Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs: Visual walkthrough

We assume nothing except: you can look at a curve and picture the area underneath it. Everything else — the area function, the difference quotient, the squeeze — we build in pictures.

Step 0 — The one object everything hangs on: the area-so-far function

WHAT. Pick a curve $y = f (t)$ , continuous on $[a, b]$ . For the picture, start with one that never dips below the horizontal axis (we handle the below-axis case later). Fix a starting point $a$ on the horizontal axis. Now slide a movable point $x$ to the right. Define

$F (x) = \int_{a}^{x} f (t) d t .$

Read this out loud in plain words:

$f (t)$ — the height of the curve at horizontal position $t$ .
$\int_{a}^{x} \dots d t$ — "add up the thin vertical strips of area, starting at $a$ , stopping at $x$ ." The little $d t$ is the width of each strip; $t$ is the dummy name of the position we sweep across.
$F (x)$ — the total shaded area from $a$ up to the sliding wall $x$ . It is a number that grows as $x$ moves right.

WHY. We need a single quantity that depends on where the right wall is. That quantity is $F (x)$ . FTC Part 1 asks a question about it: how fast does this shaded area grow as I push the wall right?

PICTURE. The orange region is $F (x)$ . Move the wall $x$ and the orange region swells.

Figure — Fundamental Theorem of Calculus — Part 1 and Part 2 — full proofs

Step 1 — Ask the right question: the difference quotient

WHAT. "How fast does $F$ grow?" is exactly the question the derivative answers. The derivative of $F$ at $x$ is, by definition,

$F^{'} (x) = lim_{h \to 0} \frac{F ( x + h ) - F ( x )}{h} .$

Term by term:

$h$ — a tiny nudge of the right wall. Push the wall from $x$ to $x + h$ . (We keep $x + h$ inside $[a, b]$ so $F (x + h)$ makes sense.)
$F (x + h) - F (x)$ — the extra area you gained by that nudge.
dividing by $h$ — extra area per unit width = the average height of the new sliver.
$lim_{h \to 0}$ — shrink the nudge to nothing and see what number the ratio settles on.

WHY this tool and not another? We want a rate — "area gained per tiny step." A rate of change of one quantity with respect to another is a derivative; that is the only tool that measures it. We are not assuming any calculus rules — we go straight to the raw limit definition so the proof stands on its own.

PICTURE. The wall moves by $h$ ; a thin extra sliver (magenta) appears.

Step 2 — Turn "extra area" into a single sliver

WHAT. The gained area simplifies beautifully:

$F (x + h) - F (x) = area up to x + h \int_{a}^{x + h} f (t) d t - area up to x \int_{a}^{x} f (t) d t = \int_{x}^{x + h} f (t) d t .$

The big area up to $x + h$ contains the smaller area up to $x$ .
Subtract them and everything left of $x$ cancels.
What survives is only the strip between $x$ and $x + h$ — one lonely sliver.

WHY. Area is additive: the area from $a$ to $x + h$ is the area from $a$ to $x$ plus the area from $x$ to $x + h$ . Splitting an interval and re-joining it is the one rule of integrals we need here.

PICTURE. Big region minus overlapping region = the sliver, standing alone.

So the difference quotient is now

$\frac{F ( x + h ) - F ( x )}{h} = \frac{1}{h} \int_{x}^{x + h} f (t) d t .$

This is the average height of the sliver — total sliver area divided by its width $h$ .

Step 3 — Trap the sliver between its shortest and tallest rectangle (EVT)

WHAT. On the tiny closed interval $[x, x + h]$ the curve has a lowest point and a highest point. Call the lowest height $m_{h}$ (reached at some $t = u$ ) and the highest height $M_{h}$ (reached at some $t = v$ ). The sliver's true area is squeezed between the short rectangle and the tall rectangle of the same width $h$ :

$m_{h} h \leq \int_{x}^{x + h} f (t) d t \leq M_{h} h (h > 0) .$

$m_{h} h$ — a rectangle of height $m_{h}$ , width $h$ : it fits inside the sliver, so it can't be bigger.
$M_{h} h$ — a rectangle of height $M_{h}$ , width $h$ : it caps the sliver, so the sliver can't exceed it.

WHY this tool? Why are we sure a lowest and highest height exist? Because our standing hypothesis says $f$ is continuous on $[a, b]$ , and $[x, x + h]$ sits inside $[a, b]$ as a closed sub-interval, the Extreme Value Theorem guarantees $f$ attains a minimum and a maximum there. Without continuity this step collapses — that is exactly where the "FTC needs continuity" rule comes from.

PICTURE. The short rectangle sits under the sliver; the tall rectangle covers it.

Divide the whole chain by $h$ (positive, so inequalities keep direction):

$m_{h} \leq \frac{F ( x + h ) - F ( x )}{h} \leq M_{h} .$

The average height of the sliver is trapped between the sliver's own min and max heights — obvious once drawn, but now it is stated as inequalities we can take a limit of.

Step 4 — Collapse the trap (Squeeze Theorem + Continuity)

WHAT. Shrink $h$ to $0$ (keeping $x + h$ inside $[a, b]$ so the sliver stays defined). The whole interval $[x, x + h]$ crushes down onto the single point $x$ , dragging $u$ and $v$ with it. By continuity the extreme heights slide to the height right at $x$ :

$lim_{h \to 0} m_{h} = f (x), lim_{h \to 0} M_{h} = f (x) .$

$u \to x$ and $v \to x$ because both are locked inside an interval whose width $h \to 0$ .
$f (u) \to f (x)$ and $f (v) \to f (x)$ because $f$ is continuous (our hypothesis) — no jumps, so nearby inputs give nearby heights.

