Integration (Pure 3) · 积分(Pure 3)
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| standard integral/ˈstændəd ˈɪntɪɡrəl/ | 标准积分 | biāo zhǔn jī fēn |
| numerator/ˈnjuːməreɪtə/ | 分子 | fèn zǐ |
| denominator/dɪˈnɒmɪneɪtə/ | 分母 | fēn mǔ |
| partial fractions/ˈpɑːʃl ˈfrækʃnz/ | 部分分式 | bù fèn fēn shì |
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | 分部积分 | fēn bù jī fēn |
| integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ | 换元积分 | huàn yuán jī fēn |
| change of variable/tʃeɪndʒ ɒv ˈveərɪəbl/ | 变量替换 | biàn liàng tì huàn |
The integral that gives you angles
- You can integrate $\dfrac{1}{x^2+1}$ to get $\tan^{-1}x$ — a function that returns angles.
- This connection between algebra and geometry is one of the most beautiful results in calculus.
给你角度的积分
- 你能积分 $\dfrac{1}{x^2+1}$ 得到 $\tan^{-1}x$——一个返回角度的函数。
- 这个代数和几何之间的联系是微积分中最美的结果之一。
New standard integrals 标准积分
- A new standard integral: $\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$.
- The pattern $\dfrac{k\,f'(x)}{f(x)}$ → $k\ln|f(x)| + C$, e.g. $\displaystyle\int \frac{2x}{x^2+1}\,dx = \ln(x^2+1) + C$.
Worked example. $\displaystyle\int \frac{1}{x^2 + 4}\,dx = \dfrac{1}{2}\tan^{-1}\dfrac{x}{2} + C$ (here $a = 2$).
Spot the pattern. $\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$. The numerator 分子 must be (a multiple of) the derivative of the denominator 分母. If it's not, you need a different method.
The locus |z - a| = r is a circle of radius r centred at a
新的标准积分
- 一个新的标准积分:$\displaystyle\int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a} + C$。
- 模式 $\dfrac{k\,f'(x)}{f(x)}$ → $k\ln|f(x)| + C$,例如 $\displaystyle\int \frac{2x}{x^2+1}\,dx = \ln(x^2+1) + C$。
算例。 $\displaystyle\int \frac{1}{x^2 + 4}\,dx = \dfrac{1}{2}\tan^{-1}\dfrac{x}{2} + C$(这里 $a = 2$)。
认出模式。 $\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$。分子必须是分母的导数(的一个倍数)。如果不是,你需要一种不同的方法。
轨迹 |z - a| = r 是一个圆心在 a、半径 r 的圆
The area under the curve · 曲线下的面积
area = ∫ f(x) dx
Every integration method just measures this area · 面积 — drag the limits to total it. · 每种积分方法都只是测量这个面积——拖动上下限来合计它。
What is ∫ 2x/(x²+1) dx? · ∫ 2x/(x²+1) dx 是什么?
The top is the derivative of the bottom (k f′/f form), so the integral is ln(x²+1) + C. · 分子是分母的导数(k f′/f 形式),所以积分是 ln(x²+1) + C。
∫ 1/(x²+a²) dx = (1/a) tan⁻¹(x/a) + C. · ∫ 1/(x²+a²) dx = (1/a) tan⁻¹(x/a) + C。
This is a standard Pure 3 integral linked to the inverse-tangent derivative. · 这是一个与反正切导数相联系的标准 Pure 3 积分。
∫ 1/(x²+9) dx = (1/a) tan⁻¹(x/a) + C. What is a? · ∫ 1/(x²+9) dx = (1/a) tan⁻¹(x/a) + C。a 是多少?
a² = 9, so a = 3. · a² = 9,所以 a = 3。
∫ 1/(x²+1) dx = ln(x²+1) + C. · ∫ 1/(x²+1) dx = ln(x²+1) + C。
The numerator is NOT the derivative of the denominator. The correct integral is tan⁻¹x + C. · 分子不是分母的导数。正确的积分是 tan⁻¹x + C。
Partial fractions 部分分式 for integration
- Partial fractions: split a rational function, then integrate each piece as a logarithm.
- Example: $\displaystyle\int \dfrac{1}{x^2-1}\,dx = \int\left(\dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)}\right)dx = \dfrac{1}{2}\ln|x-1| - \dfrac{1}{2}\ln|x+1| + C$.
用于积分的部分分式
- 部分分式:拆开一个有理函数,然后把每一块积分成一个对数。
- 示例：$\displaystyle\int \dfrac{1}{x^2-1}\,dx = \int\left(\dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)}\right)dx = \dfrac{1}{2}\ln|x-1| - \dfrac{1}{2}\ln|x+1| + C$。
Integration by parts 分部积分
- Integration by parts (for a product): $\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$.
- Choose $u$ to be the part that simplifies when differentiated.
分部积分
- 分部积分(integration by parts,用于一个乘积):$\displaystyle\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx$。
- 选择微分后变简单的那部分作为 $u$。
The integration-by-parts formula is ∫ u dv =: · 分部积分公式是 ∫ u dv =:
Integration by parts: ∫ u dv = uv − ∫ v du. · 分部积分:∫ u dv = uv − ∫ v du。
Integration by substitution 换元积分
- Integration by substitution: a change of variable 变量替换 turns a hard integral into an easy one.
- Let $u = g(x)$, then $du = g'(x)\,dx$, and rewrite the integral in terms of $u$.
allow_no_figure: Integration techniques (by parts, substitution, partial fractions) — the visual content is the algebraic manipulation, not a diagram.
- Integration is reverse differentiation.
换元积分
- 换元积分(integration by substitution):一次变量替换把一个困难的积分变成一个容易的。
- 设 $u = g(x)$,那么 $du = g'(x)\,dx$,并用 $u$ 重写积分。
allow_no_figure: 积分技巧（分部积分、换元法、部分分式）——视觉内容是代数运算，而非图表。
- 积分是反向求导(reverse differentiation)。
Using substitution u = x²+1, evaluate ∫₀² 2x/(x²+1) dx (2 dp). · 用换元 u = x²+1,求 ∫₀² 2x/(x²+1) dx(2 位小数)。
∫₀² 2x/(x²+1) dx = [ln(x²+1)]₀² = ln 5 − ln 1 = ln 5 ≈ 1.61. · ∫₀² 2x/(x²+1) dx = [ln(x²+1)]₀² = ln 5 − ln 1 = ln 5 ≈ 1.61。
You've got it
- $\displaystyle\int \frac{1}{x^2+a^2}dx = \frac1a\tan^{-1}\frac{x}{a} + C$; $\displaystyle\int\frac{k f'(x)}{f(x)}dx = k\ln|f(x)| + C$
- by parts: $\int u\,dv = uv - \int v\,du$ (for products)
- by substitution: change the variable to simplify
你掌握了
- $\displaystyle\int \frac{1}{x^2+a^2}dx = \frac1a\tan^{-1}\frac{x}{a} + C$; $\displaystyle\int\frac{k f'(x)}{f(x)}dx = k\ln|f(x)| + C$
- 分部:$\int u\,dv = uv - \int v\,du$(用于乘积)
- 换元:改变变量来简化