Applications: Using Integration

Velocity is the derivative of displacement, so displacement is an integral of velocity. That integral counts a minute of sinking against a minute of rising, which is exactly what is wanted for the depth at the end, and exactly what is not wanted for the tether. For the tether the interval has to be cut at every instant the velocity is zero, because those are the only places the vehicle can change direction, and the three pieces are then added with their signs taken off.

1. Displacement is an antiderivative of velocity. Integrating term by term, s = ∫ (1.5n² − 12n + 18)dn = 0.5n³ − 6n² + 18n, taking s = 0 at the start of the maneuver.
2. (a) Over the whole eight minutes the displacement is s(8) − s(0) = 256 − 384 + 144 = 16 meters, and it is positive, so the vehicle finishes 16 meters higher than it began. Its depth is 40 − 16 = 24 meters.
3. The vehicle can only turn where its velocity is zero. Factorizing, v = 1.5(n − 2)(n − 6), so v = 0 at n = 2 and at n = 6, and between those two times v is negative: the vehicle is sinking.
4. Take the three stretches separately. From 0 to 2, s(2) − s(0) = 4 − 24 + 36 = 16 meters up. From 2 to 6, s(6) − s(2) = 0 − 16 = −16, which is 16 meters down. From 6 to 8, s(8) − s(6) = 16 − 0 = 16 meters up.
5. (b) The tether moves 16 + 16 + 16 = 48 meters in all. Check: putting the signs back gives 16 − 16 + 16 = 16 meters, the displacement of part (a), so the three pieces and the single integral agree.

answer(a) it is 24 meters below the surface, having risen 16 meters in all; (b) the winch moves 48 meters of tether

techniqueTotal Distance Traveled · Velocity and Displacement by Integration

examsO-Level

Each curve is a rate, so the money each one stands for over a stretch of output is the area under it, and the profit gained over that stretch is the area between them. The gap between the curves is itself a function, and where that gap is zero the two rates are equal: those outputs are the limits of the region, and they are also the outputs where the profit stops rising and starts falling.

1. The gap between the two rates is r − c = (30 + 6x − x²) − (37 − 2x) = −x² + 8x − 7, in thousand dollars per hundred fittings.
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.
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.
4. (b) The money gained between the two outputs is the area between the curves, ∫₁⁷(−x² + 8x − 7)dx = [−x³3 + 4x² − 7x]₁⁷.
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.

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

techniqueArea Between Two Curves

examsH2

A solid of revolution is built from slices taken at right angles to its axis, and each slice is a circle whose radius the question gives. Where the solid is filled the slice is a disc; where it has been hollowed out the slice is a washer, a disc with a smaller disc removed, and its area is the difference of the two areas rather than the area of the difference of the radii. Integrating the slice area up the height adds the slices back into a volume.

1. A slice of the cavity at height y is a disc of radius √y + 4, so its area is π(√y + 4)² = π(y + 4) square centimeters.
2. (a) The capacity is π∫₀¹²(y + 4)dy = π[y²2 + 4y]₀¹² = π(72 + 48) = 120π, which is 377 cubic centimeters to the nearest cubic centimeter.
3. A slice of the wall at the same height is a washer: a disc of radius √y + 9 with a disc of radius √y + 4 taken out. Its area is π[(y + 9) − (y + 4)] = 5π square centimeters, the same at every height.
4. (b) The wood in the wall is therefore π∫₀¹²5 dy = 5π × 12 = 60π, which is 188 cubic centimeters to the nearest cubic centimeter.
5. Check: the block the vase was turned from is π∫₀¹²(y + 9)dy = π(72 + 108) = 180π, and the cavity and the wall add back to it, since 120π + 60π = 180π.

answer(a) the capacity is 120π, which is 377 cubic centimeters to the nearest cubic centimeter; (b) the wall holds 60π, which is 188 cubic centimeters of wood

techniqueVolumes of Revolution · Volumes by Washers

examsH2

The solid is not a prism, because its cross-section changes from rib to rib, so no single cross-section can simply be multiplied by a length. What is fixed is the shape of each rib, and the floor fixes how wide each rib has to be. Writing the area of a rib in terms of its position turns the solid into a stack of slabs, and integrating that area across the floor adds the slabs up.

