Area Between Curves

The area trapped between two curves is \(\int_a^b (\text{top} - \text{bottom})\,dx\) — each thin vertical strip has height (top \(-\) bottom), and the integral sums the strips. Where the curves cross, top and bottom trade places, so you split the integral there.

By the end you'll be able to find where two curves intersect, decide which one is on top on each piece, and set up and evaluate \(\int (\text{top} - \text{bottom})\,dx\) — or \(\int (\text{right} - \text{left})\,dy\) when horizontal strips are simpler.

Predict: as the strip approaches a point where the curves cross, what happens to (top \(-\) bottom) — and which curve is on top just past the crossing? Scrub the slider to check.

The line \(y = x + 2\) and the parabola \(y = x^2\) cross at \(x = -1\) and \(x = 2\) (the marked dots). Between the crossings the region is tinted blue because the line is on top; outside them it's tinted orange because the parabola takes over. Drag the strip across the picture and watch its height (top \(-\) bottom) shrink to zero at each crossing, then grow again with the roles swapped. Hover or focus a crossing dot for its exact coordinates.

At x = 0.50: top = line (2.50), bottom = parabola (0.25) · strip height top − bottom = 2.25

y = x + 2 vs y = x² — shaded between-region, crossings, and one movable strip
y = x + 2 (line) y = x² (parabola) crossings: roles switch

The area between two curves is \(\int (\text{top} - \text{bottom})\,dx\) (or \(\int (\text{right} - \text{left})\,dy\)), integrating the gap over the interval where the curves bound a region and splitting at intersection points where the roles switch.

Strips first, formula second

Everything here comes from one picture: slice the region into thin vertical strips of width \(dx\). A strip at position \(x\) reaches from the bottom curve up to the top curve, so its height is \(f(x) - g(x)\) where \(f\) is on top, and its area is the integrand times \(dx\). Summing all the strips gives

$$A = \int_a^b \big(f(x) - g(x)\big)\,dx.$$

That's exactly what the demo shows: the line minus the parabola is the height of the strip you scrubbed, and the shaded region is the sum of all such strips.

Visual

Why "top minus bottom" and not just \(f - g\) all the way across? Because a single subtraction only stays positive while \(f\) actually is on top. At each crossing the gap passes through zero and flips sign — you saw the strip height hit 0 at \(x = -1\) and \(x = 2\) and the tint change color. Integrating one fixed difference straight through a crossing makes the pieces cancel instead of add. The robust procedure: (1) solve \(f(x) = g(x)\) to find every intersection, (2) test a point in each subinterval to see who's on top, (3) integrate (top \(-\) bottom) on each piece and add the positive results. Writing \(A = \int |f - g|\,dx\) says the same thing — evaluating the absolute value forces the identical case-split.

Applied

This setup is the workhorse behind consumer-surplus areas in economics, the material in a cross-section between an outer and inner profile in engineering, and — next lesson — volumes: revolve the gap and each strip becomes a washer. Sometimes the region is easier to slice sideways. If the boundaries are naturally \(x = \text{(something in } y\text{)}\), use horizontal strips: \(A = \int_c^d (\text{right} - \text{left})\,dy\) — same idea, rotated 90°. A very tall, thin region bounded by two functions of \(y\) is the classic cue.

Worked example

Find the area of the region bounded by \(y = x^2\) and \(y = x + 2\) — the demo's region.

Intersections: \(x^2 = x + 2 \iff x^2 - x - 2 = 0 \iff (x-2)(x+1) = 0\), so \(x = -1, 2\). Who's on top? Test \(x = 0\): the line gives \(2\), the parabola gives \(0\), so the line is on top on \((-1, 2)\). Then

$$A = \int_{-1}^{2} \big[(x + 2) - x^2\big]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}.$$

At \(x = 2\): \(2 + 4 - \tfrac{8}{3} = \tfrac{10}{3}\). At \(x = -1\): \(\tfrac12 - 2 + \tfrac13 = -\tfrac{7}{6}\). So \(A = \tfrac{10}{3} - \left(-\tfrac{7}{6}\right) = \tfrac{27}{6} = \) \(\tfrac{9}{2}\).

Your turn

Find the area of the region bounded by \(y = 4\) and \(y = x^2\). Intersections: \(x^2 = 4\) gives \(x = -2, 2\). Test \(x = 0\): \(4 > 0\), so the horizontal line is on top. Set up: \(A = \int_{-2}^{2} (4 - x^2)\,dx = \left[4x - \tfrac{x^3}{3}\right]_{-2}^{2}\). Now finish it: \(A = \left(8 - \tfrac{8}{3}\right) - \big(\) ____ \(\big) = \) ____

Reveal the answer

At \(x = -2\): \(4(-2) - \tfrac{(-2)^3}{3} = -8 + \tfrac{8}{3} = -\tfrac{16}{3}\). So \(A = \tfrac{16}{3} - \left(-\tfrac{16}{3}\right) = \) \(\tfrac{32}{3}\). Sanity check against the picture: the bounding box is \(4 \times 4 = 16\), and \(\tfrac{32}{3} \approx 10.7\) is a believable two-thirds of it (a parabolic segment is always exactly two-thirds of its bounding box).

More info — the absolute-value view, and when no antiderivative exists

Another way to say the whole lesson: \(A = \int_a^b |f(x) - g(x)|\,dx\). The absolute value guarantees every strip counts positively, but to actually evaluate it you must find where \(f - g\) changes sign and split there — exactly the intersection-hunting procedure above, so most people just track top and bottom by hand. And if the gap \(f - g\) has no elementary antiderivative at all (say one curve is \(e^{-x^2}\)), the area is still perfectly well defined — approximate the integral numerically with trapezoids or Simpson's rule, as in the previous section. Paul's Notes (Dive deeper below) walks several multi-curve splits in full.

Check your understanding

Question 1 of 4

Find the area of the region bounded by \(y = 2x\) and \(y = x^2\).

Question 2 of 4

The curves \(y = \sin x\) and \(y = 0\) cross at \(x = \pi\) on \([0, 2\pi]\). Why does \(\int_0^{2\pi} \sin x\,dx = 0\) fail to give the total area between them?

Question 3 of 4

Two curves bound a region on \([0, 2]\), but the gap has no elementary antiderivative. You measure the strip height \(h = \text{top} - \text{bottom}\) at three points: \(h(0) = 2\), \(h(1) = 3\), \(h(2) = 2\). Using the trapezoidal rule with \(n = 2\) (\(\Delta x = 1\)), estimate the area between the curves.

Question 4 of 4

The region bounded by \(x = y^2\) and \(x = y + 2\) is easiest to compute with which setup?

Recap

  • Area between curves: \(A = \int_a^b (\text{top} - \text{bottom})\,dx\) — the integrand is the height of a thin vertical strip.
  • Find intersections by solving \(f(x) = g(x)\); they set the limits and split the interval wherever the curves cross.
  • Where curves cross, top and bottom swap — integrate each piece separately and add positive results, or you'll cancel area instead of adding it.
  • When boundaries are naturally functions of \(y\), slice horizontally: \(A = \int_c^d (\text{right} - \text{left})\,dy\).

Dive deeper

Sources

  • Area Between Curves