Area Between Two Curves

Integrate the gap: upper minus lower.

The region between two curves

The line y = 3x and the parabola y = x² cross twice, and between the crossings they enclose a region. Its area is not the area under either curve, but the part of the area under the upper curve that is not under the lower one.

xy

y = 3x and y = x² cross at the dots, (0, 0) and (3, 9). Between them the line is the upper curve and the parabola the lower.

Integrate the gap

Cut the region into thin vertical strips. A strip at x runs from the lower curve up to the upper one, so its height is upper − lower and its area is (upper − lower) dx. Adding the strips, the area is the integral of upper − lower.

This is the same as the area under the upper curve less the area under the lower, done in one integral.

The crossings are the limits

The region begins and ends where the curves meet, so solve for the crossings before integrating. x² = 3x gives x² − 3x = 0, so x(x − 3) = 0 and x = 0 or x = 3.

To see which curve is on top, try one x between the crossings. At x = 1, 3x = 3 and x² = 1, so the line is above. The area is the integral of 3x − x² from 0 to 3, which is 3x²/2 − x³/3 at 3, less its value at 0: 27/2 − 9 = 9/2.

For y = x and y = x², x² = x gives x = 0 and x = 1, and at x = ½ the line is higher. The area is the integral of x − x² from 0 to 1: 1/2 − 1/3 = 1/6. As a check, the region sits inside the triangle under y = x from 0 to 1, whose area is 1/2.

Below the axis too

Neither curve has to be above the x-axis, because a strip’s height is a difference of two heights. For f(x) = 4 − x² and g(x) = 2x + 1, 4 − x² = 2x + 1 gives x² + 2x − 3 = 0, so (x + 3)(x − 1) = 0 and x = −3 or x = 1. At x = 0, f = 4 and g = 1, so f is on top.

The gap is f − g = 3 − 2x − x², and its integral from −3 to 1 is 3x − x² − x³/3 at 1, less its value at −3: 5/3 − (−9) = 32/3, although part of the region lies below the x-axis.

f(x) = 4 − x²g(x) = 2x + 1x = −1f(x) = 3g(x) = −1f − g = 4−4−3−2−112−4−224

f − g > 0: f is the upper curve here, so the strip counts positive in ∫ (f − g) dx

Drag the strip to where the curves cross

f(x) = 4 − x² and g(x) = 2x + 1, with one strip. At x = −1, f = 3 and g = −1, so the strip is 4 tall although g is below the axis. Drag it to x = −3 or x = 1, where f − g = 0: those are the limits. Past them f − g is negative, because the curves have swapped.

When the curves swap over

y = x³ and y = x cross where x³ = x, that is x(x − 1)(x + 1) = 0, at x = −1, 0 and 1. Between −1 and 0 the cubic is on top: at x = −½ the cubic is at −1/8 and the line lower, at −½. Between 0 and 1 the line is on top: at x = ½ the line is at ½ and the cubic at 1/8.

So each piece needs its own subtraction. From −1 to 0 the area is the integral of x³ − x, which is 0 − (1/4 − 1/2) = 1/4. From 0 to 1 it is the integral of x − x³, which is 1/2 − 1/4 = 1/4. The total area is 1/4 + 1/4 = 1/2.

One integral of x − x³ from −1 to 1 gives −1/4 + 1/4 = 0, because on the left piece x − x³ is negative and cancels the right piece.

xy

y = x³ and y = x cross at the three dots. Between −1 and 0 the cubic is on top; between 0 and 1 the line is. Each piece has area 1/4.

The usual mistakes

Lower minus upper. The integral of x² − x from 0 to 1 is −1/6, the area with its sign reversed.

Adding the two areas. 1/2 + 1/3 = 5/6 counts the region under y = x² as well; the gap subtracts it.

Leaving out the lower curve. 1/2 is the area under y = x alone.

One integral across a swap. For y = x³ and y = x from −1 to 1 it gives 0; the area is 1/2.

