(x) = \int_a^x f(u) du$ for $a \leq x \leq b$. Then $F$ is continuous on $[a,b]$, differentiable on $(a,b)$ and moreover, $F'(x) =f(x)$.
->
-> Part 2: Now suppose $f$ is any integrable function on a closed interval $[a,b]$ and $F(x)$ is *any* differentiable function on $[a,b]$ with $F'(x) = f(x)$. Then $\int_a^b f(x)dx=F(b)-F(a)$.
+Part 1: Let $f$ be a continuous function on a closed interval $[a,b]$ and define $F(x) = \int_a^x f(u) du$ for $a \leq x \leq b$. Then $F$ is continuous on $[a,b]$, differentiable on $(a,b)$ and moreover, $F'(x) =f(x)$.
 
+Part 2: Now suppose $f$ is any integrable function on a closed interval $[a,b]$ and $F(x)$ is *any* differentiable function on $[a,b]$ with $F'(x) = f(x)$. Then $\int_a^b f(x)dx=F(b)-F(a)$.
+
+:::
 
 
 :::{.callout-note}
@@ -366,25 +369,27 @@ This statement is nothing more than the derivative formula $[cf(x) + dg(x)]' = c
   * The antiderivative of the polynomial $p(x) = a_n x^n + \cdots + a_1 x + a_0$ follows from the linearity of the integral and the general power rule:
 
 
-
+$$
 \begin{align*}
 \int (a_n x^n + \cdots + a_1 x + a_0) dx
 &= \int a_nx^n dx + \cdots + \int a_1 x dx + \int a_0 dx                   \\
 &= a_n \int x^n dx + \cdots +  a_1 \int x dx + a_0 \int dx                   \\
 &= a_n\frac{x^{n+1}}{n+1} + \cdots +  a_1 \frac{x^2}{2} +  a_0 \frac{x}{1}.
 \end{align*}
+$$
 
 
   * More generally, a [Laurent](https://en.wikipedia.org/wiki/Laurent_polynomial) polynomial allows for terms with negative powers. These too can be handled by the above. For example
 
 
-
+$$
 \begin{align*}
 \int (\frac{2}{x} + 2 + 2x) dx
 &= \int \frac{2}{x} dx + \int 2 dx + \int 2x dx \\
 &= 2\int \frac{1}{x} dx + 2 \int dx + 2 \int xdx\\
 &= 2\log(x) + 2x + 2\frac{x^2}{2}.
 \end{align*}
+$$
 
 
   * Consider this integral:
@@ -645,13 +650,14 @@ Under assumptions that the $X$ are identical and independent, the largest value,
 This problem is constructed to take advantage of the FTC, and we have:
 
 
-
+$$
 \begin{align*}
 \left[P(M \leq a)\right]'
 &= \left[F(a)^n\right]'\\
 &= n \cdot F(a)^{n-1} \left[F(a)\right]'\\
 &= n F(a)^{n-1}f(a)
 \end{align*}
+$$
 
 
 ##### Example

quarto/integrals/integration_by_parts.qmd

@@ -95,7 +95,7 @@ An illustration can clarify.
 Consider the integral $\int_0^\pi x\sin(x) dx$. If we let $u=x$ and $dv=\sin(x) dx$, then $du = 1dx$ and $v=-\cos(x)$. The above then says:
 
 
-
+$$
 \begin{align*}
 \int_0^\pi x\sin(x) dx &= \int_0^\pi u dv\\
 &= uv\big|_0^\pi - \int_0^\pi v du\\
@@ -104,6 +104,7 @@ Consider the integral $\int_0^\pi x\sin(x) dx$. If we let $u=x$ and $dv=\sin(x)
 &= \pi + \sin(x)\big|_0^\pi\\
 &= \pi.
 \end{align*}
+$$
 
 
 The technique means one part is differentiated and one part integrated. The art is to break the integrand up into a piece that gets easier through differentiation and a piece that doesn't get much harder through integration.
@@ -129,7 +130,7 @@ $$
 Putting together gives:
 
