Work done by variable force — integration: Visual walkthrough

Step 1 — What "work" means when the push never changes

WHAT. We start with the simplest possible situation: a constant push. You shove a box with a steady force $F$ (measured in newtons, N — a unit of "how hard you push") across a distance $d$ (in metres, m). The work done is defined as $W = F \times d .$

Let me name every piece right where it sits:

$F$ — the steady push, same number the whole way (N).
$d$ — how far the box travelled while you pushed (m).
$W$ — the work, the "total effort delivered" (unit: newton·metre = joule, J).

WHY this first. Everything harder is built out of this. We need a rock-solid mental image of $W = F d$ before we let $F$ wobble.

PICTURE. Put the push $F$ on the vertical axis and position $x$ on the horizontal axis. A constant push is a flat horizontal line. The work $W = F \times d$ is exactly the area of the rectangle underneath: height $F$ , width $d$ . Look at the red rectangle.

Figure — Work done by variable force — integration

Step 2 — The trouble: real forces change as you move

WHAT. Now let the push depend on where you are. We write this as $F (x)$ , read "F of x" — meaning "the force when the box is at position $x$ ." At $x = 1$ the force might be $5$ N; at $x = 2$ it might be $9$ N. Same box, different push at different places.

Term-by-term:

$x$ — the position of the box along its path (m).
$F (x)$ — the push evaluated at that position. The little " $(x)$ " is not multiplication; it means "look up the force for this $x$ ."

WHY this matters. The rectangle from Step 1 needs one height. But now the top edge slopes and curves — there is no single height. The formula $W = F d$ has no honest number to use for $F$ . It is broken.

PICTURE. Same axes as before, but the top edge is now a rising curve. Ask yourself: "what is the area under a curved roof?" A single rectangle can't cover it — it either pokes out or leaves gaps (red shows the mismatch).

Step 3 — The one honest move: zoom in until the curve looks flat

WHAT. Pick a tiny sliver of the path, from $x$ to $x + Δ x$ . Here $Δ x$ (read "delta x", the Greek capital delta) just means a small width in $x$ — a little step. Over a step that small, the curve barely moves, so the force is practically one number, $F (x)$ .

The work done on that single sliver is a thin rectangle: $Δ W \approx F (x) Δ x .$

Every symbol:

$Δ x$ — the tiny width of this one step (m).
$F (x)$ — the force at the start of the step, used as the (almost) constant height (N).
$Δ W$ — the small bit of work done crossing that step (J).
$\approx$ — "approximately equal," because the curve isn't perfectly flat over $Δ x$ — but nearly.

WHY this is allowed. This is the key idea of the whole chapter: a curve, zoomed in far enough, looks straight. On a narrow enough slice, $F (x)$ hardly changes, so Step 1's rectangle is legal on that slice. We traded one impossible problem (curved roof) for many easy ones (tiny flat-topped rectangles).

PICTURE. Zoom into a single thin strip under the curve. Its top is almost flat; it is a genuine little rectangle of height $F (x)$ and width $Δ x$ (shown in red). Its area is the work for that step.

Step 4 — Tile the whole path with slivers and add them up

WHAT. Do Step 3 everywhere. Chop the path from the start $x_{A}$ to the end $x_{B}$ into $N$ slivers. Number them $i = 1, 2, \dots, N$ . Sliver $i$ has width $Δ x_{i}$ and force $F (x_{i})$ , so its work is $F (x_{i}) Δ x_{i}$ . Add every sliver: $W \approx \sum_{i = 1}^{N} F (x_{i}) Δ x_{i} .$

Reading the new symbols:

$x_{A}, x_{B}$ — start and end positions of the box (m).
$N$ — how many slices we cut the path into (a plain counting number).
$i$ — the label of a slice, running $1$ to $N$ .
$i = 1 \sum N$ — the summation sign; "add up the thing on the right for $i = 1$ , then $2$ , all the way to $N$ ."
$F (x_{i}) Δ x_{i}$ — the area of the $i$ -th rectangle: its height times its width.

WHY. Total work is the total effort, and effort is additive: doing the whole path is doing sliver 1, then sliver 2, then sliver 3… So we just sum the sliver-rectangles. This total is called a Riemann sum — a staircase of rectangles hugging the curve.

PICTURE. A staircase of rectangles filling the region under the curve. Each red step is one $F (x_{i}) Δ x_{i}$ . Together they approximate the true curved area — but you can still see little triangular gaps between the staircase and the curve.

