Integration (Pure 3)
Introduced| English | Meaning |
|---|---|
| standard integral/ˈstændəd ˈɪntɪɡrəl/ | standard integral |
| numerator/ˈnjuːməreɪtə/ | The upper part of a fraction, showing how many of the denominator's equal parts are represented. |
| denominator/dɪˈnɒmɪneɪtə/ | The lower part of a fraction, indicating the divisor or the number of equal parts in a whole. |
| partial fractions/ˈpɑːʃl ˈfrækʃnz/ | Simpler rational expressions whose sum equals a given rational expression. |
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | An integration method derived from the product rule, transferring differentiation from one factor to another. |
| integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ | Changing the variable of integration, including its differential and any limits, to simplify an integral. |
| change of variable/tʃeɪndʒ ɒv ˈveərɪəbl/ | change of variable |
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.
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
The area under the curve
area = ∫ f(x) dx
Every integration method just measures this area — drag the limits to total it.
What is ∫ 2x/(x²+1) dx?
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The top is the derivative of the bottom (k f′/f form), so the integral is ln(x²+1) + C.
∫ 1/(x²+a²) dx = (1/a) tan⁻¹(x/a) + C.
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This is a standard Pure 3 integral linked to the inverse-tangent derivative.
∫ 1/(x²+9) dx = (1/a) tan⁻¹(x/a) + C. What is a?
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a² = 9, so a = 3.
∫ 1/(x²+1) dx = ln(x²+1) + C.
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The numerator is NOT the derivative of the denominator. The correct integral is 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: $\displaystyle\int \dfrac{1}{x^2-1}\,dx = \int\left(\dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)}\right)dx = \dfrac{1}{2}\ln|x-1| - \dfrac{1}{2}\ln|x+1| + C$.
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.
The integration-by-parts formula is ∫ u dv =:
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Integration by parts: ∫ 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.
Using substitution u = x²+1, evaluate ∫₀² 2x/(x²+1) dx (2 dp).
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∫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