 
-
+$$
 \begin{align*}
 \int_1^2 x \log(x) dx
 &= (\log(x) \cdot \frac{x^2}{2}) \big|_1^2 - \int_1^2 \frac{x^2}{2} \frac{1}{x} dx\\
@@ -137,6 +138,7 @@ Putting together gives:
 &= 2\log(2) - (1 - \frac{1}{4}) \\
 &= 2\log(2) - \frac{3}{4}.
 \end{align*}
+$$
 
 
 ##### Example
@@ -145,14 +147,15 @@ Putting together gives:
 This related problem, $\int \log(x) dx$, uses the same idea, though perhaps harder to see at first glance, as setting `dv=dx` is almost too simple to try:
 
 
-
+$$
 \begin{align*}
 u &= \log(x) & dv &= dx\\
 du &= \frac{1}{x}dx & v &= x
 \end{align*}
+$$
 
 
-
+$$
 \begin{align*}
 \int \log(x) dx
 &= \int u dv\\
@@ -161,6 +164,7 @@ du &= \frac{1}{x}dx & v &= x
 &= x \log(x) - \int dx\\
 &= x \log(x) - x
 \end{align*}
+$$
 
 
 Were this a definite integral problem, we would have written:
@@ -244,13 +248,14 @@ $$
 Positive integer powers of trigonometric functions can be addressed by this technique. Consider $\int \cos(x)^n dx$. We let $u=\cos(x)^{n-1}$ and $dv=\cos(x) dx$. Then $du = (n-1)\cos(x)^{n-2}(-\sin(x))dx$ and $v=\sin(x)$. So,
 
 
-
+$$
 \begin{align*}
 \int \cos(x)^n dx &= \cos(x)^{n-1} \cdot (\sin(x)) + \int (\sin(x)) ((n-1)\sin(x) \cos(x)^{n-2}) dx \\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int \sin^2(x) \cos(x)^{n-2} dx\\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int (1 - \cos(x)^2) \cos(x)^{n-2} dx\\
 &= \sin(x) \cos(x)^{n-1} + (n-1)\int \cos(x)^{n-2}dx - (n-1)\int \cos(x)^n dx.
 \end{align*}
+$$
 
 
 We can then solve for the unknown ($\int \cos(x)^{n}dx$) to get this *reduction formula*:
@@ -279,12 +284,13 @@ The visual interpretation of integration by parts breaks area into two pieces, t
 Let $uv = x f^{-1}(x)$. Then we have $[uv]' = u'v + uv' = f^{-1}(x) + x [f^{-1}(x)]'$. So, up to a constant $uv = \int [uv]'dx = \int f^{-1}(x)dx + \int  x [f^{-1}(x)]'dx$. Re-expressing gives:
 
 
-
+$$
 \begin{align*}
 \int f^{-1}(x) dx
 &= xf^{-1}(x) - \int x [f^{-1}(x)]' dx\\
 &= xf^{-1}(x) - \int f(u) du.\\
 \end{align*}
+$$
 
 
 The last line follows from the $u$-substitution: $u=f^{-1}(x)$ for then $du = [f^{-1}(x)]' dx$ and $x=f(u)$.
@@ -293,12 +299,13 @@ The last line follows from the $u$-substitution: $u=f^{-1}(x)$ for then $du = [f
 We use this to find an antiderivative for $\sin^{-1}(x)$:
 
 
-
+$$
 \begin{align*}
 \int \sin^{-1}(x) dx &= x \sin^{-1}(x) - \int \sin(u) du \\
 &=  x \sin^{-1}(x) + \cos(u) \\
 &= x \sin^{-1}(x) + \cos(\sin^{-1}(x)).
 \end{align*}
+$$
 
 
 Using right triangles to simplify, the last value $\cos(\sin^{-1}(x))$ can otherwise be written as $\sqrt{1 - x^2}$.
@@ -322,11 +329,12 @@ This [proof](http://www.math.ucsd.edu/~ebender/20B/77_Trap.pdf) for the error es
 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 $v=t+A$:
 