Step 5 — Shrink the slivers to zero: the staircase becomes the curve

WHAT. Those little gaps in Step 4 are the error. Kill them by using more, thinner slivers: let $N \to \infty$ (infinitely many) so every $Δ x_{i} \to 0$ (infinitely thin). In that limit the staircase presses perfectly onto the curve and the sum becomes the definite integral: $W = lim_{Δ x_{i} \to 0} \sum_{i = 1}^{N} F (x_{i}) Δ x_{i} = \int_{x_{A}}^{x_{B}} F (x) d x .$

Decoding the integral piece by piece — it is the summation, transformed:

$\int$ — a stretched "S" for Sum; the limit of $\sum$ .
$x_{A}$ (bottom) and $x_{B}$ (top) — the start and end of the path; they replace $i = 1$ and $i = N$ .
$F (x)$ — the height, exactly as before.
$d x$ — the infinitely thin width, what $Δ x_{i}$ became in the limit.
Put together: "add up (height $F (x)$ ) × (infinitesimal width $d x$ ) continuously from $x_{A}$ to $x_{B}$ ."

WHY this tool and not another. We needed a machine that sums infinitely many infinitely small products. Ordinary multiplication ( $F d$ ) can't — it uses one height. Ordinary addition ( $\sum$ ) leaves gaps. The definite integral is defined as exactly this limit-of-a-sum, so it is the precise tool the problem demands. (Its full machinery lives in Area under curves and the definite integral.)

PICTURE. Left half: a coarse staircase with visible gaps. Right half: many razor-thin red rectangles — the gaps have vanished and the shape is the smooth area under the curve. The staircase has become the curve's area.

Step 6 — When force and motion aren't lined up: the $cos θ$ case

WHAT. So far the push pointed straight along the motion. In general the force $F$ (an arrow with size and direction) and the tiny displacement $d r$ (a tiny move-arrow) can sit at an angle $θ$ to each other. Then a single sliver's work is $d W = F \cdot d r = F d r cos θ .$

Term-by-term:

$F$ — the force arrow (the little arrow on top means "this has a direction").
$d r$ — the tiny displacement arrow, length $d r$ .
$θ$ — the angle between the two arrows.
$cos θ$ — the fraction of the force that points along the motion (see Dot product and components of vectors).
$F \cdot d r$ — the dot product, which is built to pick out the along-the-motion part.

WHY $cos θ$ . Only the part of the push in the direction you actually move does work. Push a cart while also lifting slightly? The lifting part goes nowhere useful. $cos θ$ is exactly the "how much of the force points forward" dial: full push forward $\Rightarrow cos 0^{\circ} = 1$ ; sideways push $\Rightarrow cos 9 0^{\circ} = 0$ ; backward push $\Rightarrow cos 18 0^{\circ} = - 1$ .

PICTURE. A force arrow at angle $θ$ to a horizontal motion arrow; drop its shadow (red) onto the motion line. That red shadow of length $F cos θ$ is the only part that counts.

Step 7 — All the signs: forward, sideways, backward slivers

WHAT. The sliver work $F (x) d x$ (or $F d r cos θ$ ) can be positive, zero, or negative, and the integral just adds these signed slivers. Three cases:

Situation	$cos θ$	sliver $d W$	meaning
Push along motion	$+$	$+$	energy into the body
Push perpendicular	$0$	$0$	no energy exchanged
Push opposes motion	$-$	$-$	energy out of the body

On a graph this is: area above the $x$ -axis counts positive, area below counts negative.

WHY cover all three. The reader must never meet a scenario we skipped. A retarding force like $F (x) = - 5 x$ (from Example 4 in the parent) gives every sliver a negative sign, so its integral is negative — the force removes energy. Reversing the direction of travel swaps the integration limits and flips the whole sign.

PICTURE. One graph where the force curve dips below the axis. The area above the axis is red-positive; the area below is hatched-negative. Net work = (positive area) − (negative area).

