Fundamental Theorem Of Calculus

TFundamental theorem of calculus:

Let $I \subseteq \mathbb{R}$ be an interval, $f : I \to \mathbb{R}$ be continuous and $a \in I$ . Then the following statements are satisfied:

i) The function $F : I \to \mathbb{R}, \quad x \mapsto F(x) = \int_a^x f(t) \, dt$ is an antiderivative of $f$ .

How to use

Visaulization of the first part of the proof
Drag the green point $x$ on the $x$ -axis.
Drag the purple point $x+h$ on the $x$ -axis (this changes both the size and the sign of $h$ ).
The blue curve is $f(t)$ .
The red curve is $F(x)=\int_0^x f(t)\,dt$ .
The blue shaded area represents $F(x)=\int_0^x f(t)\,dt$ .
The orange shaded area represents $\int_{x_L}^{x_R} f(t)\,dt = F(x_R)-F(x_L),$ i.e. the area “slice” between $x$ and $x+h$ .
The red points $F(x)$ and $F(x+h)$ lie on the graph of $F$ , and the line through them is the secant line. Its slope is the difference quotient $\dfrac{F(x+h)-F(x)}{h}$ .
The green rectangle has base width $|h|=x_R-x_L$ and height $\frac{F(x_R)-F(x_L)}{x_R-x_L},$ so its area matches the orange slice.
The green point $\xi_h$ (with the point $f(\xi_h)$ on the graph) is chosen so that $f(\xi_h)=\frac{F(x+h)-F(x)}{h}.$
First choose $x$ , then move $x+h$ so that $h$ is not too small: compare the secant slope on $F$ with the height $f(\xi_h)$ . They represent the same number.
Now drag $x+h$ $x + h$ closer to $x$ $x$ (so $|h|\to 0$ $∣ h ∣ \to 0$ ):
- the orange slice becomes thinner,
- $\xi_h$ moves toward $x$ ,
- the secant line becomes closer to the tangent line at $x$ , tuiirning into a tangent line (colored in purple)
- the slope $\dfrac{F(x+h)-F(x)}{h}$ approaches $f(x)$ .

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Points shaped like <> act as sliders

Description

Link to code.

Reference: Lecture Notes Calculus 1 ( May22,2022 ) Theorem7.10 page 133

The applet visualizes the Fundamental Theorem of Calculus (I) following the proof.

Given a continuous function $f$ , we define

F(x) := \int_0^x f(t)\,dt,

(blue area). The graph of $F$ is shown together with the graph of $f$ .

To study the differentiability of $F$ , two points $x$ and $x+h$ are selected on the $x$ -axis. The difference quotient

\frac{F(x+h)-F(x)}{h}

is visualized as the slope of the secant line of $F$ between these two points.

I am reusing part of the Mean value theorem for integrals, wich is used in the proof as well as visualizing the derivative $F'(x) = f(x)$ , as in the example Secant and Tangent.

In this applet, we choose a strictly increasing and invertible function.

// f(t) = 0.1 t^3 + 1
const f = (t: number) => 0.1 * t ** 3 + 1;

// Antiderivative F(x) = 0.025 x^4 + x
const F = (x: number) => 0.025 * x ** 4 + x;

// inverse of f
const fInverse = (y: number) => Math.cbrt(10 * (y - 1));

This additional structure allows the point $\xi_h$ to be computed explicitly as

\xi_h = f^{-1}\!\left(\frac{F(x+h)-F(x)}{h}\right),

without relying on numerical methods. The point $\xi_h$ and the corresponding height $f(\xi_h)$ are displayed in the applet.

As $h$ becomes smaller, the interval $[x,x+h]$ shrinks and the point $\xi_h$ moves closer to $x$ . Consequently, the slope of the secant line converges to $f(x)$ . This visually illustrates the conclusion of the theorem:

F'(x) = f(x).