نظرية التفاضل والتكامل الأساسية ودوال التجميع
| English | العربية |
|---|---|
| Fundamental Theorem of Calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ | المبرهنة الأساسية للتفاضل والتكامل |
| accumulation function/əˌkjuːmjʊˈleɪʃn ˈfʌŋkʃn/ | دالة التراكم |
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.
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.
المساحة المتراكمة حتى x
y = t² + 1 (مدخل التكامل)
عندما تنزلق الحد العلوي يميناً، يجمع $g(x)$ مساحة أكثر؛ ومعدل نموه اللحظي يساوي بالضبط $f(x)$.
دالة من الشكل $g(x)=\int_a^x f(t)\,dt$ تُسمى دالة ____.
إنها تجمع المساحة الموجهة حتى $x$.
في $g(x)=\int_a^x f(t)\,dt$، المتغير $t$ هو...
$t$ متغير وهمي؛ $x$ هو المدخل الفعلي لـ $g$.
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.
وفقًا للمادة 1 من قانون FTC، $\dfrac{d}{dx}\int_a^x f(t)\,dt=$
مشتقة دالة التجميع هي مدخل التكامل $f(x)$.
بالنسبة لـ $g(x)=\int_2^x (t^2+1)\,dt$، أوجد $g'(3)$.
$g'(x)=x^2+1$، إذن $g'(3)=10$.
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.)
بالنسبة لـ $\dfrac{d}{dx}\int_a^{x^2} f(t)\,dt$، الإجابة هي...
قاعدة السلسلة: عوض بالحد العلوي، مضروباً في مشتقتها $2x$.
عندما يكون الحد العلوي $u(x)$، تضرب مدخل التكامل عند $u$ في $u'(x)$.
عامل قاعدة السلسلة هذا مطلوب.
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.
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.)
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).