Applications: Techniques of Integration

A flow rate is the derivative of a count, so the count is an antiderivative of the flow rate. Integrating the rate term by term gives the shape of the count but leaves one number undecided, and that number is not a loose end: a single reading at a known time fixes it. Once the count is known in full, a total over a stretch of time is the difference between two of its values.

1. The count is an antiderivative of the flow rate. Integrating term by term with the power rule, N = ∫ (6 + 0.9m − 0.015m²)dm = 6m + 0.45m² − 0.005m³ + C.
2. Fix the constant from the reading at a known time. At m = 0 every term in m vanishes, so N = C there, and the counter read 8000. Hence C = 8000 and N = 8000 + 6m + 0.45m² − 0.005m³.
3. (a) At 7:20, m = 20. Then 6 × 20 = 120, 0.45 × 400 = 180 and 0.005 × 8000 = 40, so N = 8000 + 120 + 180 − 40 = 8260. The counter reads 8260, which is 260 vehicles in the twenty minutes.
4. (b) The traffic in the hour is the change in the count, N(60) − N(0), and the 8000 cancels in the difference. With 6 × 60 = 360, 0.45 × 3600 = 1620 and 0.005 × 216000 = 1080, the change is 360 + 1620 − 1080 = 900 vehicles. Check: the flow is 6 a minute at each end of the hour and 19.5 a minute at its busiest, at m = 30; a mean of 15 a minute for 60 minutes is 900, which agrees.

answer(a) the counter reads 8260, which is 260 vehicles in the twenty minutes; (b) 900 vehicles between 7:00 and 8:00

techniqueFinding the Constant of Integration · Integrating Powers · The Antiderivative

examsO-Level

A Riemann sum replaces the area under a curve by rectangles, and the only choice is where each rectangle's height is read: at the left of its strip, at the right, or in the middle. When the curve climbs all the way across, that choice decides the direction of the error, so the left and right sums trap the true figure between them. The exact answer then comes from the integral, and the three estimates can be measured against it.

1. Six strips across twelve hours makes each strip Δ h = 2 hours wide, and every height the three rules need is already in the table. The left-hand ends are at h = 0, 2, 4, 6, 8, 10, the right-hand ends at h = 2, 4, 6, 8, 10, 12, and the midpoints at h = 1, 3, 5, 7, 9, 11.
2. The left sum takes the height at the left of each strip: ∑ P Δ h = 2(40 + 48 + 72 + 112 + 168 + 240) = 2 × 680 = 1360 kilowatt-hours.
3. The right sum takes the height at the right of each strip: 2(48 + 72 + 112 + 168 + 240 + 328) = 2 × 968 = 1936 kilowatt-hours.
4. (a) The midpoint sum takes the height in the middle: 2(42 + 58 + 90 + 138 + 202 + 282) = 2 × 812 = 1624 kilowatt-hours. The load climbs all shift, so every left rectangle sits under its strip and every right rectangle stands over it: the true energy is trapped between 1360 and 1936, and the midpoint sum lies between them.
5. (b) Exactly, ∫₀¹²(40 + 2h²)dh = [40h + 2h³3]₀¹² = 480 + 1152 = 1632 kilowatt-hours. So the left sum is 272 low, the right sum 304 high, and the midpoint sum only 8 low. Check: the right sum less the left sum is 2(328 − 40) = 576, and 1936 − 1360 = 576, as it must be.

answer(a) the left sum is 1360 kilowatt-hours and the right sum 1936, so the energy is trapped between them, with the midpoint sum 1624; (b) exactly 1632 kilowatt-hours, so the left sum is 272 low, the right sum 304 high and the midpoint sum only 8 low

techniqueLeft, Right and Midpoint Riemann Sums · The Riemann Sum in Sigma Notation · Reading the Integral Sign

The gas burnt over a stretch of time is the area under the rate curve, and the Fundamental Theorem of Calculus says that area need never be summed: find one antiderivative of the rate and take the difference of its values at the two ends. The same theorem makes a bill easy to split, because an integral over an interval is the sum of the integrals over its pieces.