 
-
+$$
 \begin{align*}
 \int_0^h f(t + x_i) dt &= uv \big|_0^h - \int_0^h v du\\
 &= f(t+x_i)(t+A)\big|_0^h - \int_0^h (t + A) f'(t + x_i) dt.
 \end{align*}
+$$
 
 
 We choose $A$ to be $-h/2$, any constant is possible, for then the term $f(t+x_i)(t+A)\big|_0^h$ becomes $(1/2)(f(x_i+h) + f(x_i)) \cdot h$, or the trapezoid approximation. This means, the error over this interval - actual minus estimate - satisfies:
@@ -339,11 +347,12 @@ $$
 For this, we *again* integrate by parts with
 
 
-
+$$
 \begin{align*}
 u  &= f'(t + x_i)  & dv &= (t + A)dt\\
 du &= f''(t + x_i) & v  &= \frac{(t + A)^2}{2} + B
 \end{align*}
+$$
 
 
 Again we added a constant of integration, $B$, to  $v$. The  error becomes:
@@ -418,13 +427,14 @@ We added a rectangle for a Riemann sum for $t_i = \pi/3$ and $t_{i+1} = \pi/3 +
 Taking this Riemann sum approach, we can approximate the area under the curve parameterized by $(u(t), v(t))$ over the time range $[t_i, t_{i+1}]$ as a rectangle with height $y(t_i)$ and base $x(t_{i}) - x(t_{i+1})$. Then we get, as expected:
 
 
-
+$$
 \begin{align*}
 A &\approx \sum_i y(t_i) \cdot (x(t_{i}) - x(t_{i+1}))\\
   &= - \sum_i y(t_i) \cdot (x(t_{i+1}) - x(t_{i}))\\
   &= - \sum_i y(t_i) \cdot \frac{x(t_{i+1}) - x(t_i)}{t_{i+1}-t_i} \cdot (t_{i+1}-t_i)\\
   &\approx -\int_a^b y(t) x'(t) dt.
 \end{align*}
+$$
 
 
 So with a counterclockwise rotation, the actual answer for the area includes a minus sign. If the area is traced out in a *clockwise* manner, there is no minus sign.

quarto/integrals/mean_value_theorem.qmd

@@ -89,13 +89,14 @@ Though not continuous, $f(x)$ is integrable as it contains only jumps. The integ
 What is the average value of the function $e^{-x}$ between $0$ and $\log(2)$?
 
 
-
+$$
 \begin{align*}
 \text{average} = \frac{1}{\log(2) - 0} \int_0^{\log(2)} e^{-x} dx\\
 &= \frac{1}{\log(2)} (-e^{-x}) \big|_0^{\log(2)}\\
 &= -\frac{1}{\log(2)} (\frac{1}{2} - 1)\\
 &= \frac{1}{2\log(2)}.
 \end{align*}
+$$
 
 
 Visualizing, we have
@@ -118,11 +119,15 @@ $$
 
 When we assume that $f(x)$ is continuous, we can describe $K$ as a value in the range of $f$:
 
+::: {.callout-note icon=false}
+## The mean value theorem for integrals
 
-> **The mean value theorem for integrals**: Let $f(x)$ be a continuous function on $[a,b]$ with $a < b$. Then there exists $c$ with $a \leq c \leq b$ with
->
-> $f(c) \cdot (b-a) = \int_a^b f(x) dx.$`
+Let $f(x)$ be a continuous function on $[a,b]$ with $a < b$. Then there exists $c$ with $a \leq c \leq b$ with
 
+$$
+f(c) \cdot (b-a) = \int_a^b f(x) dx.
+$$
+:::
 
 
 The proof comes from the intermediate value theorem and the extreme value theorem. Since $f$ is continuous on a closed interval, there exists values $m$ and $M$ with $f(c_m) = m \leq f(x) \leq M=f(c_M)$, for some $c_m$ and $c_M$ in the interval $[a,b]$. Since $m \leq f(x) \leq M$, we must have:

quarto/integrals/partial_fractions.qmd

@@ -28,13 +28,16 @@ Let $f(x) = p(x)/q(x)$, where  $p$ and $q$ are polynomial functions with real co
 
 The function $q(x)$ will factor over the real numbers. The fundamental theorem of algebra can be applied to say that $q(x)=q_1(x)^{n_1} \cdots q_k(x)^{n_k}$ where $q_i(x)$ is a linear or quadratic polynomial and $n_k$ a positive integer.
 
