Visual walkthrough — Area between curves — horizontal and vertical slices

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Step 1 — What "area between curves" even means

WHAT. We have two curves. The top one we will call $y=f(x)$; the bottom one $y=g(x)$. Here $y=f(x)$ is just shorthand for "the height of the top curve when you stand at position $x$" — $x$ is how far right you are, $y$ is how high up you are. The shaded region is everything caught between them.

WHY start here. Before any integral, we must be sure what quantity we are chasing: the area of the shaded pocket, not the area under one curve. Naming top and bottom now saves us from sign errors later.

PICTURE. The two curves and the shaded pocket between them. Notice: at any horizontal position $x$, the pocket has a top edge (on $f$) and a bottom edge (on $g$).

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Step 2 — Slice the pocket into standing rectangles

WHAT. Pick a position $x$ inside the pocket and cut out a super-thin vertical strip there. Give it a tiny width. We call that tiny width $\Delta x$ — the Greek letter $\Delta$ ("delta") is the standard flag for "a small change in," so $\Delta x$ means "a small change in the horizontal position." The strip is (almost exactly) a rectangle standing upright.

WHY slice. A pocket with curvy edges has no area formula. A rectangle does: length times width. So we trade one impossible shape for a pile of thousands of easy rectangles — this is the whole idea behind the Definite Integral as Riemann Sum.

PICTURE. One red rectangle standing inside the pocket, its width labelled $\Delta x$, its top touching $f$ and its bottom touching $g$.

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Step 3 — Measure ONE rectangle (height and width)

WHAT. The red rectangle has:

• height = top edge minus bottom edge $=f(x)-g(x)$,
• width = $\Delta x$.

So its area is

$$\underset{\text{height}=\text{top }y-\text{bottom }y}{\underbrace{(f(x)-g(x))}}\cdot \underset{\text{tiny width}}{\underbrace{\Delta x}}.$$

WHY subtract? Height is measured upward. The top of the strip sits at height $f(x)$; the bottom sits at height $g(x)$. The distance between two heights is the bigger minus the smaller, exactly like your height above a chair is (your head's height) $-$ (the chair's height). Because we declared $f\ge g$, this gap $f(x)-g(x)$ is never negative — good, since a length can't be negative.

PICTURE. A zoom on the single rectangle: a vertical measuring arrow from $g(x)$ up to $f(x)$ labelled "$f(x)-g(x)$", and a horizontal brace labelled $\Delta x$.

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Step 4 — Add up all the rectangles (the Riemann sum)

WHAT. Cover the whole pocket with $n$ of these strips, side by side, from $x=a$ to $x=b$. Number them $i=1,2,\dots ,n$. Strip $i$ sits at some sample position we label $x_i^{\ast }$ (the little star just means "the particular $x$ I picked inside strip $i$"). For simplicity give every strip the same width $\Delta x=\frac{b-a}{n}$ (the total span $b-a$ chopped into $n$ equal pieces). Adding every strip's area:

$$A\approx \sum _{i=1}^n\underset{\text{height of strip }i}{\underbrace{(f(x_i^{\ast })-g(x_i^{\ast }))}}\underset{\text{its width}}{\underbrace{\Delta x}}.$$

The symbol ${\sum }_{i=1}^n$ (capital Greek "sigma") means "add the following up, letting the counter $i$ run from $1$ to $n$." It is just a compact way to write "strip $1$ + strip $2$ + $\dots$ + strip $n$."

WHY only approximately equal? Each strip's top is slightly slanted but we treated it as flat. So we're a little off — the tops leave tiny triangular gaps or overhangs.

PICTURE. The pocket filled with many red rectangles; the little error slivers along the top edge are visible and shrinking toward the right where strips are drawn thinner.

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Step 5 — Shrink the width to zero: the integral is born

WHAT. Now make the strips thinner and more numerous: let $n\to \infty$, which forces $\Delta x\to 0$. The error slivers vanish and the approximation becomes exact. The limit of this sum is — by definition — the integral:

$$A=\underset{n\to \infty }{\lim }\sum _{i=1}^n(f(x_i^{\ast })-g(x_i^{\ast }))\Delta x={\int }_a^b(f(x)-g(x))dx.$$

Reading the right-hand side term by term:

• $\int$ = "the continuous adding-up" (a stretched $S$ for Sum),
• $a$ (bottom) and $b$ (top) = where the sweep starts and stops,
• $f(x)-g(x)$ = the height of the rectangle at position $x$ (top minus bottom),
• $dx$ = the now-infinitely-thin width (what $\Delta x$ became).

WHY this tool and not simple multiplication? The height $f(x)-g(x)$ changes as $x$ moves — you cannot just do "height × width" once. The integral is precisely the machine invented to multiply a varying height by width and total it up. Turning this limit into an actual number is the job of the Fundamental Theorem of Calculus.

PICTURE. Two panels: coarse strips (visible steps along the top) beside ultra-fine strips (top looks perfectly smooth) — the visual of $\Delta x\to 0$.

