这一节从上节的黎曼求和表达式中，推导出定积分(definite integral)的定义。
在上一节中，黎曼求和表达式中，我们将每个子区间称为partition P，将其长度的最大值称为norm（∣ ∣ p ∣ ∣ ||p|| ∣∣ p ∣∣
定积分的定义：
Let f(x) be a function defined on closed internal [a, b]. We say that a number I is the definite integral of f over [a, b] and that I is the limit of the Riemann sums ∑ k = 1 n f ( x k ) Δ x k \sum_{k = 1}^n f(x_k)\Delta x_k ∑ k = 1 n f ( x k ) Δ x k
Given any number ϵ > 0 \epsilon > 0 ϵ > 0 δ > 0 \delta > 0 δ > 0 P = { x 0 , x 1 , . . . , x n } P = \{x_0, x_1, ..., x_n\} P = { x 0 , x 1 , ... , x n } ∣ ∣ P ∣ ∣ < δ ||P|| < \delta ∣∣ P ∣∣ < δ c k c_k c k [ x k − 1 , x k ] [x_{k - 1}, x_k] [ x k − 1 , x k ]
∣ ∑ k = 1 n f ( c k ) Δ x k − I ∣ < ϵ |\sum_{k = 1}^n f(c_k)\Delta x_k - I | < \epsilon ∣ k = 1 ∑ n f ( c k ) Δ x k − I ∣ < ϵ 说大白话就是，I是黎曼求和表达式的极限，这个极限就是函数f在区间[a, b]上的定积分。
更常见的表示符号是莱布尼茨引入的，将黎曼求和改写成：
∫ a b f ( x ) d x \int_a^b f(x) dx ∫ a b f ( x ) d x 离散的f ( c k ) f(c_k) f ( c k ) Δ x k \Delta x_k Δ x k
"the integral from a to b of ƒ of x dee x" or
真够绕的。。。
当极限存在时，称为函数f在区间[a,b]上的黎曼求和收敛(converge)于定积分I，函数f在区间[a, b]可积(integrable)。用公式表述为：
lim ∣ ∣ p ∣ ∣ → 0 ∑ k = 1 n f ( c k ) Δ x k = I = ∫ a b f ( x ) d x \lim_{||p|| \to 0} \sum_{k = 1}^n f(c_k)\Delta x_k = I = \int_a^b f(x)dx ∣∣ p ∣∣ → 0 lim k = 1 ∑ n f ( c k ) Δ x k = I = ∫ a b f ( x ) d x 或
lim n → ∞ ∑ k = 1 n f ( c k ) Δ x = I = ∫ a b f ( x ) d x \lim_{n \to \infty} \sum_{k = 1}^n f(c_k)\Delta x = I = \int_a^b f(x)dx n → ∞ lim k = 1 ∑ n f ( c k ) Δ x = I = ∫ a b f ( x ) d x 定积分存在的判定
连续函数是可积的，函数在连续区间存在定积分 。
那么不连续的区间是否就不可积呢？未必。这里引用原文一句话："For integrability to fail, a function needs to be sufficiently discontinuous so that the region between its graph and the x-axis cannot be approximated well by increasingly thin rectangles."
书上举了一个例子：
f ( x ) = { 1 , if x is rational 0 , if x is irrational f(x) =
\begin{cases}
1, & \text{if x is rational} \\
0, & \text{if x is irrational} \\
\end{cases} f ( x ) = { 1 , 0 , if x is rational if x is irrational 假设我们去求区间[0, 1]的黎曼求和，表达式如下：
∑ k = 1 n f ( c k ) Δ x k \sum_{k = 1}^n f(c_k)\Delta x_k k = 1 ∑ n f ( c k ) Δ x k 对于任意的子区间[ x k − 1 , x k ] [x_{k - 1}, x_k] [ x k − 1 , x k ] f ( c k ) f(c_k) f ( c k )
∑ k = 1 n f ( c k ) Δ x k = ∑ k = 1 n ( 1 ) Δ x k = ( 1 ) ⋅ ∑ k = 1 n Δ x k = 1 ⋅ 1 = 1 \sum_{k = 1}^n f(c_k)\Delta x_k = \sum_{k = 1}^n (1) \Delta x_k = (1) \cdot \sum_{k = 1}^n \Delta x_k = 1 \cdot 1 = 1 k = 1 ∑ n f ( c k ) Δ x k = k = 1 ∑ n ( 1 ) Δ x k = ( 1 ) ⋅ k = 1 ∑ n Δ x k = 1 ⋅ 1 = 1 注意∑ k = 1 n Δ x k \sum_{k = 1}^n \Delta x_k ∑ k = 1 n Δ x k
如果我们取f ( c k ) f(c_k) f ( c k )
