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--- math104-s21:s:vpak [2021/05/11 06:18]
68.186.63.173 [Summary of Material]
+++ math104-s21:s:vpak [2026/02/21 14:41] (current)
@@ 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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