Area Below the Axis

The integral keeps score with signs.

An arch above the axis

Differentiating −cos x gives sin x, so −cos x is an antiderivative of sin x. The integral of sin x from 0 to π is therefore (−cos π) − (−cos 0) = 1 + 1 = 2.

Between 0 and π the curve y = sin x stays above the x-axis, so every sliver has a positive height and a positive width, and the integral 2 is the area of the arch.

xy

The arch of y = sin x from 0 to π, above the x-axis. Its area is 2, and so is its integral.

An arch below the axis

From π to 2π the curve hangs below the axis. The integral is (−cos 2π) − (−cos π) = −1 − 1 = −2.

The minus sign comes from the slivers. Each one is f(x) dx, a height times a width. The width dx is positive, because the sweep runs left to right, but every height sin x between π and 2π is negative, so every sliver counts negative and so does their sum. The region still has an area of 2; the integral reports it as −2, below the axis.

xy

The arch of y = sin x from π to 2π, below the x-axis. Its area is 2, and its integral is −2.

A signed count

Integrate over the whole cycle, from 0 to 2π: (−cos 2π) − (−cos 0) = −1 + 1 = 0. The +2 above and the −2 below cancel. The integral is a signed count: area above the axis adds, and area below takes away.

The same happens on a straight line. Under y = x − 1 from 0 to 2 there are two triangles, each with base 1 and height 1, so each has area ½. The one from 0 to 1 is below the axis and the one from 1 to 2 above. An antiderivative is x²/2 − x, and (2 − 2) − (0 − 0) = 0: the triangles cancel.

xy

Both arches of y = sin x from 0 to 2π. The integral over the whole cycle is 2 + (−2) = 0, although 4 units of area are shaded.

The area itself

When a question asks for the area, the signs must not cancel. Find where the curve crosses the x-axis, integrate each piece between the crossings on its own, drop the sign of each piece, and add. For sin x from 0 to 2π the pieces are 2 and −2, so the area is 2 + 2 = 4. For y = x − 1 from 0 to 2 the pieces are −½ and ½, so the area is ½ + ½ = 1.

Take y = x² − 2x from 0 to 4. It crosses the axis where x(x − 2) = 0, at x = 0 and x = 2, and between them it is below the axis. An antiderivative is F(x) = x³/3 − x², with F(0) = 0, F(2) = 8/3 − 4 = −4/3 and F(4) = 64/3 − 16 = 16/3.

The piece from 0 to 2 is F(2) − F(0) = −4/3, and the piece from 2 to 4 is F(4) − F(2) = 16/3 + 4/3 = 20/3. The integral from 0 to 4 is −4/3 + 20/3 = 16/3, but the area is 4/3 + 20/3 = 24/3 = 8.

xy

y = x² − 2x from 0 to 4, crossing the x-axis at the dots, x = 0 and x = 2. The shaded piece below the axis has area 4/3 and the piece above has area 20/3, so the area is 8, while the integral is 20/3 − 4/3 = 16/3.

The usual mistakes

Giving the integral when the area is asked for. Over a full cycle of sin x the integral is 0, but the area is 4.

Taking the size of the whole integral as the area. For y = x² − 2x from 0 to 4 the integral is 16/3, but the area is 8: inside one integral the piece below the axis has already been taken away.

Giving an area as a negative number. A region below the axis whose integral is −5 has area 5; the minus only says where it sits.

Counting only one piece. An integral of +3 followed by −3 covers an area of 3 + 3 = 6, not 3.

A tidal channel

In the application below, water runs into a basin on the flood and out on the ebb, and a meter reads the flow as positive in and negative out. The integral of the flow is the change in the water held; the water that passed through the channel needs the split at slack water.

Worked example: A Tidal Basin's Channel: Why the Net Change in Water Is Not the Water That Moved

Question A tidal basin is joined to the sea by a single channel. A meter in the channel reads f = 100cos(π h6) cubic meters per hour, where h is the number of hours after the flood runs strongest; a positive reading is water running into the basin and a negative reading water running out. (a) What is the net change in the water held by the basin between h = 0 and h = 9? (b) How much water passed through the channel altogether over those nine hours? Give each answer to three significant figures.