∑ k = 1 n f ( c k ) Δ x k = ∑ k = 1 n ( 0 ) Δ x k = ( 0 ) ⋅ ∑ k = 1 n Δ x k = 0 ⋅ 1 = 0 \sum_{k = 1}^n f(c_k)\Delta x_k = \sum_{k = 1}^n (0) \Delta x_k = (0) \cdot \sum_{k = 1}^n \Delta x_k = 0 \cdot 1 = 0 k = 1 ∑ n f ( c k ) Δ x k = k = 1 ∑ n ( 0 ) Δ x k = ( 0 ) ⋅ k = 1 ∑ n Δ x k = 0 ⋅ 1 = 0 显然，黎曼求和在这个函数的[0, 1]区间是不收敛的，也就是不可积。
定积分的重要性质
f和g都可积，那么它们具有以下性质：
现在再回到5.1讨论的一些具体问题，在这里就可以把它们和定积分联系起来了。
连续函数和x轴之间的面积就是就是此函数在连续区间的定积分 ，即：
A = ∫ a b f ( x ) d x A = \int_a^b f(x)dx A = ∫ a b f ( x ) d x 不管a和b分布于x轴的各个位置，以上定义都是成立的，只要将面积概率扩大化到有负面积即可。
连续函数在某区间的平均值，就是此函数在这个区间的不定积分除以区间长度
a v e r a g e = 1 b − a ∫ a b f ( x ) d x average = \frac{1}{b - a} \int_a^b f(x) dx a v er a g e = b − a 1 ∫ a b f ( x ) d x 回顾5.1节求墙体平均温度的式子：
t a v g = ∑ k = 1 n t ( x k ) n = ∑ k = 1 n t ( x k ) ⋅ Δ x n ⋅ Δ x t_{avg} = \frac{\sum_{k = 1}^n t(x_k)}{n} = \frac{\sum_{k = 1}^n t(x_k) \cdot \Delta x}{n \cdot \Delta x} t a vg = n ∑ k = 1 n t ( x k ) = n ⋅ Δ x ∑ k = 1 n t ( x k ) ⋅ Δ x 假设所求的是墙体[a, b]区间的平均温度，那么上式就转化为：
t a v g = 1 b − a ∫ a b t ( x ) d x t_{avg} = \frac{1}{b - a} \int_a^b t(x)dx t a vg = b − a 1 ∫ a b t ( x ) d x 练习
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n c k 2 Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n c_k^2 \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n c k 2 Δ x k
∫ 0 2 x 2 d x \int_0^2 x^2 dx ∫ 0 2 x 2 d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n 2 c k 3 Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n 2c_k^3 \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n 2 c k 3 Δ x k
∫ − 1 0 2 x 3 d x \int_{-1}^0 2x^3 dx ∫ − 1 0 2 x 3 d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n ( c k 2 − 3 c k ) Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n (c_k^2 - 3c_k) \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n ( c k 2 − 3 c k ) Δ x k
∫ − 7 5 ( x 2 − 3 x ) d x \int_{-7}^5 (x^2 -3x) dx ∫ − 7 5 ( x 2 − 3 x ) d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n ( 1 c k ) Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n (\frac{1}{c_k})\Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n ( c k 1 ) Δ x k
∫ 1 4 ( 1 x ) d x \int_1^4 (\frac{1}{x}) dx ∫ 1 4 ( x 1 ) d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n 1 1 − c k Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n \frac{1}{1 - c_k} \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n 1 − c k 1 Δ x k
∫ 2 3 1 1 − x d x \int_2^3 \frac{1}{1 - x} dx ∫ 2 3 1 − x 1 d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n 4 − c k 2 Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n \sqrt{4 - c_k^2} \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n 4 − c k 2 Δ x k
