Antiderivative — definition, family of solutions (+C): Visual walkthrough

Before we begin, three plain-word anchors so no symbol arrives unearned:

Step 1 — A slope-report is just an arrow at every point

WHAT. We take a slope-report $f (x)$ and draw it not as a graph but as a field of little tilted arrows — one short line segment at each point of the plane, tilted to the slope $f$ demands there.

WHY. This is the honest picture of what an antiderivative is given. We are not told heights; we are told tilts. Drawing the tilts first stops us from secretly assuming a height we were never handed.

PICTURE. Look at the figure: at every grid point sits a short pastel dash. Its steepness is fixed by $f$ ; its vertical position is NOT fixed by anything yet — that freedom is the whole story.

Figure — Antiderivative — definition, family of solutions (+C)

$f (x)$ — the slope the arrow at horizontal position $x$ must have.
Each dash — one instruction: "a curve passing here must be this steep."

Step 2 — Threading one curve through the arrows

WHAT. We now trace a single curve $F$ that stays tangent (flush) to every arrow it passes. Wherever it goes, its ruler-tilt equals the local arrow.

WHY. A curve tangent to the field everywhere is exactly a curve whose slope-report is $f$ — i.e. $F^{'} (x) = f (x)$ . That equation is the definition of an antiderivative, now drawn instead of written.

PICTURE. The bold lavender curve hugs the arrows. Notice we had to choose where to start it — we dropped it in at one height and let the arrows steer the rest.

$F^{'} (x) = f (x)$

$F^{'} (x)$ — the tilt of our bold curve at $x$ (its ruler).
$f (x)$ — the tilt the arrow field demanded at $x$ .
$=$ — "curve matches the field": that is what makes $F$ an antiderivative.

Step 3 — Slide it up: the slope never changes

WHAT. Take the bold curve and shift it straight up by a fixed amount $C$ . Call the new curve $G (x) = F (x) + C$ . Check its tilt against the arrows.

WHY. Sliding vertically moves height but drags no tilt with it — a raised hill is the same shape. To see this in symbols we use the single fact that the slope of a flat, constant line is $0$ : $G^{'} (x) = \frac{d}{d x} (F (x) + C) = F^{'} (x) + = 0 \frac{d}{d x} (C) = f (x) .$

PICTURE. Two curves, coral above lavender, offset by the vertical gap $C$ . At the marked $x$ their little rulers are parallel — same tilt, different height. The raised copy still hugs the same arrows.

$C$ — a constant, the fixed vertical gap (the amount we slid).
$\frac{d}{d x} (C) = 0$ — a horizontal line has zero slope, so shifting adds nothing to the tilt.
$G^{'} (x) = f (x)$ — the shifted curve is also a valid antiderivative.

So there are at least as many antiderivatives as there are choices of $C$ — infinitely many.

Step 4 — Are there sneaky other curves, not just vertical slides?

WHAT. Suppose someone claims a second antiderivative $G$ that is not a plain vertical copy of $F$ . We test that claim by subtracting: define the gap function $H (x) = G (x) - F (x)$ .

WHY. If $G$ and $F$ both match the field, their difference should carry no tilt at all. Subtracting cancels the shared slope and isolates whatever "extra shape" $G$ might secretly have: $H^{'} (x) = G^{'} (x) - F^{'} (x) = f (x) - f (x) = 0 for all x .$

PICTURE. Top panel: two candidate curves. Bottom panel: their gap $H$ , drawn as a green track whose ruler is flat everywhere — slope $0$ at every $x$ .

$H (x) = G (x) - F (x)$ — the vertical distance between the two candidates at each $x$ .
$H^{'} (x) = 0$ — that distance has zero slope; it never tilts.

Step 5 — Zero slope everywhere forces a flat line (the MVT step)

WHAT. We prove the gap $H$ cannot secretly rise or fall — it must be a perfectly flat, constant line.

WHY. This is the one place we need a named theorem. The Mean Value Theorem says: between any two points, a smooth curve must somewhere have a ruler tilted at the average steepness between them. But every ruler on $H$ reads $0$ . So the average steepness between any two points is $0$ — meaning the two points sit at the same height. Same height for every pair ⇒ the whole thing is one level line.