+::: {.callout-note icon=false}
+## Partial Fraction Decomposition
 
-> **Partial Fraction Decomposition**: There are unique polynomials $a_{ij}$ with degree $a_{ij} <$ degree $q_i$ such that
->
-> $$
-> \frac{p(x)}{q(x)} = a(x) + \sum_{i=1}^k \sum_{j=1}^{n_i} \frac{a_{ij}(x)}{q_i(x)^j}.
-> $$
+There are unique polynomials $a_{ij}$ with degree $a_{ij} <$ degree $q_i$ such that
 
+$$
+\frac{p(x)}{q(x)} = a(x) + \sum_{i=1}^k \sum_{j=1}^{n_i} \frac{a_{ij}(x)}{q_i(x)^j}.
+$$
+
+:::
 
 
 The method is attributed to John Bernoulli, one of the prolific Bernoulli brothers who put a stamp on several areas of math. This Bernoulli was a mentor to Euler.
@@ -109,7 +112,7 @@ What remains is to establish that we can take $A(x) = a(x)\cdot P(x)$ with a deg
 In Proposition 3.8 of [Bradley](http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf) and Cook we can see how. Recall the division algorithm, for example, says there are $q_k$ and $r_k$ with $A=q\cdot q_k + r_k$ where the degree of $r_k$ is less than that of $q$, which is linear or quadratic. This is repeatedly applied below:
 
 
-
+$$
 \begin{align*}
 \frac{A}{q^k} &= \frac{q\cdot q_k + r_k}{q^k}\\
 &= \frac{r_k}{q^k} + \frac{q_k}{q^{k-1}}\\
@@ -119,6 +122,7 @@ In Proposition 3.8 of [Bradley](http://www.m-hikari.com/imf/imf-2012/29-32-2012/
 &= \cdots\\
 &= \frac{r_k}{q^k} + \frac{r_{k-1}}{q^{k-1}} + \cdots + q_1.
 \end{align*}
+$$
 
 
 So the term $A(x)/q(x)^k$ can be expressed in terms of a sum where the numerators or each term have degree less than $q(x)$, as expected by the statement of the theorem.
@@ -208,13 +212,14 @@ integrate(B/((a*x)^2 - 1)^4, x)
 In [Bronstein](http://www-sop.inria.fr/cafe/Manuel.Bronstein/publications/issac98.pdf) this characterization can be found - "This method, which dates back to Newton, Leibniz and Bernoulli, should not be used in practice, yet it remains the method found in most calculus texts and is often taught. Its major drawback is the factorization of the denominator of the integrand over the real or complex numbers." We can also find the following formulas which formalize the above exploratory calculations ($j>1$ and $b^2 - 4c < 0$ below):
 
 
-
+$$
 \begin{align*}
 \int \frac{A}{(x-a)^j} &= \frac{A}{1-j}\frac{1}{(x-a)^{j-1}}\\
 \int \frac{A}{x-a}     &= A\log(x-a)\\
 \int \frac{Bx+C}{x^2 + bx + c}     &= \frac{B}{2} \log(x^2 + bx + c) + \frac{2C-bB}{\sqrt{4c-b^2}}\cdot \arctan\left(\frac{2x+b}{\sqrt{4c-b^2}}\right)\\
 \int \frac{Bx+C}{(x^2 + bx + c)^j} &= \frac{B' x + C'}{(x^2 + bx + c)^{j-1}} + \int \frac{C''}{(x^2 + bx + c)^{j-1}}
 \end{align*}
+$$
 
 
 The first returns a rational function; the second yields a logarithm term; the third yields a logarithm and an arctangent term; while the last, which has explicit constants available, provides a reduction that can be recursively applied;
@@ -288,7 +293,7 @@ The answers found can become quite involved. [Corless](https://arxiv.org/pdf/171
 ex = (x^2 - 1) / (x^4 + 5x^2 + 7)
 ```
 
-But the integral is something best suited to a computer algebra system:
+But the integral is something best suited for a computer algebra system:
 
 
 ```{julia}
@@ -482,11 +487,12 @@ How to see that these give rise to real answers on integration is the point of t
 Breaking the terms up over $a$ and $b$ we have:
 