1. At a distance x from the center, the round floor is b = 2√9 − x² meters wide, because the floor's edge is the circle x² + y² = 9.
2. The rib there is an isosceles triangle on that base with both sides rising at 45^°, so its height is half its base, √9 − x². Its area is A = 12 × 2√9 − x² × √9 − x² = 9 − x² square meters. At the center that is 9 square meters, a rib 6 meters wide and 3 meters high.
3. (a) The air under the canvas is ∫₋₃³(9 − x²)dx = [9x − x³3]₋₃³ = (27 − 9) − (−27 + 9) = 36 cubic meters.
4. (b) Over a square floor 6 meters wide every rib has the same base of 6 meters, so A = 9 square meters throughout and the marquee is a prism: 9 × 6 = 54 cubic meters.
5. That is 54 − 36 = 18 cubic meters more, a half as much again. Check: A = 9 − x² is a parabola with roots at ± 3 reaching 9 in the middle, and the mean value of such a parabola is two thirds of its greatest value, so the mean cross-section is 6 square meters and the volume is 6 × 6 = 36.

answer(a) 36 cubic meters of air; (b) the square floor holds 54 cubic meters, which is 18 cubic meters more

techniqueVolumes with Known Cross Sections

A short piece of any curve is near enough a straight line, with a horizontal side and a vertical side, so Pythagoras' theorem gives its length as the horizontal step times the square root of one plus the gradient squared. Adding those pieces up is an integral along the run. The gradient on its own integrates to the drop, which is a different question, and the two are easy to confuse.

1. A short piece of the surface has horizontal side dx and vertical side dydxdx, so by Pythagoras' theorem its length is √1 + (dydx)² dx. Here (dydx)² = x96, so the length is L = ∫₀⁵⁴√1 + x96 dx.
2. Integrate: ∫√1 + x96 dx = 96 × 23(1 + x96)³² = 64(1 + x96)³².
3. (a) At x = 54, 1 + 5496 = 2516, whose three-halves power is 12564, so the bracket is 64 × 12564 = 125; at x = 0 it is 64. The graded surface is 125 − 64 = 61 meters long.
4. (b) The drop is the integral of the gradient itself: ∫₀⁵⁴√x96 dx = 23 × 54³²√96 = 23 × 54 × √5496 = 23 × 54 × 34 = 27 meters.
5. A straight ramp from the lip to the landing point is √54² + 27² = √3645 = 27√5 = 60.37 meters, so the graded surface is 61 − 60.37 = 0.63 meters longer. Check: the gradient nowhere exceeds 34, so the surface cannot be longer than 54 × √1 + 916 = 54 × 1.25 = 67.5 meters, nor shorter than the straight ramp, and 61 lies between.

answer(a) the graded surface is 61 meters long; (b) it drops 27 meters, and a straight ramp between the same two points is 27√5 = 60.37 meters, so the graded surface is only 0.63 meters longer

techniqueArc Length

The mean value of a function over an interval is its integral divided by the width of that interval: it is the height of the rectangle with the same area. A handful of readings averaged is a different thing, because each reading is treated as though it stood for an equal share of the width, and the two readings at the ends stand for only half a share each. The trapezium rule is what puts that right without any more measuring.

1. The mean value of a function over an interval is its integral divided by the width: h = 116∫₋₈⁸(8 − x²8)dx.
2. Integrating, ∫₋₈⁸(8 − x²8)dx = [8x − x³24]₋₈⁸ = (64 − 643) − (−64 + 643) = 128 − 1283 = 2563, which is the area of the cross-section in square meters.
3. (a) Dividing by the width, h = 2563 × 116 = 163 = 5.33 meters to two decimal places. That is two thirds of the 8 meter crown, which is what the mean of a parabola over the whole of its arch must be.
4. (b) The five readings, at x = −8, −4, 0, 4 and 8, are 0, 6, 8, 6 and 0 meters, and their average is 205 = 4 meters. That is well short, because the two readings at the walls stand for only half a strip of floor each and yet carry a full share of the average.
5. The trapezium rule gives each end reading half weight, which is the share it really stands for: the area is 4(02 + 6 + 8 + 6 + 02) = 4 × 20 = 80 square meters, so the mean height is 8016 = 5 meters. Check: 80 is short of the true 2563 = 85.33 only because each chord cuts inside a roof that curves.

answer(a) the mean height is 5.33 meters, that is 16/3 meters; (b) the average of the five readings is 4 meters, while the trapezium rule on the same readings gives an area of 80 square meters and a mean height of 5 meters

techniqueThe Mean Value of a Function · Numerical Integration

An integral to infinity is not a sum of infinitely many things but a limit: integrate to a finite time, keep that time as a letter, and watch what the answer does as the time grows. Both thrusts here fall away to nothing, so neither is obviously hopeless, and the thing that decides the matter is how fast each one falls. An exponential falls fast enough that the tail adds almost nothing; a reciprocal does not.