PICTURE. Pick two points on $H$ . The dashed "average-slope" chord between them is drawn — and because all rulers are flat, that chord is horizontal, forcing the right point down to the left point's height. Slide the pair anywhere: always horizontal.

$H^{'} (x) = 0 on interval I ⟹ H (x) = C .$

The dashed chord — the average slope the MVT promises must be achieved somewhere.
Horizontal chord — average slope $0$ , so no height change is possible.
$H (x) = C$ — the gap is a single constant.

Putting Step 4 and Step 5 together: $G (x) - F (x) = C ⟹ G (x) = F (x) + C .$ Any "other" antiderivative was a vertical slide all along. The family is exactly $F + C$ — no more, no less.

Step 6 — The degenerate case: what if the domain has a hole?

WHAT. Everything above quietly assumed one connected interval $I$ . Now break the ground: remove a point, so the domain is two separate stretches (the classic case is $f (x) = \frac{1}{x}$ , undefined at $x = 0$ ).

WHY. The MVT of Step 5 only compares two points if you can travel between them without leaving the domain. Across a gap there is no chord to draw — so the constant on the left piece and the constant on the right piece need not agree. Each connected island gets its own $C$ .

PICTURE. A domain split by a shaded forbidden strip. Left island: a curve raised by $C_{1}$ . Right island: an independently raised curve by $C_{2}$ . Both hug the same arrows on their own turf; nothing ties their heights together.

$\int \frac{1}{x} d x = {ln (- x) + C_{1}, ln (x) + C_{2}, x < 0 x > 0.$

shaded strip — the hole ( $x = 0$ ), where no curve and no chord may cross.
$C_{1}, C_{2}$ — two independent constants, one per island.
$ln (- x)$ / $ln (x)$ — the antiderivative shape on each side; see Logarithmic & Exponential Integrals for why the log appears exactly at the power-rule exception $n = - 1$ .

Step 7 — Choosing which curve: a single point pins $C$

WHAT. Out of the infinite stack, demand the curve pass through one known point $F (x_{0}) = y_{0}$ .

WHY. The stack has exactly one free knob, $C$ (Steps 3–5). One equation determines one unknown, so a single point selects a single member. This is the whole idea behind an initial value problem.

PICTURE. The pastel stack of parallel curves, with one coral point $(x_{0}, y_{0})$ marked. Only one curve threads that dot; the rest are dimmed.

The one-picture summary

WHAT. One figure compresses the whole walkthrough: the arrow field (Step 1), the infinite stack of parallel curves all tangent to it (Steps 2–5), the vertical gap $C$ labelled between two of them (Step 3), and one highlighted member chosen by a point (Step 7).

arrow field — the given slope-report $f$ .
the stack — the family $F (x) + C$ ; every curve shares the same tilt at each $x$ .
the gap $C$ — the one and only freedom, proven complete by Steps 4–5.
the coral dot — an initial condition selecting a single curve.

Recall Feynman retelling — the whole walkthrough in plain words

Somebody tells you the tilt of a hill at every step but never the height you started at. First you scatter those tilts as tiny dashes across the field (Step 1). You thread a curve that leans exactly along them (Step 2). Then you notice: lift that whole curve straight up and it still leans the same way — because a flat lift adds no tilt (Step 3). Could a totally different-shaped curve also fit? Subtract it from yours; the leftover has zero tilt everywhere (Step 4), and a thing with zero tilt everywhere on one stretch cannot go up or down — it's dead flat, a constant (Step 5, the Mean Value Theorem). So the "different" curve was just yours slid up by some fixed $C$ . The only trap: if the ground has a gap you can't walk across, the two sides can float at different heights, one $C$ each (Step 6). Finally, tell me the height at just one spot and I pick the exact curve from the infinite stack (Step 7). That unknown starting height, and nothing else, is the $+ C$ .

Connections

Parent topic — Antiderivative & the family +C
Mean Value Theorem — the engine of Step 5.
Fundamental Theorem of Calculus — where a chosen curve meets areas.
Power Rule (Integration) & Logarithmic & Exponential Integrals — the $n = - 1$ hole in Step 6.
Differential Equations — Initial Value Problems — Step 7 formalised.
Definite Integral — where $+ C$ harmlessly cancels.