积分直觉与连续分布Integral Intuition & Continuous Distributions
取值连续变化时，为什么区间面积才是概率？Why is interval area the probability when values vary continuously?
前置知识：Prerequisites: B02：指数与对数；B16：期望与方差。B02: exponentials and logarithms; B16: expectation and variance.
复习前置：Review prerequisites: B02 · 指数与对数：增长和尺度Exponentials & Logarithms: Growth and Scale · B16 · 期望、方差与样本统计Expectation, Variance & Sample Statistics
学完后你能：After this lesson you can: 用柱面积近似区间概率，并区分密度与单点概率。Approximate interval probability by bar areas and distinguish density from point probability.
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改变柱子的数量与区间右端点，比较小矩形面积之和与密度曲线下的面积。Change the number of bins and the interval endpoint; compare the sum of rectangle areas with the area under the density curve.
从观察到解释From observation to explanation
从柱子面积开始Start with bar area
柱高是密度、柱宽是区间宽度，二者相乘才是柱子代表的概率。若柱高 2、宽 0.1，面积为 0.2；高度超过 1 并不意味着概率超过 1。Bar height is density and width is interval width; their product is the probability represented by that bar. A height of 2 and width 0.1 gives area 0.2. Density above 1 does not imply probability above 1.
把小条加起来Add the strips
本实验使用 0≤x≤1 上的密度 p(x)=2x，区间 [0,t] 下方是一个三角形，面积为 t²。把区间分成更窄的矩形，就能近似这个面积；全部柱面积为 1。This lab uses density p(x)=2x on 0≤x≤1. The area under [0,t] is a triangle of area t². Narrower rectangles approximate this area; all bar areas add to 1.
连续不等于均匀Continuous does not mean uniform
有密度的连续分布在单点的概率为 0，区间才有非零面积。正态分布是另一种连续分布：均值 μ 移动中心，标准差 σ>0 改变宽度；面积仍为 1。A continuous distribution with a density assigns probability 0 to a single point; an interval can have nonzero area. A normal distribution is another continuous distribution: its mean μ moves the center and standard deviation σ>0 changes the width; total area remains 1.
公式与符号Formula and symbols
X 是连续随机变量，p(x) 是非负概率密度，a、b 是区间端点；积分表示把很窄的小条面积相加，dx 是小条宽度。全范围面积必须为 1。X is a continuous random variable, p(x) is a nonnegative density, and a and b are interval endpoints. The integral adds the areas of narrow strips; dx denotes a small width. Total area must equal 1.
留意这个区别Watch for this
密度高度不是单点概率；连续分布也不一定均匀。柱宽改变时不能只比较柱高。Density height is not point probability, and a continuous distribution need not be uniform. When bin width changes, compare areas, not heights alone.
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