The Fundamental Theorem of Calculus and Accumulation Functions · 微积分基本定理与累积函数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| Fundamental Theorem of Calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ | 微积分基本定理 | wēi jī fēn jī běn dìng lǐ |
| accumulation function/əˌkjuːmjʊˈleɪʃn ˈfʌŋkʃn/ | 累积函数 | lěi jī hán shù |
The bridge between area and slope
- Differentiation (slopes) and integration (areas) look unrelated — but they are inverses.
- The Fundamental Theorem of Calculus 微积分基本定理 (FTC) is that stunning connection.
- Its first part is about a function defined by an integral, called an accumulation function 累积函数.
- Take its derivative and — magically — you get the integrand back.
面积与斜率之间的桥
- 求导(斜率)与积分(面积)看似无关——但它们互为逆运算。
- 微积分基本定理(FTC)就是这个惊人的联系。
- 它的第一部分讲一个由积分定义的函数,叫累积函数。
- 对它求导,竟然——神奇地——把被积函数还原回来。
The accumulation function
- Fix a lower limit $a$ and let the upper limit vary: $$g(x)=\int_a^x f(t)\,dt$$
- $g(x)$ is the accumulated (signed) area under $f$ from $a$ up to $x$.
- As $x$ moves right, $g$ collects more area; the variable of integration $t$ is just a placeholder.
- $g$ is a genuine function of $x$ — you can graph it, differentiate it, and analyze it.
累积函数
- 固定下限 $a$,让上限变化:$$g(x)=\int_a^x f(t)\,dt$$
- $g(x)$ 是 $f$ 从 $a$ 到 $x$ 累积的(带符号)面积。
- 当 $x$ 右移,$g$ 收集更多面积;积分变量 $t$ 只是一个占位符。
- $g$ 是 $x$ 的真正函数——你能画它、求导、分析它。
Area accumulating up to x · 累积到 x 的面积
y = t² + 1 (integrand) · y = t² + 1（被积函数）
As the upper limit slides right, $g(x)$ collects more area; its instantaneous growth rate is exactly $f(x)$. · 随着上限向右滑动，$g(x)$ 收集了更多面积；其瞬时增长率恰好等于 $f(x)$。
A function of the form $g(x)=\int_a^x f(t)\,dt$ is called an ____ function. · 形如 $g(x)=\int_a^x f(t)\,dt$ 的函数称为 ____ 函数。
It accumulates signed area up to $x$. · 它在 $x$ 处累积有向面积。
In $g(x)=\int_a^x f(t)\,dt$, the variable $t$ is... · 在 $g(x)=\int_a^x f(t)\,dt$ 中，变量 $t$ 是...
$t$ is a dummy variable; $x$ is the actual input of $g$. · $t$ 是哑变量；$x$ 是 $g$ 的实际输入。
FTC Part 1: the derivative undoes the integral
- The first part of the FTC says:
-
$$g'(x)=\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$
- In words: the derivative of the accumulation function is the integrand itself.
- Rate of area accumulation at $x$ = the height $f(x)$ there. Beautifully simple.
FTC 第一部分:求导撤销积分
- FTC 的第一部分说:
-
$$g'(x)=\frac{d}{dx}\int_a^x f(t)\,dt = f(x)$$
- 用文字说:累积函数的导数就是被积函数本身。
- 在 $x$ 处面积累积的速率 = 那里的高度 $f(x)$。美丽而简单。
By FTC Part 1, $\dfrac{d}{dx}\int_a^x f(t)\,dt=$ · 根据FTC第1部分，$\dfrac{d}{dx}\int_a^x f(t)\,dt=$
The derivative of the accumulation function is the integrand $f(x)$. · 累积函数的导数即为被积函数 $f(x)$。
For · 支持 $g(x)=\int_2^x (t^2+1)\,dt$, find $g'(3)$. · 对于 $g(x)=\int_2^x (t^2+1)\,dt$，求 $g'(3)$。
$g'(x)=x^2+1$, so $g'(3)=10$. · $g'(x)=x^2+1$，所以 $g'(3)=10$。
With a chain-rule twist
- If the upper limit is a function $u(x)$, the chain rule kicks in:
- $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\cdot u'(x)$.
- Plug the upper limit into $f$, then multiply by the derivative of that upper limit.
- (A variable lower limit flips the sign, since swapping limits negates the integral.)
带一点链式法则
- 若上限是一个函数 $u(x)$,链式法则就登场:
- $\dfrac{d}{dx}\int_a^{u(x)} f(t)\,dt = f(u(x))\cdot u'(x)$。
- 把上限代入 $f$,再乘以那个上限的导数。
- (变的下限会翻转符号,因为交换积分限使积分变号。)
For · 支持 $\dfrac{d}{dx}\int_a^{x^2} f(t)\,dt$, the answer is... · 对于 $\dfrac{d}{dx}\int_a^{x^2} f(t)\,dt$，答案是...
Chain rule: plug in the upper limit, times its derivative $2x$. · 链式法则：代入上限，再乘以它的导数 $2x$。
When the upper limit is $u(x)$, you multiply the integrand-at-$u$ by $u'(x)$. · 当上限为 $u(x)$ 时，你将 $u$ 处的被积函数乘以 $u'(x)$。
That chain-rule factor is required. · 这个链式法则因子是必需的。
FTC Part 1 gives $g'(x)=f(x)$ only when $x$ is the plain upper limit. If the upper limit is $u(x)$ (like $x^2$), you must multiply by $u'(x)$: $\frac{d}{dx}\int_a^{x^2} f(t)\,dt = f(x^2)\cdot 2x$. Forgetting that chain-rule factor is the classic mistake.
FTC 第一部分给出 $g'(x)=f(x)$ 仅当 $x$ 是普通上限。若上限是 $u(x)$(如 $x^2$),你必须乘以 $u'(x)$:$\frac{d}{dx}\int_a^{x^2} f(t)\,dt = f(x^2)\cdot 2x$。忘掉那个链式法则因子是经典错误。
Let $g(x)=\displaystyle\int_2^x (t^2+1)\,dt$. Find $g'(x)$ and $g'(3)$.
- By FTC Part 1: $g'(x)=x^2+1$ (just the integrand with $t\to x$).
- $g'(3)=3^2+1=10$.
- (No need to compute the integral itself — the derivative reads it straight off.)
设 $g(x)=\displaystyle\int_2^x (t^2+1)\,dt$。求 $g'(x)$ 与 $g'(3)$。
- 由 FTC 第一部分:$g'(x)=x^2+1$(把被积函数中 $t\to x$)。
- $g'(3)=3^2+1=10$。
- (无需真正算出积分——求导直接读出它。)
An accumulation function $g(x)=\int_a^x f(t)\,dt$ collects signed area up to $x$. FTC Part 1: $g'(x)=f(x)$ — differentiating an accumulation function returns the integrand. If the upper limit is $u(x)$, multiply by $u'(x)$ (chain rule).
累积函数 $g(x)=\int_a^x f(t)\,dt$ 收集到 $x$ 为止的带符号面积。FTC 第一部分:$g'(x)=f(x)$——对累积函数求导返回被积函数。若上限是 $u(x)$,乘以 $u'(x)$(链式法则)。