∫ 0 1 4 − x 2 d x \int_0^1 \sqrt{4 - x^2} dx ∫ 0 1 4 − x 2 d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n ( sec c k ) Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n (\sec c_k) \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n ( sec c k ) Δ x k
∫ − π / 4 0 ( sec x ) d x \int_{-\pi/4}^0 (\sec x) dx ∫ − π /4 0 ( sec x ) d x
lim ∣ ∣ P ∣ ∣ → p ∑ k = 1 n ( tan c k ) Δ x k \lim_{||P|| \to p} \sum_{k = 1}^n (\tan c_k) \Delta x_k lim ∣∣ P ∣∣ → p ∑ k = 1 n ( tan c k ) Δ x k
∫ 0 π / 4 ( tan x ) d x \int_0^{\pi/4} (\tan x) dx ∫ 0 π /4 ( tan x ) d x
∫ 1 2 f ( x ) d x = − 4 , ∫ 1 5 f ( x ) d x = 6 , ∫ 1 5 g ( x ) d x = 8 \int_1^2 f(x)dx = -4, \ \int_1^5 f(x)dx = 6, \ \int_1^5 g(x)dx = 8 ∫ 1 2 f ( x ) d x = − 4 , ∫ 1 5 f ( x ) d x = 6 , ∫ 1 5 g ( x ) d x = 8
a. ∫ 2 2 g ( x ) d x = 0 \int_2^2 g(x)dx = 0 ∫ 2 2 g ( x ) d x = 0
b. ∫ 5 1 g ( x ) d x = − 8 \int_5^1 g(x)dx = -8 ∫ 5 1 g ( x ) d x = − 8
c. ∫ 1 2 3 f ( x ) d x = − 12 \int_1^2 3f(x)dx = -12 ∫ 1 2 3 f ( x ) d x = − 12
d. ∫ 2 5 f ( x ) d x = ∫ 1 5 f ( x ) d x − ∫ 1 2 f ( x ) d x = 10 \int_2^5 f(x)dx = \int_1^5 f(x)dx - \int_1^2 f(x)dx = 10 ∫ 2 5 f ( x ) d x = ∫ 1 5 f ( x ) d x − ∫ 1 2 f ( x ) d x = 10
e. ∫ 1 5 [ f ( x ) − g ( x ) ] d x = ∫ 1 5 f ( x ) d x − ∫ 1 5 g ( x ) d x = − 2 \int_1^5 [f(x) - g(x)]dx = \int_1^5 f(x)dx - \int_1^5 g(x)dx = -2 ∫ 1 5 [ f ( x ) − g ( x )] d x = ∫ 1 5 f ( x ) d x − ∫ 1 5 g ( x ) d x = − 2
f. ∫ 1 5 [ 4 f ( x ) − g ( x ) ] d x = 24 − 8 = 16 \int_1^5 [4f(x) - g(x)]dx = 24 - 8 = 16 ∫ 1 5 [ 4 f ( x ) − g ( x )] d x = 24 − 8 = 16
∫ 1 9 f ( x ) d x = − 1 , ∫ 7 9 f ( x ) d x = 5 , ∫ 7 9 h ( x ) d x = 4 \int_1^9 f(x)dx = -1, \ \int_7^9 f(x)dx = 5, \ \int_7^9 h(x)dx = 4 ∫ 1 9 f ( x ) d x = − 1 , ∫ 7 9 f ( x ) d x = 5 , ∫ 7 9 h ( x ) d x = 4
a. ∫ 1 9 − 2 f ( x ) d x = 2 \int_1^9 -2f(x)dx = 2 ∫ 1 9 − 2 f ( x ) d x = 2
b. ∫ 7 9 [ f ( x ) + h ( x ) ] d x = 9 \int_7^9 [f(x) + h(x)]dx = 9 ∫ 7 9 [ f ( x ) + h ( x )] d x = 9
c. ∫ 7 9 [ 2 f ( x ) − 3 h ( x ) ] d x = 10 − 12 = − 2 \int_7^9 [2f(x) - 3h(x)]dx = 10 - 12 = -2 ∫ 7 9 [ 2 f ( x ) − 3 h ( x )] d x = 10 − 12 = − 2
d. ∫ 9 1 f ( x ) d x = 1 \int_9^1 f(x)dx = 1 ∫ 9 1 f ( x ) d x = 1
e. ∫ 1 7 f ( x ) d x = ∫ 1 9 f ( x ) d x − ∫ 7 9 f ( x ) d x = − 6 \int_1^7 f(x)dx = \int_1^9 f(x)dx - \int_7^9 f(x)dx = -6 ∫ 1 7 f ( x ) d x = ∫ 1 9 f ( x ) d x − ∫ 7 9 f ( x ) d x = − 6
f. ∫ 9 7 [ h ( x ) − f ( x ) ] d x = ∫ 7 9 f ( x ) d x − ∫ 7 9 h ( x ) d x = 1 \int_9^7 [h(x) - f(x)]dx = \int_7^9 f(x)dx - \int_7^9 h(x)dx = 1 ∫ 9 7 [ h ( x ) − f ( x )] d x = ∫ 7 9 f ( x ) d x − ∫ 7 9 h ( x ) d x = 1