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Step 6 — Where do $a$ and $b$ come from? (the pinch points)

WHAT. The limits $a$ and $b$ are the $x$-values where the two curves meet. To find them, set the two heights equal, $f(x)=g(x)$, and solve — often a quadratic or polynomial, see Solving Quadratic & Polynomial Equations.

WHY. At a meeting point the top and bottom coincide, so the pocket has zero height there — it is pinched shut. That is exactly the natural left and right edge of the region.

PICTURE. The curves crossing at two dots; the region pinched to nothing at each dot, fat in the middle.

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Step 7 — The dangerous case: curves that cross inside

WHAT. Suppose the curves swap roles partway: $f$ is on top for a while, then $g$ takes over past a crossing point $x=c$. If you blindly integrate ${\int }_a^b(f-g)dx$, the part where $g$ is now on top contributes a negative amount, which cancels part of the real area.

WHY it fails. $f-g$ is positive on the left piece and negative on the right piece; the integral adds them with signs, so they eat each other. Area, though, is always positive — no cancellation allowed.

FIX. Split at the crossing $c$ and always write (top $-$ bottom) on each piece:

$$A={\int }_a^c(f-g)dx+{\int }_c^b(g-f)dx={\int }_a^b|f-g|dx.$$

The $|\cdots |$ ("absolute value") means "make it positive," which is just a shorthand for "top minus bottom, whichever is on top right now."

PICTURE. Curves crossing at $c$; left region shaded where $f$ is up, right region shaded where $g$ is up — two pockets, each measured with its own correct top-minus-bottom.

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Step 8 — Rotate your head: horizontal slices ($dy$)

WHAT. Some regions are ugly for standing strips because the top curve itself changes as you sweep right (like a sideways parabola $x=y^2$, which for one $x$ gives two $y$'s). Cure: lay the rectangles down. Now a strip has a fixed height $y$, a tiny thickness $\Delta y$ ("a small change in vertical position"), a right edge on a curve $x=R(y)$ and a left edge on $x=L(y)$. Here $R(y)$ means "how far right the right curve is when you stand at height $y$." Repeat Steps 2–5 rotated $90^{\circ }$:

$$A={\int }_c^d(==R(y)-L(y)==)dy.$$

• $R(y)-L(y)$ = length of a lying-down rectangle (right $x$ minus left $x$),
• $dy$ = its tiny thickness,
• $c,d$ = the $y$-values where the curves meet (solve in $y$ now).

WHY. The geometry never changed — only the direction we sweep. To rewrite $y=f(x)$ as $x=(\text{something in }y)$ you use Inverse Functions. This is the same slice-measure-add pattern we will reuse one dimension up in Volumes by Slicing & Disks.

PICTURE. The same pocket, now filled with horizontal red rectangles; one is highlighted with a left-to-right measuring arrow "$R(y)-L(y)$" and thickness $\Delta y$.

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The one-picture summary

Everything collapses to a single mantra and a single diagram: choose a slicing direction, measure one rectangle, and let the integral add infinitely many.

Reading the summary figure. The figure has two panels showing the same pocket:

• Left panel (vertical slices). The grey shape is the pocket from Steps 1–6. The single red rectangle standing up is the tile of Step 3: its height is top $-$ bottom $=f(x)-g(x)$ and its width is $\Delta x$. Sweeping such tiles left→right and adding them is the integral ${\int }_a^b(f-g)dx$ printed above it.
• Right panel (horizontal slices). The identical pocket, now filled by a lying-down red rectangle from Step 8: its length is right $-$ left $=R(y)-L(y)$ and its thickness is $\Delta y$. Sweeping bottom→top gives ${\int }_c^d(R-L)dy$.
• The punchline the two panels make visible: same grey region, two sweep directions, one area — the direction of slicing can never change the answer.

Recall Feynman: tell the whole walkthrough to a 12-year-old

Picture two wavy lines with a coloured pocket between them. You can't measure the pocket directly, so you slice it into a bunch of super-thin sticks. If the sticks stand up, each one's height is "how far the top line is above the bottom line right there" and its width is tiny — call that width $dx$. Multiply height by width for one stick, then add up every stick: that adding-up is what the $\int$ symbol does. The edges of the pocket (where you start and stop adding) are just the spots where the two lines touch and pinch the pocket shut. If the lines cross in the middle, one stick's "top" becomes the "bottom," so split the job at the crossing and always take the taller minus the shorter. And if the pocket is easier to fill with sticks lying down, do that instead — measure length "right line minus left line" with thickness $dy$. Same pocket, same area, just sticks turned sideways.

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Connections

• Definite Integral as Riemann Sum — Steps 4–5 are this idea in action.
• Fundamental Theorem of Calculus — evaluates the integral once it's set up.
• Inverse Functions — rewrites $y=f(x)$ as $x=R(y)$ for horizontal slices (Step 8).
• Solving Quadratic & Polynomial Equations — finds the pinch points (Step 6).
• Volumes by Slicing & Disks — the same slice-measure-add pattern, one dimension higher.