Step 8 — The spring: watching the triangle appear

WHAT. Now run the machine on a real force. A spring you stretch pulls back, and to hold it you must apply $F (x) = k x$ (from Hooke's law and spring potential energy). Here $k$ is the spring constant (stiffness, N/m) and $x$ is the stretch. Work to stretch from $0$ to $x_{0}$ : $W = \int_{0}^{x_{0}} k x d x = k [\frac{x ^{2}}{2}]_{0}^{x_{0}} = \frac{1}{2} k x_{0}^{2} .$

Term-by-term:

$k$ — how stiff the spring is; bigger $k$ = harder to stretch.
$k x$ — the force at stretch $x$ ; a straight line through the origin with slope $k$ .
$\frac{1}{2} k x_{0}^{2}$ — the answer; note it is the area of a triangle of base $x_{0}$ and height $k x_{0}$ : $\frac{1}{2} (base) (height) = \frac{1}{2} x_{0} (k x_{0})$ .

WHY it's a clean triangle. Because $F = k x$ is a straight slanted line, the area under it from $0$ to $x_{0}$ is a right triangle — no fancy integral needed to see the answer, and the integral confirms it.

PICTURE. The line $F = k x$ rising from the origin; the red triangle underneath, base $x_{0}$ , height $k x_{0}$ . Its area $= \frac{1}{2} k x_{0}^{2}$ is the work.

The one-picture summary

Everything on one canvas: a flat rectangle ( $W = F d$ ) fails on a curved roof; we tile the curve with thin red rectangles; shrink them to nothing; the sum-sign $\sum$ becomes the integral-sign $\int$ ; and the final smooth red area under the $F$ – $x$ curve is the work $W = \int_{x_{A}}^{x_{B}} F (x) d x$ .

Recall Feynman retelling — the whole walkthrough in plain words

You want the total effort to move a box, but the push keeps changing along the way. If the push were steady, you'd just do push × distance — that's the area of a plain rectangle in a "push vs. distance" picture. But a changing push makes the top of that picture a wiggly curve, and a rectangle can't cover a wiggly shape.

So you cheat cleverly: take a tiny step. Over a step that small, the push barely changes, so on that step it is basically a rectangle — height = the push right there, width = the tiny step. That's a sliver of effort. Now do this for the whole trip: chop the path into a staircase of thin rectangles and add all their little areas. That's close, but there are tiny gaps between the staircase and the curve.

Kill the gaps by making the steps thinner and thinner until they're infinitely thin and infinitely many. The staircase melts perfectly onto the curve, and the giant addition of "push × tiny step" gets a special name and symbol: the integral, $\int F d x$ . It equals the exact area under the push-vs-distance curve — the total work.

Two footnotes: if the push points sideways to your motion, only its forward shadow ( $F cos θ$ ) counts; a purely sideways push does no work. And if the push fights your motion, its slivers count negative — it's draining energy instead of adding it.

Connections

Parent topic — full note
Work done by a constant force — the flat-roof special case, Step 1.
Area under curves and the definite integral — the limit-of-sums machinery of Step 5.
Hooke's law and spring potential energy — the spring triangle of Step 8.
Dot product and components of vectors — justifies the $cos θ$ shadow in Step 6.
Work–Energy Theorem — this same integral equals the change in kinetic energy.
Conservative forces and potential energy — when this integral is path-independent.

Concept Map

Recall

Work as area — even for a constant force, what is

W = F d

geometrically?

The area of a rectangle of height

F

and width

d

under the force–position graph.

Why does a single rectangle fail once force varies?

A rectangle needs one height, but

F (x)

has no single value along the path — the top edge is a curve.

What lets us treat a sliver as a rectangle?

Over an infinitely small width

Δ x

the curve looks flat, so

F (x)

is practically constant there.

What is a Riemann sum here?

\sum_{i = 1}^{N} F (x_{i}) Δ x_{i}

— the total area of the staircase of thin rectangles.

What does taking

Δ x_{i} \to 0

do to the staircase?

The gaps vanish and the staircase becomes the exact area under the curve — the sum becomes

\int F d x

What does

d x

mean inside the integral?

The infinitely thin sliver width that

Δ x_{i}

became in the limit.

Why does a perpendicular force do zero work?

d W = F d r cos θ

with

θ = 9 0^{\circ}

, and

cos 9 0^{\circ} = 0

What sign does a sliver have when force opposes motion?

Negative (

cos θ < 0

), so that part of the area lies below the axis and removes energy.

Why is the spring result

\frac{1}{2} k x_{0}^{2}

a triangle?

F = k x

is a straight line from the origin; the area under it to

x_{0}

is a triangle of base

x_{0}

and height

k x_{0}