Integration · 积分
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| integration/ˌɪntɪˈɡreɪʃn/ | 积分 | jī fēn |
| definite integral/ˈdefɪnət ˈɪntɪɡrəl/ | 定积分 | dìng jī fēn |
| fundamental theorem of calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ | 微积分基本定理 | wēi jī fēn jī běn dìng lǐ |
| constant of integration/ˈkɒnstənt ɒv ˌɪntɪˈɡreɪʃn/ | 积分常数 | jī fēn cháng shù |
| integration by substitution/ˌɪntɪˈɡreɪʃn baɪ ˌsʌbstɪˈtjuːʃn/ | 换元积分法 | huàn yuán jī fēn fǎ |
Signed accumulation can cancel
- Integration 积分 can recover an antiderivative or calculate signed accumulation. A definite integral is not automatically a nonnegative geometric area.
- For a continuous integrand, the fundamental theorem of calculus 微积分基本定理 connects a definite integral to an antiderivative evaluated at its bounds.
有符号累积可以相互抵消
- 积分 可以恢复原函数或计算有符号累积。定积分并不自动等于非负的几何面积。
- 对于连续被积函数，微积分基本定理 将定积分与被积函数在上下限处的原函数值联系起来。
Keep the antiderivative family
- An indefinite integral needs a constant of integration 积分常数 because differentiating any constant gives zero. The power formula is $\int x^n dx=x^{n+1}/(n+1)+C$ for $n\ne-1$ on a valid domain.
- The reciprocal exception is $\int1/x\,dx=\ln|x|+C$ on an interval not crossing zero. A positive or negative domain needs attention before choosing a logarithm expression.
保留原函数族
- 不定积分需要积分常数，因为常数的导数为零。幂公式在有效定义域内为 $\int x^n dx=x^{n+1}/(n+1)+C$（针对 $n\ne-1$）。
- 倒数例外情况为 $\int1/x\,dx=\ln|x|+C$（在不跨越零点的区间上）。选择对数表达式前需注意定义域的正负性。
Integrate x³ with respect to x. Include the constant. · 对 x³ 关于 x 积分。包含常数。
Add one to the power and divide by the new power, then add c because any constant differentiates to zero. · 指数加一并除以新指数，然后加上 c，因为任何常数的导数为零。
A definite integral needs a constant of integration. · 定积分不需要积分常数。
The constant cancels in F(b) − F(a). Only an indefinite integral needs it — and there it is compulsory. · 常数在 F(b) − F(a) 中抵消。只有不定积分需要它——且在那里是强制的。
In one English sentence, explain why an indefinite integral needs "+ c". · 用一句英语解释为何不定积分需要 "+ c"。
Example: "Any constant differentiates to zero, so the original constant cannot be recovered from the derivative." · 示例：“任何常数导数为零，故原始常数无法从导数中恢复。”
Evaluate upper minus lower
- A definite integral 定积分 gives $\int_a^bf(x)dx=F(b)-F(a)$ when the antiderivative method applies. Substitute both bounds into the same F.
- C cancels between those evaluations. For geometric area, split where the curve changes sign or where upper and lower curves exchange order.
用上减下求值
- 当适用原函数法时，定积分 给出 $\int_a^bf(x)dx=F(b)-F(a)$。将两个界限代入同一个 F 中。
- C 在两次求值中相互抵消。对于几何面积，需在曲线变号处或上下曲线交换顺序处分段计算。
The area under the curve · 曲线下方的面积
$\int_a^b f(x)\,dx$
A definite integral is the signed area between the curve and the axis. · 定积分是曲线与轴之间的有向面积。
Evaluate the integral of x² from 0 to 3. · 计算 x² 从 0 到 3 的积分值。
[x³/3] from 0 to 3 = 27/3 − 0 = 9. · [x³/3] 从 0 到 3 = 27/3 − 0 = 9。
Zero integral, positive area. For $y=x$ from negative 1 to 1, $I=[x^2/2]_{-1}^1=0$. Geometric area is $A=-\int_{-1}^0x\,dx+\int_0^1x\,dx=1$. Each triangular region has area one half; their signed contributions cancel.
积分为零，面积为正。 对于 $y=x$ 从 -1 到 1，↫$I=[x^2/2]_{-1}^1=0$。几何面积为 $A=-\int_{-1}^0x\,dx+\int_0^1x\,dx=1$。每个三角形区域面积为二分之一；它们的有符号贡献相互抵消。
Substitute both the function and differential
- Integration by substitution 换元积分法 reverses the chain rule. With $u=x^2+1$, $du=2x\,dx$, so $\int2x(x^2+1)^5dx=\int u^5du$.
- For a definite integral, transform its bounds to u or return to x before evaluating original bounds. Do not combine u expressions with x bounds.
同时代入函数与微分
- 换元积分法 是链式法则的逆运算。令 $u=x^2+1$，则 $du=2x\,dx$，故 $\int2x(x^2+1)^5dx=\int u^5du$。
- 对于定积分，将其界限转换为 u 或返回 x 后再评估原始界限。不要将 u 的表达式与 x 的界限混合使用。
Which substitution simplifies the integral of 2x(x² + 1)⁵? · 哪种代换可简化 2x(x² + 1)⁵ 的积分？
Its derivative 2x is already present, so du = 2x dx and the integral becomes u⁵ du. · 其导数 2x 已存在，故 du = 2x dx，积分变为 u⁵ du。
Differentiate an antiderivative to check it. Also check the domain, integration constant and bounds; differentiation alone does not prove that a definite integral or geometric area was evaluated correctly. Sheet 3.2 uses velocity to distinguish displacement, total distance and final position.
通过求导原函数来验证。 还需检查定义域、积分常数和界限；仅靠求导不能证明定积分或几何面积的评估正确。第3.2页利用速度区分位移、总路程和最终位置。
How do you check an integration answer reliably? · 如何可靠地检查积分答案？
Differentiate a proposed antiderivative and compare it with the original integrand on its domain. Also check the constant, bounds and sign or geometric-area interpretation; differentiation alone does not validate those steps. · 对假设的原函数求导，并将其与原被积函数在其定义域上比较。还需检查常数项、上下限以及符号或几何面积解释；仅凭微分不足以验证这些步骤。
An antiderivative valid on an interval cannot automatically justify evaluating across a singularity. Check the integrand's domain before applying upper-minus-lower. An initial position must be added when an integral gives only a displacement.
在某个区间上有效的原函数不能自动证明可以跨越奇点求值。应用上减下之前，请先检查被积函数的定义域。若积分仅给出位移，则必须加上初始位置。