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--- math104-s21:s:vpak [2021/05/10 23:18]
68.186.63.173 [Summary of Material]
+++ math104-s21:s:vpak [2022/01/11 10:57] (current)
pzhou ↷ Page moved from math104-2021sp:s:vpak to math104-s21:s:vpak
@@ Line 132: / Line 132: @@
   * If a$< $s$ <$b, f is bounded, f is continuous at s, and  $\alpha$ (x)  $=$  I(x-s) where I is the //unit step function//, then  $\int_a^b fd{\alpha}$   $=$  f(s)
   * Suppose  $\alpha$  increases monotonically,  $\alpha$ ' is integrable on [a,b], and f is bounded real function on [a,b]. Then f is integrable with respect to  $\alpha$  if and only if f $\alpha$ ' is integrable: \\  $\int_a^b fd{\alpha}$   $=$  $\int_a^b f(x){\alpha}'(x)d(x)$
+  * Let f be integrable on [a,b] and for a $\leq$ x $\leq$ b, let F(x)  $=$   $\int_a^x f(t)dt$ , then \\ (1) F(x) is continuous on [a,b] \\ (2) if f(x) is continuous at p  $\in$  [a,b], then F(x) is differentiable at p, with F'(p)  $=$  f(p)
+**Fundamental Theorem of Calculus.** Let  $f$  be integrable on  $[a,b]$  and  $F$  be a differentiable function on [a,b] such that $F'(x)=f(x) $, then$ \int_a^b f(x)dx=F(b)-F(a)$
+Let  $\alpha$  be increasing weight function on  $[a,b]$ . Suppose  $f_n$  is integrable, and  $f_n$   $\to$   $f$  uniformly on  $[a,b]$ . Then  $f$  is integrable, and \\
+ $\int_a^b fd{\alpha}$   $=$   $\lim\limits_{n \to \infin}$   $\int_a^b f_{n}d{\alpha}$
+Suppose { $f_n$ } is a sequence of differentiable functions on  $[a,b]$  such that  $f_n$   $\to$   $g$  uniformly and there exists p  $\in$   $[a,b]$  where { $f_n(p)$ } converges. Then  $f_n$  converges to some  $f$  uniformly, and \\
+$f'(x)=g(x)=\lim\limits_{n \to \infin}f'_{n}(x)$ \\
+Note $f'_{n}(x)$ may not be continuous.
+==== Questions ====
+**1. What **

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