 
-
+$$
 \begin{align*}
 I &= \frac{a}{x - (\alpha + i \beta)} + \frac{a}{x - (\alpha - i \beta)} \\
 II &= i\frac{b}{x - (\alpha + i \beta)} - i\frac{b}{x - (\alpha - i \beta)}
 \end{align*}
+$$
 
 
 Integrating $I$ leads to two logarithmic terms, which are combined to give:

quarto/integrals/substitution.qmd

@@ -41,13 +41,14 @@ $$
 So,
 
 
-
+$$
 \begin{align*}
 \int_a^b g(u(t)) \cdot u'(t) dt &= \int_a^b (G \circ u)'(t) dt\\
 &= (G\circ u)(b) - (G\circ u)(a) \quad\text{(the FTC, part II)}\\
 &= G(u(b)) - G(u(a)) \\
 &= \int_{u(a)}^{u(b)} g(x) dx. \quad\text{(the FTC part II)}
 \end{align*}
+$$
 
 
 That is, this substitution formula applies:
@@ -181,7 +182,7 @@ when $-1 \leq x \leq 1$ and $0$ otherwise. The area under $f$ is just $1$ - the
 Let $u(x) = (x-c)/h$ and $g(x) = (1/h) \cdot f(u(x))$. Then, as $du = 1/h dx$
 
 
-
+$$
 \begin{align*}
 \int_{c-h}^{c+h} g(x) dx
 &= \int_{c-h}^{c+h} \frac{1}{h} f(u(x)) dx\\
@@ -189,6 +190,7 @@ Let $u(x) = (x-c)/h$ and $g(x) = (1/h) \cdot f(u(x))$. Then, as $du = 1/h dx$
 &= \int_{-1}^1 f(u) du\\
 &= 1.
 \end{align*}
+$$
 
 
 So the area of this transformed function is still $1$. The shifting by $c$ we know doesn't effect the area, the scaling by $h$ inside of $f$ does, but is balanced out by the multiplication by $1/h$ outside of $f$.
@@ -248,13 +250,14 @@ $$
 But $u^3/3 - 4u/3 = (1/3) \cdot u(u-2)(u+2)$, so between $-2$ and $0$ it is positive and between $0$ and $1$ negative, so this integral is:
 
 
-
+$$
 \begin{align*}
 \int_{-2}^0 (u^3/3 - 4u/3 ) du + \int_{0}^1 -(u^3/3 - 4u/3) du
 &= (\frac{u^4}{12} - \frac{4}{3}\frac{u^2}{2}) \big|_{-2}^0 - (\frac{u^4}{12} - \frac{4}{3}\frac{u^2}{2}) \big|_{0}^1\\
 &= \frac{4}{3} - -\frac{7}{12}\\
 &= \frac{23}{12}.
 \end{align*}
+$$
 
 
 ##### Example
@@ -270,13 +273,14 @@ $$
 Integrals involving this function are typically transformed by substitution. For example:
 
 
-
+$$
 \begin{align*}
 \int_a^b f(x; \mu, \sigma) dx
 &= \int_a^b  \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} \exp(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2) dx \\
 &= \int_{u(a)}^{u(b)} \frac{1}{\sqrt{2\pi}} \exp(-\frac{1}{2}u^2) du \\
 &= \int_{u(a)}^{u(b)} f(u; 0, 1) du,
 \end{align*}
+$$
 
 
 where $u = (x-\mu)/\sigma$, so $du = (1/\sigma) dx$.
@@ -295,12 +299,13 @@ $$
 A further change of variables by $t = u/\sqrt{2}$ (with $\sqrt{2}dt = du$) gives:
 