1. Integrate the first thrust to a finite time T: ∫₀^T240e^−0.002ndn = [−120000e^−0.002n]₀^T = 120000(1 − e^−0.002T) millinewton-seconds.
2. (a) As T grows, e^−0.002T → 0, so the total settles at 120000 millinewton-seconds, that is 120 newton-seconds. At T of a million seconds and again at T of a hundred million the figure reads 120000 to every place a calculator shows: the tail beyond a million seconds is smaller than any of them.
3. Nine tenths of that total has arrived when 1 − e^−0.002n = 0.9, that is when e^−0.002n = 0.1. Taking logarithms, 0.002n = ln 10, so n = 500ln 10 = 1151 seconds, about nineteen minutes.
4. (b) Integrate the second thrust to a finite time in the same way: ∫₀^T2401 + 0.002ndn = [120000ln(1 + 0.002n)]₀^T = 120000ln(1 + 0.002T).
5. At T = 10⁶ that is 120000ln 2001 = 912000 millinewton-seconds, and at T = 10⁸ it is 120000ln 200001 = 1465000. The answer has not settled; it has gone up by half as much again. A logarithm has no limit, so the second design has no total impulse: whatever total is asked for, it is reached eventually.
6. Check: both thrusts fall to nothing, so falling away is not what decides it. At T = 10¹⁶ the second design would have delivered about 3680000 millinewton-seconds and would still be climbing.

answer(a) the total impulse settles at 120000 millinewton-seconds, that is 120 newton-seconds, and nine tenths of it has arrived by 1151 seconds; (b) the second design gives about 912000 millinewton-seconds by a million seconds and about 1465000 by a hundred million, so it has no total impulse at all

techniqueImproper Integrals

There is no formula for a river bed, only the soundings, so the area has to be estimated from them. The trapezium rule joins neighboring soundings by a straight chord; Simpson's rule takes the soundings two strips at a time and lays a parabola through the three, which follows a curving bed far more closely. The two rules use exactly the same nine numbers, so any difference between them is a difference of method, not of measurement.

1. Eight strips of 3 meters span the 24 meter width, and eight is even, so Simpson's rule applies. It weights the two bank soundings 1, the odd-numbered soundings 4 and the interior even-numbered ones 2, and multiplies by a third of the strip width.
2. (a) The area is 33[0 + 4(1.75 + 3.75 + 3.75 + 1.75) + 2(3.00 + 4.00 + 3.00) + 0] = 4 × 11 + 2 × 10 = 44 + 20 = 64 square meters.
3. The flow is the area of the cross-section times the mean speed of the water: 64 × 0.5 = 32 cubic meters per second.
4. (b) The trapezium rule gives every interior sounding weight 1 and halves the two at the banks: 3(02 + 1.75 + 3.00 + 3.75 + 4.00 + 3.75 + 3.00 + 1.75 + 02) = 3 × 21 = 63 square meters, a flow of 31.5 cubic meters per second.
5. The trapezium rule is the smaller by 1 square meter. Between soundings the bed dips away below the straight chord drawn across it, so every trapezium leaves out a sliver, and the eight slivers come to that square meter. Check: Simpson's mean depth is 6424 = 2.67 meters, two thirds of the deepest sounding of 4.00 meters, which is what a bed shaped like a parabola gives, and Simpson's rule is exact on a parabola.

answer(a) the cross-section is 64 square meters and the flow is 32 cubic meters per second; (b) the trapezium rule gives 63 square meters and a flow of 31.5 cubic meters per second, 1 square meter the smaller, because every straight chord cuts inside a bed that dips away beneath it

techniqueSimpson's Rule · Numerical Integration

A sentence about a rate becomes a differential equation as soon as the rate is written as a derivative, the words after proportional to are written out, and a sign is chosen to match what is happening. This equation separates, because every h can be moved to one side and every n to the other. Two readings are then needed and both are given: one fixes the constant of integration, the other fixes the constant of proportionality.

1. Let h be the depth in centimeters and n the number of minutes since the tap was opened. The depth falls at a rate proportional to √h, so dhdn = −k√h with k > 0, the minus sign because the depth is falling.
2. Separate the variables and integrate: ∫ h⁻¹²dh = −k∫ dn, so 2√h = −kn + C, which is tidier written as √h = A − k2n.
3. At n = 0 the depth is 64, so √64 = 8 = A. At n = 5 the depth is 49, so 7 = 8 − 5k2, giving k2 = 0.2. Hence √h = 8 − 0.2n, that is h = (8 − 0.2n)².
4. (a) At n = 20, √h = 8 − 4 = 4, so the depth is h = 4² = 16 centimeters.
5. (b) The butt is empty when h = 0, that is when 8 − 0.2n = 0, so n = 40 minutes. Check: the rule gives (8 − 1)² = 49 at n = 5, as it must, and the emptying slows down as it should, losing 15 centimeters in the first five minutes and only 1 centimeter in the last five.

answer(a) the depth is 16 centimeters; (b) the butt is empty 40 minutes after the tap was opened

techniqueSeparable Differential Equations · Forming Differential Equations

examsH2

Separating the variables here gives a relation between c and n that is perfectly correct and perfectly useless for reading a value off, because it cannot be rearranged to give c. The slope field still shows what every solution does, and Euler's method walks along it: from the point you are at, follow the slope the equation gives for one step, and start again from wherever that lands you. Halving the step and comparing is how the size of the error is judged.