  1. 1.An antiderivative of 100cos(π h6) is 600πsin(π h6), because differentiating the sine brings the π6 back down and 100 ÷ π6 = 600π.

    −100−500501000369h, hours after the strongest floodcubic meters per hourantiderivative: 600/pi times sin(pi h / 6)
    −100−500501000369h, hours after the strongest floodcubic meters per hourantiderivative: 600/pi times sin(pi h / 6)
    An antiderivative of 100cos(π h6) is 600πsin(π h6).

  2. 2.(a) The net change is ∫09 f dh = 600π[sin(π h6)]09 = 600π(−1 − 0) = −191. The basin ends the nine hours with 191 cubic meters less water than it began with.

    −100−500501000369h, hours after the strongest floodcubic meters per hournet: 191 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191
    −100−500501000369h, hours after the strongest floodcubic meters per hournet: 191 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191
    (a) ∫09 f dh = 600π(−1 − 0) = −191: the basin holds 191 cubic meters less than it did.

  3. 3.The flow changes sign at slack water, where cos(π h6) = 0, that is at h = 3. Before then the reading is positive and the basin fills; after it the reading is negative and the basin empties.

    −100−500501000369h, hours after the strongest floodcubic meters per hourslack waternet: 191 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 3
    −100−500501000369h, hours after the strongest floodcubic meters per hourslack waternet: 191 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 3
    The flow changes sign at slack water, where cos(π h6) = 0, that is at h = 3.

  4. 4.Split the interval there. ∫03 f dh = 600π(1 − 0) = 191 cubic meters in on the flood, and ∫39 f dh = 600π(−1 − 1) = −1200π = −382, which is 382 cubic meters out on the ebb.

    −100−500501000369h, hours after the strongest floodcubic meters per hourslack water191 in382 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 30 to 3: 191 in; 3 to 9: 382 out
    −100−500501000369h, hours after the strongest floodcubic meters per hourslack water191 in382 outantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 30 to 3: 191 in; 3 to 9: 382 out
    Split there: ∫03 f dh = 191 in on the flood, and ∫39 f dh = −382, which is 382 out on the ebb.

  5. 5.(b) The water that actually passed through the channel is the two pieces added without their signs: 191 + 382 = 573 cubic meters. Check: with the signs kept, 191 − 382 = −191, the net change found in part (a), so the two answers agree with each other.

    −100−500501000369h, hours after the strongest floodcubic meters per hourslack water191 in382 out573 in allantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 30 to 3: 191 in; 3 to 9: 382 out(b) 191 + 382 = 573 through the channelwhile 191 − 382 = −191, the net change
    −100−500501000369h, hours after the strongest floodcubic meters per hourslack water191 in382 out573 in allantiderivative: 600/pi times sin(pi h / 6)(a) 600/pi × (−1 − 0) = −191the flow changes sign at h = 30 to 3: 191 in; 3 to 9: 382 out(b) 191 + 382 = 573 through the channelwhile 191 − 382 = −191, the net change
    (b) The water that passed through is 191 + 382 = 573 cubic meters, while the signed total is 191 − 382 = −191.

Answer: (a) the basin ends with 191 cubic meters less water, 191 having run in on the flood and 382 out on the ebb; (b) 573 cubic meters passed through the channel

Common mistakes

  • Reading the net change as the water that moved. The integral counts the ebb as negative, so it returns −191 cubic meters, which is the change in what the basin holds. The water that passed through the channel is 573 cubic meters, three times as much, and no single integral from 0 to 9 gives it.
  • Taking the size of the whole integral as the area, and calling the answer 191 cubic meters. The size must be taken piece by piece, after the interval has been split at the sign change: inside one integral the flood and the ebb cancel, and once they have canceled neither can be recovered.

More techniques of integration problems, worked step by step →

Practice Area Below the Axis in the app