 
-
+$$
 \begin{align*}
 \int_a^b f(x; \mu, \sigma) dx &=
 \int_{t(u(a))}^{t(u(b))} \frac{\sqrt{2}}{\sqrt{2\pi}} \exp(-t^2) dt\\
 &= \frac{1}{2} \int_{t(u(a))}^{t(u(b))} \frac{2}{\sqrt{\pi}} \exp(-t^2) dt
 \end{align*}
+$$
 
 
 Up to a factor of $1/2$ this is `erf`.
@@ -309,13 +314,14 @@ Up to a factor of $1/2$ this is `erf`.
 So we would have, for example, with $\mu=1$,$\sigma=2$ and $a=1$ and $b=3$ that:
 
 
-
+$$
 \begin{align*}
 t(u(a)) &= (1 - 1)/2/\sqrt{2} = 0\\
 t(u(b)) &= (3 - 1)/2/\sqrt{2} = \frac{1}{\sqrt{2}}\\
 \int_1^3 f(x; 1, 2)
 &= \frac{1}{2} \int_0^{1/\sqrt{2}} \frac{2}{\sqrt{\pi}} \exp(-t^2) dt.
 \end{align*}
+$$
 
 
 Or
@@ -488,7 +494,7 @@ integrate(1 / (a^2  + (b*x)^2), x)
 The expression $1-x^2$ can be attacked by the substitution $\sin(u) =x$ as then $1-x^2 = 1-\sin(u)^2 = \cos(u)^2$. Here we see this substitution being used successfully:
 
 
-
+$$
 \begin{align*}
 \int \frac{1}{\sqrt{9 - x^2}} dx &= \int \frac{1}{\sqrt{9 - (3\sin(u))^2}} \cdot 3\cos(u) du\\
 &=\int \frac{1}{3\sqrt{1 - \sin(u)^2}}\cdot3\cos(u) du \\
@@ -496,6 +502,7 @@ The expression $1-x^2$ can be attacked by the substitution $\sin(u) =x$ as then
 &= u \\
 &= \sin^{-1}(x/3).
 \end{align*}
+$$
 
 
 Further substitution allows the following integral to be solved for an antiderivative:
@@ -513,23 +520,25 @@ integrate(1 / sqrt(a^2 - b^2*x^2), x)
 The expression $x^2 - 1$ is a bit different, this lends itself to $\sec(u) = x$ for a substitution, for $\sec(u)^2 - 1 = \tan(u)^2$. For example, we try $\sec(u) = x$ to integrate:
 
 
-
+$$
 \begin{align*}
 \int \frac{1}{\sqrt{x^2 - 1}} dx &= \int \frac{1}{\sqrt{\sec(u)^2 - 1}} \cdot \sec(u)\tan(u) du\\
 &=\int \frac{1}{\tan(u)}\sec(u)\tan(u) du\\
 &= \int \sec(u) du.
 \end{align*}
+$$
 
 
 This doesn't seem that helpful, but the antiderivative to $\sec(u)$ is $\log\lvert (\sec(u) + \tan(u))\rvert$, so we can proceed to get:
 
 
-
+$$
 \begin{align*}
 \int \frac{1}{\sqrt{x^2 - 1}} dx &= \int \sec(u) du\\
 &= \log\lvert (\sec(u) + \tan(u))\rvert\\
 &= \log\lvert x + \sqrt{x^2-1} \rvert.
 \end{align*}
+$$
 
 
 SymPy gives a different representation using the arccosine:
@@ -566,13 +575,14 @@ $$
 The identify $\cos(u)^2 = (1 +  \cos(2u))/2$ makes this tractable:
 
 
-
+$$
 \begin{align*}
 4ab \int \cos(u)^2 du
 &= 4ab\int_0^{\pi/2}(\frac{1}{2} + \frac{\cos(2u)}{2}) du\\
 &= 4ab(\frac{1}{2}u + \frac{\sin(2u)}{4})\big|_0^{\pi/2}\\
 &= 4ab (\pi/4 + 0) = \pi ab.
 \end{align*}
+$$
 
 
 Keeping in mind that that a circle with radius $a$ is an ellipse with $b=a$, we see that this gives the correct answer for a circle.

