The Fundamental Theorem of Calculus and Accumulation Functions
Introduced| English | Meaning |
|---|---|
| Fundamental Theorem of Calculus/ˌfʌndəˈmentl ˈθɪərəm ɒv ˈkælkjʊləs/ | The theorem linking differentiation and integration: accumulation differentiates to its continuous integrand, and definite integrals can be evaluated using antiderivatives. |
| accumulation function/əˌkjuːmjʊˈleɪʃn ˈfʌŋkʃn/ | A function defined by accumulating a quantity, commonly through an integral from a fixed lower limit to a variable upper limit. |
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.
Area accumulating up to x
y = t² + 1 (integrand)
As the upper limit slides right, $g(x)$ collects more area; its instantaneous growth rate is exactly $f(x)$.
A function of the form $g(x)=\int_a^x f(t)\,dt$ is called an ____ function.
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It accumulates signed area up to $x$.
In $g(x)=\int_a^x f(t)\,dt$, the variable $t$ is...
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$t$ is a dummy variable; $x$ is the actual input of $g$.
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.
By FTC Part 1, $\dfrac{d}{dx}\int_a^x f(t)\,dt=$
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The derivative of the accumulation function is the integrand $f(x)$.
For $g(x)=\int_2^x (t^2+1)\,dt$, find $g'(3)$.
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$g'(x)=x^2+1$, so $g'(3)=10$.
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.)
For $\dfrac{d}{dx}\int_a^{x^2} f(t)\,dt$, the answer is...
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Chain rule: plug in the upper limit, times its derivative $2x$.
When the upper limit is $u(x)$, you multiply the integrand-at-$u$ by $u'(x)$.
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That chain-rule factor is required.
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).
Write from an accumulation derivative of 4
- g(x) is the integral of t squared from 0 to x.
- The derivative is wanted at 2.
What is g prime of 2, and what extra factor appears if the upper limit is x squared?
The integrand t squared at 2 is 4.
g(x) = integral from 0 to x of t squared dt. g prime of 2 is what?
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The integrand at 2 is 4.