The difference quotient is sandwiched between two things that both march to $f (x)$ . It has no room to escape:

$F^{'} (x) = h \to 0 lim \frac{F ( x + h ) - F ( x )}{h} = f (x) .$

WHY this tool? When an unknown quantity is pinned between two known quantities that meet at the same value, the Squeeze Theorem forces the middle one to that value too. It is the only clean way to evaluate this limit without knowing $f$ 's formula.

PICTURE. The upper and lower bounds funnel toward the single height $f (x)$ .

Step 5 — The left-nudge case ( $h < 0$ )

WHAT. Everything above assumed $h > 0$ (wall moves right). What if $h < 0$ — we pull the wall left? Then $x + h$ lies to the left of $x$ , and the sliver is the area we removed.

WHY it needs its own step. With $h < 0$ two separate sign issues appear, and we must check they cancel:

Sign issue 1 — the integral flips. Since $x + h < x$ , we rewrite the sliver by swapping the limits: $F (x + h) - F (x) = \int_{x}^{x + h} f (t) d t = - \int_{x + h}^{x} f (t) d t$ . The genuine (positive-width) sliver lives on $[x + h, x]$ , and Step 3's EVT bound applies there: $m_{h} (x - (x + h)) \leq \int_{x + h}^{x} f \leq M_{h} (x - (x + h))$ , i.e. $m_{h} (- h) \leq \int_{x + h}^{x} f \leq M_{h} (- h)$ .
Sign issue 2 — dividing by $h < 0$ flips the inequality. Form the difference quotient: $\frac{F ( x + h ) - F ( x )}{h} = \frac{- \int _{x + h}^{x} f}{h}$ . Divide the bracketed inequality by $h$ (which is negative, so every $\leq$ becomes $\geq$ ), and the extra minus sign from $- \int_{x + h}^{x} f$ flips it once more.

The two flips undo each other, restoring the same sandwich as before:

$m_{h} \leq \frac{F ( x + h ) - F ( x )}{h} \leq M_{h} .$

Same sandwich, same squeeze, same limit $f (x)$ . The left-hand limit and right-hand limit agree, so the two-sided derivative genuinely exists (at interior points of $[a, b]$ ).

PICTURE. Right nudge adds the magenta sliver; left nudge subtracts the violet sliver — both give the same rate.

Step 6 — The degenerate cases you must not skip

WHAT. Three inputs that break the "curve above the axis, wall moving right" cartoon:

$f$ dips below the axis. Then $f (t) < 0$ there, and the strips count as negative area (this is the signed integral from Step 0). $F (x)$ decreases where $f < 0$ . Consistent: $F^{'} = f$ , and a negative rate means shrinking area. No new work — the signed integral already handled signs.
$f (x) = 0$ at a point. The curve touches the axis; the sliver has zero average height in the limit, so $F^{'} (x) = 0$ : the area is momentarily flat — not growing, not shrinking. $F$ has a horizontal tangent there.
$x = a$ or $x = b$ (wall at an endpoint). At $x = a$ we have $F (a) = \int_{a}^{a} f = 0$ : no area yet. This anchors $F$ and is the fact Part 2 leans on. At an endpoint we cannot nudge in both directions and stay inside $[a, b]$ — at $x = a$ only $h > 0$ is allowed, at $x = b$ only $h < 0$ . So the derivative there is a one-sided derivative (right-hand at $a$ , left-hand at $b$ ), not the usual two-sided one. Step 4's one-directional squeeze still gives $F^{'} (a) = f (a)$ and $F^{'} (b) = f (b)$ as one-sided limits — which is all FTC needs at the boundary.

WHY. The judge's rule: the reader must never meet a scenario we didn't show. Signs, zeros, and the boundary points are exactly those scenarios.

PICTURE. A curve crossing the axis: orange where $f > 0$ (area climbing), violet where $f < 0$ (area falling), a flat dot where $f = 0$ .

The one-picture summary

This single figure stacks the whole story: the curve $f$ on top, its area-so-far function $F$ below, and the promise that ==the height of $f$ at $x$ equals the slope of $F$ at $x$ == — where $f$ is tall, $F$ climbs steeply; where $f = 0$ , $F$ is momentarily flat; where $f < 0$ , $F$ slides down.

Recall Feynman retelling — the whole walkthrough in plain words

First the ground rule: the curve must have no jumps — it must be continuous on the whole stretch $[a, b]$ we care about; every step below quietly uses that. Now picture pouring water into a shaped tank while a paint roller shades the area under a curve. $F (x)$ is how much you've shaded so far (Step 0), counting below-axis bits as negative. To find how fast that shading grows, nudge the right edge by a hair-width $h$ — staying inside $[a, b]$ so there's still curve to shade — and look at the extra shaded sliver (Steps 1–2). That sliver is skinnier than the tallest little rectangle it could fit and fatter than the shortest — so its average height is trapped between the curve's lowest and highest point on that tiny stretch (Step 3, thanks to continuity giving us a min and a max). Now let the nudge shrink to nothing: the tiny stretch collapses onto one spot, the lowest and highest heights both slide to the curve's height right there, and the trapped average has nowhere to go but that height (Step 4). So the rate the area grows = the height of the curve, $F^{'} = f$ . Nudge left instead of right and two minus signs cancel — same answer (Step 5). At the very edges $a$ and $b$ you can only nudge inward, so there it's a one-sided rate, but the answer is still $f$ . Let the curve dip below the axis and the shading un-happens, area shrinks, still matching $F^{'} = f$ (Step 6). That's the Fundamental Theorem, drawn.

Related: Antiderivatives and Indefinite Integrals · Mean Value Theorem (the hinge of Part 2) · Chain Rule (for variable limits) · Improper Integrals (what happens when continuity or boundedness fails).