quarto/integrals/surface_area.qmd

@@ -50,19 +50,24 @@ revolution, there is an easier way. (Photo credit to
 [firepanjewellery](http://firepanjewellery.com/).)
 ](./figures/gehry-hendrix.jpg)
 
-> The surface area generated by rotating the graph of $f(x)$ between $a$ and $b$  about the $x$-axis is given by the integral
->
-> $$
-> \int_a^b 2\pi f(x) \cdot \sqrt{1 + f'(x)^2} dx.
-> $$
->
-> If the curve is parameterized by $(g(t), f(t))$ between $a$ and $b$ then the surface area is
->
-> $$
-> \int_a^b 2\pi f(t) \cdot \sqrt{g'(t)^2 + f'(t)^2} dx.
-> $$
->
-> These formulas do not add in the surface area of either of the ends.
+
+::: {.callout-note icon=false}
+## Surface area of a rotated curve
+
+The surface area generated by rotating the graph of $f(x)$ between $a$ and $b$  about the $x$-axis is given by the integral
+
+$$
+\int_a^b 2\pi f(x) \cdot \sqrt{1 + f'(x)^2} dx.
+$$
+
+If the curve is parameterized by $(g(t), f(t))$ between $a$ and $b$ then the surface area is
+
+$$
+\int_a^b 2\pi f(t) \cdot \sqrt{g'(t)^2 + f'(t)^2} dx.
+$$
+
+These formulas do not add in the surface area of either of the ends.
+:::
 
 
 
@@ -129,7 +134,7 @@ Lets see that the surface area of an open cone follows from this formula, even t
 A cone can be envisioned as rotating the function $f(x) = x\tan(\theta)$ between $0$ and $h$ around the $x$ axis. This integral yields the surface area:
 
 
-
+$$
 \begin{align*}
 \int_0^h 2\pi f(x) \sqrt{1 + f'(x)^2}dx
 &= \int_0^h 2\pi x \tan(\theta) \sqrt{1 + \tan(\theta)^2}dx   \\
@@ -137,6 +142,7 @@ A cone can be envisioned as rotating the function $f(x) = x\tan(\theta)$ between
 &= \pi \tan(\theta) \sec(\theta) h^2                          \\
 &= \pi r^2 / \sin(\theta).
 \end{align*}
+$$
 
 
 (There are many ways to express this, we used $r$ and $\theta$ to match the work above. If the cone is parameterized by a height $h$ and radius $r$, then the surface area of the sides is $\pi r\sqrt{h^2 + r^2}$. If the base is included, there is an additional $\pi r^2$ term.)
@@ -322,13 +328,14 @@ plot(g, f, 0, 1pi)
 The integrand simplifies to $8\sqrt{2}\pi \sin(t) (1 + \cos(t))^{3/2}$. This lends itself to $u$-substitution with $u=\cos(t)$.
 
 
-
+$$
 \begin{align*}
 \int_0^\pi 8\sqrt{2}\pi \sin(t) (1 + \cos(t))^{3/2}
 &= 8\sqrt{2}\pi \int_1^{-1} (1 + u)^{3/2} (-1) du\\
 &= 8\sqrt{2}\pi (2/5) (1+u)^{5/2} \big|_{-1}^1\\
 &= 8\sqrt{2}\pi (2/5) 2^{5/2} = \frac{2^7 \pi}{5}.
 \end{align*}
+$$
 
 
 ## The first Theorem of Pappus
@@ -378,11 +385,12 @@ surface(ws..., legend=false, zlims=(-12,12))
 The surface area of sphere will be SA$=2\pi \rho (\pi r) = 2 \pi^2 r \cdot \rho$. What is $\rho$? The centroid of an arc formula can be derived in a manner similar to that of the centroid of a region. The formulas are:
 
 
-
+$$
 \begin{align*}
 \text{cm}_x &= \frac{1}{L} \int_a^b g(t) \sqrt{g'(t)^2 + f'(t)^2} dt\\
 \text{cm}_y &= \frac{1}{L} \int_a^b f(t) \sqrt{g'(t)^2 + f'(t)^2} dt.
 \end{align*}
+$$
 
 
 Here, $L$ is the arc length of the curve.