align fix; theorem style; condition number

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