Taking limits from the axes. The limits are the x-values where the two curves meet, not where either meets an axis.

Money in against money out

In the application below, the two curves are a foundry’s rate of income and its rate of cost. Where income is the upper curve, each extra hundred fittings adds money, and the area between the curves from one crossing to the other is the money gained.

Worked example: A Foundry's Money In Against Its Money Out: The Profit Read as the Area Between Two Rate Curves

Question A foundry can cast up to 900 fittings a day. At an output of x hundred fittings a day, money comes in at a rate of r = 30 + 6x − x2 thousand dollars for each extra hundred fittings, and goes out at a rate of c = 37 − 2x thousand dollars for each extra hundred. (a) Find the two outputs at which the two rates are equal. (b) Find the output that makes the foundry the most money, and how much more it makes there than at an output of 100 fittings a day.

  1. 1.The gap between the two rates is r − c = (30 + 6x − x2) − (37 − 2x) = −x2 + 8x − 7, in thousand dollars per hundred fittings.

    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7
    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7
    The gap between the rates is r − c = (30 + 6x − x2) − (37 − 2x) = −x2 + 8x − 7.

  2. 2.(a) The rates are equal where the gap is zero: x2 − 8x + 7 = 0, so (x − 1)(x − 7) = 0 and x = 1 or x = 7. The two outputs are 100 fittings a day and 700 fittings a day.

    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7
    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7
    (a) The rates are equal where the gap is zero: (x − 1)(x − 7) = 0, so x = 1 and x = 7.

  3. 3.Between those outputs the gap is positive, since at x = 4 it is −16 + 32 − 7 = 9. So every extra hundred fittings up to x = 7 adds money, and past x = 7 each one takes money away: the best output is 700 a day.

    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9
    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9
    Between them the gap is positive — at x = 4 it is 9 — so the profit climbs all the way to x = 7.

  4. 4.(b) The money gained between the two outputs is the area between the curves, ∫17(−x2 + 8x − 7)dx = [−x33 + 4x2 − 7x]17.

    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9the profit gained is the area between the curves
    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney outgap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9the profit gained is the area between the curves
    The profit gained between the two outputs is the area of the shaded region, ∫17(−x2 + 8x − 7)dx.

  5. 5.At x = 7 that is −3433 + 196 − 49 = 983, and at x = 1 it is −13 + 4 − 7 = −103. The difference is 1083 = 36 thousand dollars a day. Check: the gap is a parabola with roots 1 and 7 reaching 9 at the middle, and the mean height of such a parabola is two thirds of its greatest height, so the area is 6 × 6 = 36.

    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney out36gap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9the profit gained is the area between the curves(b) 98/3 + 10/3 = 36 thousand dollars a day
    01020304001479x, hundred fittings a daythousand dollars per hundredmoney inmoney out36gap = money in − money out = −x2+ 8x − 7(a) the gap is 0 at x = 1 and at x = 7the gap is widest at x = 4, where it is 9the profit gained is the area between the curves(b) 98/3 + 10/3 = 36 thousand dollars a day
    (b) That comes to 983 + 103 = 36 thousand dollars a day, the gain from running at 700 rather than 100.

Answer: (a) the rates are equal at x = 1 and x = 7, that is at 100 and at 700 fittings a day; (b) 700 fittings a day, which makes 36 thousand dollars a day more than 100 fittings a day

Common mistakes

  • Integrating the gap from x = 0 instead of from x = 1. Below an output of 100 fittings the costs are the greater, so that stretch of the region lies below the axis and takes 103 thousand dollars off the total, leaving 983 instead of 36. The limits of a region between two curves are the outputs where they cross.
  • Multiplying the widest gap by the width of the region, 9 × 6 = 54. That is the area of a rectangle, and the gap is a curve which is 9 only at x = 4 and 0 at both ends. The integral is what adds the varying gap up.

More using integration problems, worked step by step →

Practice Area Between Two Curves in the app