Riemann Sum and Definite Integral

Partition 

Definition 1. For a closed interval $[a, b]$, we subdivide the interval into subintervals, not necessarily of equal widths, as choosing $n-1$ points $\{ x_1, ..., x_{n-1} \}$ between $a$ and $b$ that are in increasing order, so that $x_0 = a < x_1 < \cdots < x_{n-1} < b = x_n$. The set of all of these points, $$P = \{ x_0, x_1, ..., x_{n-1}, x_n \}$$ is called a partition of $[a, b]$. And we denote the width of the $k$th subinterval by $\Delta x_k$ which means that $\Delta x_k = x_k - x_{k-1}$. 

Norm of a Partition 

Definition 2. We define the norm of a partition $P$, written $||P||$, to be the largest of all the subinterval widths.

Riemann Sum

Definition 3. Let $f$ be a function defined on $[a, b]$. For given partition $P$, we can select some point $c_k$ chosen in the $k$th subinterval $[x_{k-1}, x_k]$. Then we call the sum $$S_P = \sum_{k=1}^n f(c_k) \Delta x_k$$ a Riemann sum for $f$ on the interval $[a, b]$. 

Definite Integral

Definition 4. Let $f$ be a function defined on a closed interval $[a, b]$. We say that a number $J$ is the definite integral of $f$ over $[a, b]$ and that $J$ is the limit of the Riemann sums $\sum_{k=1}^n f(c_k) \Delta x_k$ if the following condition is satisfied:
Given any number $\epsilon > 0$ there is a corresponding number $\delta > 0$ such that for every partition $P = \{ x_0, ..., x_n \}$ of $[a, b]$ with $||P|| < \delta$ and any choice of $c_k$ in $[x_{k-1}, x_k]$, we have $$\Bigl \vert \sum_{k=1}^n f(c_k) \Delta x_k - J \Bigl \vert < \epsilon.$$  
When the limit exists we write $$J = \lim_{||P|| \rightarrow 0} \sum_{k=1}^n f(c_k) \Delta x_k = \int_a^b f(x) dx,$$ and we say that the definite integral exists or $f$ is intergrable over $[a, b]$. 

Epsilon-Delta Method와 같은 방식이다. 우리에겐 subinterval을 어떻게 잡을거냐, 다시 말해 partition을 어떻게 잡을거냐, 그리고 각 subinterval 안에 point $c_k$를 어떻게 잡을거냐 하는 문제가 있다. 그러나 임의의 $\epsilon$보다 $J$와 Riemann sum의 차이를 작게 만드는 partition과 point를 무조건 잡아낼 수 있다면 우리는 definite integral이 존재한다고 말한다. 

 

A Formula for the Riemann Sum with Equal-Width Subintervals

$$\int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{k=1}^n f \left( a + k \frac{b-a}{n} \right) \left( \frac{b-a}{n} \right)$$

Definition 4는 어떻게 definite integral을 계산하는지에 대해서는 얘기하지 않는다. 따라서 partition을 잡는 가장 보편적인 방법인 equal-width subinterval로 택하고 각 subinterval의 point를 endpoint로 잡으면 explicit하게 integral을 계산할 수 있다. 

 

Integrability of Continuous Functions

Theorem 1. If a function $f$ is continuous over $[a, b]$, or if $f$ has at most finitely many jump discontinuities there, then the definite integral $\int_{a}^b f(x) dx$ exists and $f$ is integrable over $[a, b]$. 

모든 연속함수와 유한 개수 불연속 점을 가지고 있는 함수는 integrable하다.

Properties of Definite Integrals

Theorem 2. When $f$ and $g$ are integrable over $[a, b]$, the definite integral satisfies the following rules:
(1) $\int_a^b f(x) dx = -\int_b^a f(x) dx$
(2) $\int_a^a f(x) dx = 0$
(3) $\int_a^b (kf(x) \pm g(x)) dx = k\int_a^b f(x) dx + \int_a^b g(x) dx$
(4) $\int_a^b f(x) dx + \int_b^c f(x) dx = \int_a^c f(x) dx$
(5) If $f$ has maximum value $max f$ and minimum value $min f$ on $[a, b]$, then $$(min f) \cdot (b-a) \leq \int_a^b f(x) dx \leq (max f) \cdot (b-a)$$
(6) If $f(x) \geq g(x)$ on $[a, b]$ then $\int_a^b f(x) dx \geq \int_a^b g(x) dx$

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Reference is here:
https://product.kyobobook.co.kr/detail/S000003155860

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