1. Separating gives ∫4 + ccdc = −∫ 6 dn, that is 4ln c + c = K − 6n. It is a true relation between c and n, but no rearrangement of it gives c on its own, so a value has to be stepped out instead of read off.
2. Euler's method steps along the slope field: c_next = c + h × f, where f = −6c4 + c is the slope the equation gives at the point reached so far.
3. (a) With h = 0.5: at c = 8, f = −4812 = −4, so c → 8 − 2 = 6; at c = 6, f = −3610 = −3.6, so c → 6 − 1.8 = 4.2; at c = 4.2, f = −3.073, so c → 2.663; at c = 2.663, f = −2.398, so c → 1.464. After two hours the estimate is 1.46 milligrams per liter.
4. (b) Eight steps of h = 0.25, worked the same way, give 1.667, so the estimate is 1.67 milligrams per liter.
5. A step small enough for the answer to settle gives 1.852. Euler's method reads low both times, because it holds the slope of the start of each step for the whole of that step, and the slope eases as the ammonia falls: each step takes off more than it should. Halving the step roughly halves the error, from 0.39 to 0.19, which is what a first-order method does.
6. Check: at n = 0 the relation gives K = 4ln 8 + 8 = 16.318, so at n = 2 it asks for 4ln c + c = 16.318 − 12 = 4.318. Putting c = 1.852 in gives 4ln 1.852 + 1.852 = 4.318, so the settled value is right and it is Euler's two estimates that are short.

answer(a) about 1.46 milligrams per liter; (b) about 1.67 milligrams per liter, and Euler's method reads low both times, since a step small enough to settle gives 1.852

techniqueEuler’s Method · Slope Fields

The body on its spring and damper obeys a second-order linear equation, and the road's push is its right side. Such an equation is solved in two parts that are added: the complementary function makes the left side zero and carries the two constants, and the particular integral is one solution built to give the right side. The starting conditions can fix the constants only once the two parts are added, and in the long run the damper wears the complementary function away, leaving the particular integral as the motion the road keeps up.

1. Let X be the height in meters. Newton's second law, with the spring and the damper both pulling back, is 400d²Xdt² = 300sin 5t − 1600dXdt − 16000X. Putting x = 1000X millimeters and dividing every term by 400 gives d²xdt² + 4dxdt + 40x = 750sin 5t, with x = 0 and dxdt = 0 at t = 0, since the body was resting.
2. The complementary function solves the equation with 0 on the right. Its auxiliary equation is r² + 4r + 40 = 0, so r = −4 ± √16 − 1602 = −2 ± 6i, and the complementary function is e^−2t(Acos 6t + Bsin 6t): a swing of 6 radians a second inside a factor that shrinks by e⁻² every second.
3. For the particular integral, copy the shape of the right side with a cosine as well as a sine, because the damper turns one into the other: try x = pcos 5t + qsin 5t. Then d²xdt² = −25x and dxdt = −5psin 5t + 5qcos 5t, so the left side is (15p + 20q)cos 5t + (15q − 20p)sin 5t. Matching it to 750sin 5t gives 15p + 20q = 0 and 15q − 20p = 750, so q = −34p and −1254p = 750, that is p = −24 and q = 18.
4. (a) The general solution is the two parts added: x = e^−2t(Acos 6t + Bsin 6t) − 24cos 5t + 18sin 5t. At t = 0, x = A − 24 = 0, so A = 24; and dxdt = −2A + 6B + 90 = 0, so 6B = 48 − 90 and B = −7. Hence x = e^−2t(24cos 6t − 7sin 6t) − 24cos 5t + 18sin 5t millimeters.
5. (b) By t = 3 the factor e^−2t is below 0.003, so the complementary function has died away and the body moves as −24cos 5t + 18sin 5t, whose amplitude is √24² + 18² = √900 = 30 millimeters. A steady push of 300 newtons puts 750 on the right in place of 750sin 5t; its particular integral is the constant x = 75040 = 18.75 millimeters. So the undulations swing the body 3018.75 = 1.6 times as far as the same force held steady. Check: putting the particular integral back gives 15(−24) + 20(18) = 0 and 15(18) − 20(−24) = 750, as it must.

answer(a) x = e^−2t(24cos 6t − 7sin 6t) − 24cos 5t + 18sin 5t millimeters; (b) the amplitude settles at 30 millimeters, 1.6 times the 18.75 millimeters at which a steady push of 300 newtons would hold the body

techniqueComplementary Function and Particular Integral · The Auxiliary Equation