Applying Properties of Definite Integrals · 应用定积分的性质
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| linearity/lɪˈnɪərɪti/ | 线性 | xiàn xìng |
Rules that make integrals easier
- Definite integrals obey clean algebra — the same friendly rules limits and derivatives have.
- These properties let you split, combine, and rearrange integrals without recomputing areas.
- Most follow directly from "integral = signed area."
- Learn the handful here and hard-looking integrals often simplify in one step.
让积分更容易的规则
- 定积分遵守干净的代数——与极限和导数一样友好的规则。
- 这些性质让你拆分、合并、重排积分,而无需重新计算面积。
- 大多数直接来自"积分 = 带符号面积"。
- 学会这里这几条,看似难的积分常常一步就化简。
$\int fg\,dx$ equals $\left(\int f\,dx\right)\left(\int g\,dx\right)$. · $\int fg\,dx$ 等于 $\left(\int f\,dx\right)\left(\int g\,dx\right)$。
There is no product rule for integrals. · 积分没有乘法法则。
Linearity: split sums and pull out constants
- Constant multiple: $\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$.
- Sum / difference: $\displaystyle\int_a^b \big(f\pm g\big)\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$.
- Together these are linearity 线性 — integrate term by term, coefficients out front.
- Just like differentiation, you handle a combination one piece at a time.
线性性:拆开和、提出常数
- 常数倍: $\displaystyle\int_a^b k\,f(x)\,dx = k\int_a^b f(x)\,dx$。
- 和 / 差: $\displaystyle\int_a^b \big(f\pm g\big)\,dx = \int_a^b f\,dx \pm \int_a^b g\,dx$。
- 它们合起来是线性性——逐项积分,系数提到前面。
- 和求导一样,你把组合一块一块地处理。
Given $\int_0^4 f\,dx=10$ and $\int_0^4 g\,dx=3$, find $\int_0^4 (2f-g)\,dx$. · 已知 $\int_0^4 f\,dx=10$ 和 $\int_0^4 g\,dx=3$，求 $\int_0^4 (2f-g)\,dx$。
$2(10)-3=17$.
Zero width and swapping limits
- Zero width: $\displaystyle\int_a^a f(x)\,dx = 0$ — no interval, no area.
- Swap the limits: $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$ — reversing direction negates it.
- So the order of the limits carries a sign.
- These two keep bookkeeping consistent when limits move around.
零宽度与交换积分限
- 零宽度: $\displaystyle\int_a^a f(x)\,dx = 0$——没有区间,没有面积。
- 交换积分限: $\displaystyle\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$——反向使它变号。
- 所以积分限的顺序带着一个符号。
- 当积分限移动时,这两条保持记法一致。
What is $\int_5^5 f(x)\,dx$ for any $f$? · 什么是 $\int_5^5 f(x)\,dx$ 对于任何 $f$?
Zero width interval → $0$. · 宽度为零的区间 → $0$。
If $\int_1^4 f\,dx=7$, then $\int_4^1 f\,dx=$ · 若 $\int_1^4 f\,dx=7$，则 $\int_4^1 f\,dx=$
Swapping limits negates: $-7$. · 交换上下限改变符号：$-7$。
Select all · 所有 valid definite-integral properties. · 选择所有有效的定积分性质。
The first three hold; the product "rule" is false. · 前三项成立；乘积“法则”是错误的。
Splitting an interval
- Adjacent intervals: $\displaystyle\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$.
- You can break one integral into pieces at any interior point $b$ and add them.
- This is essential for piecewise integrands and for using given sub-areas.
- It also runs backward: combine two adjacent integrals into one.
拆分区间
- 相邻区间: $\displaystyle\int_a^c f\,dx = \int_a^b f\,dx + \int_b^c f\,dx$。
- 你可以在任意内部点 $b$ 把一个积分拆成几块并相加。
- 这对分段被积函数和使用已知子面积很关键。
- 它也能反向:把两个相邻积分合并成一个。
Split the area at an interior point · 在内点处分隔面积
y = x²
The area from $a$ to · 到 $c$ equals the area from $a$ to · 到 $b$ plus $b$ to · 到 $c$ — integrals split over adjacent intervals. · 从 $a$ 到 $c$ 的面积等于从 $a$ 到 $b$ 的面积加上从 $b$ 到 $c$ 的面积——积分可在相邻区间上拆分。
If $\int_0^2 f\,dx=5$ and $\int_2^6 f\,dx=8$, find $\int_0^6 f\,dx$. · 若 $\int_0^2 f\,dx=5$ 且 $\int_2^6 f\,dx=8$，求 $\int_0^6 f\,dx$。
Adjacent intervals add: $5+8=13$. · 相邻区间相加：$5+8=13$。
There is no "product rule" for integrals: $\int fg\,dx \neq \left(\int f\,dx\right)\left(\int g\,dx\right)$. Linearity only splits across sums and constant multiples, never products or quotients. And remember swapping the limits flips the sign — a very easy point to drop.
积分没有"乘积法则":$\int fg\,dx \neq \left(\int f\,dx\right)\left(\int g\,dx\right)$。线性性只在和与常数倍上拆分,绝不在乘积或商上。并记住交换积分限会翻转符号——一个极易漏掉的点。
Given $\displaystyle\int_0^4 f\,dx=10$ and $\displaystyle\int_0^4 g\,dx=3$, find $\displaystyle\int_0^4\big(2f-g\big)\,dx$.
- Linearity: $\displaystyle\int_0^4 2f\,dx - \int_0^4 g\,dx = 2\int_0^4 f\,dx - \int_0^4 g\,dx$.
- $=2(10)-3=17$.
- No need to know $f$ or $g$ themselves — the properties do it.
已知 $\displaystyle\int_0^4 f\,dx=10$、$\displaystyle\int_0^4 g\,dx=3$,求 $\displaystyle\int_0^4\big(2f-g\big)\,dx$。
- 线性性:$\displaystyle\int_0^4 2f\,dx - \int_0^4 g\,dx = 2\int_0^4 f\,dx - \int_0^4 g\,dx$。
- $=2(10)-3=17$。
- 无需知道 $f$ 或 $g$ 本身——性质就搞定了。
Definite-integral properties: linearity ($\int kf = k\int f$; $\int (f\pm g)=\int f\pm\int g$), zero width ($\int_a^a f=0$), swapping limits negates ($\int_b^a f=-\int_a^b f$), and splitting ($\int_a^c=\int_a^b+\int_b^c$). There is no product rule for integrals.
定积分性质:线性性($\int kf = k\int f$;$\int (f\pm g)=\int f\pm\int g$)、零宽度($\int_a^a f=0$)、交换积分限变号($\int_b^a f=-\int_a^b f$)、拆分($\int_a^c=\int_a^b+\int_b^c$)。积分没有乘积法则。