1. The gas burnt between 6:00 and noon is ∫₀⁶(3h² + 8h + 5)dh. Rather than add up slivers, find one antiderivative G of the rate; the theorem then gives the area as G(6) − G(0).
2. Integrating term by term, G(h) = h³ + 4h² + 5h. Any constant may be added to G, but it cancels in the difference, so take the constant to be zero.
3. (a) G(6) = 216 + 144 + 30 = 390 and G(0) = 0, so the boiler burns 390 − 0 = 390 cubic meters between 6:00 and noon.
4. The tariff changes at h = 4. There G(4) = 64 + 64 + 20 = 148, so the gas burnt before 10:00 is G(4) − G(0) = 148 cubic meters.
5. (b) After 10:00 the gas burnt is ∫₄⁶ g dh = G(6) − G(4) = 390 − 148 = 242 cubic meters. Check: 148 + 242 = 390, the whole, which is the rule that ∫₀⁴ and ∫₄⁶ together make ∫₀⁶. Reading that last stretch backwards would give ∫₆⁴ g dh = −242, the same size with its sign turned round.

answer(a) 390 cubic meters; (b) 148 cubic meters before 10:00 and 242 after, and 148 + 242 = 390

techniqueThe Fundamental Theorem of Calculus · Properties of the Definite Integral · The Definite Integral

examsO-Level

An integral keeps score with signs. Where the meter reads negative the curve lies below the axis and the integral counts that stretch as a loss, so one integral over the whole nine hours gives the net change in what the basin holds. The water that actually passed through the channel is a different quantity: to get it, break the interval at the instant the flow changes sign and add the two pieces without their signs.

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

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

techniqueArea Below the Axis · Integrating Trigonometric Functions · Properties of the Definite Integral

examsO-Level · H2

When the inside of a function is a straight line in the variable, nothing about the integration changes except a single constant factor. Integrate as though the inside were the variable, then divide by the coefficient of the variable inside, because differentiating back would multiply by it. The two pulses here have different coefficients, so the same correction is read twice, with a different divisor each time.

1. The inside of the first sine is π(n−3)4, a straight line in n whose coefficient of n is π4. Integrating a sine gives minus a cosine, and the whole result is then divided by that coefficient.
2. So ∫ 18sin(π(n−3)4)dn = −18 × 4πcos(π(n−3)4) + C = −72πcos(π(n−3)4) + C.
3. (a) At n = 7 the inside is π and the cosine is −1; at n = 3 the inside is 0 and the cosine is 1. The displacement is 72π − (−72π) = 144π = 45.8 millimeters.
4. The second pulse has π2 inside, twice as large, so the divisor is twice as large too: ∫ 6sin(π(n−9)2)dn = −6 × 2πcos(π(n−9)2) + C = −12πcos(π(n−9)2) + C.
5. (b) At n = 11 the inside is π and at n = 9 it is 0, so the second displacement is 12π + 12π = 24π = 7.64 millimeters, and the ground moves 144π + 24π = 168π = 53.5 millimeters in all. Check: the second pulse is a third of the height and half the width of the first, so a sixth of the displacement, and 45.8 ÷ 6 = 7.64.

answer(a) 144π = 45.8 millimeters; (b) 24π = 7.64 millimeters from the second pulse, and 168π = 53.5 millimeters in all

techniqueIntegrating f(ax + b) · Integration by Substitution · Integrating Trigonometric Functions

examsO-Level

Substitution is the chain rule read backwards. The awkward part of this integrand is the expression under the root, so that is what the new variable is set equal to; its differential then shows that the numerator is exactly what is needed to complete the trade. Once the integral is written in the new variable it is a single power, and carrying the limits across saves changing back at the end.

1. The volume is ∫ q dn. The awkward part is n² + 144 under the root, so put u = n² + 144. Its differential is du = 2n dn, so n dn = 12du, and an n dn is precisely what the numerator offers.
2. The integral becomes ∫30√u × 12du = 15∫ u⁻¹²du = 15 × 2u¹² = 30√u = 30√n² + 144.
3. Carry the limits over rather than changing back. At n = 0, u = 144; at n = 5, u = 169; at n = 9, u = 225. Their square roots are 12, 13 and 15, all whole numbers.
4. (a) ∫₀⁵ q dn = 30(√169 − √144) = 30(13 − 12) = 30 cubic meters.
5. (b) ∫₅⁹ q dn = 30(√225 − √169) = 30(15 − 13) = 60 cubic meters, twice as much in four minutes as in the first five, because the gate stands further open. Check: 30 + 60 = 90, and ∫₀⁹ q dn = 30(15 − 12) = 90 cubic meters, so the two parts add to the whole.

answer(a) 30 cubic meters; (b) 60 cubic meters, and 90 cubic meters over the nine minutes altogether

techniqueIntegration by Substitution · Differentials · Properties of the Definite Integral

examsH2

Nothing differentiates to a power times an exponential in one step, so the integral has to be traded for an easier one. Integration by parts does that trade, and the choice of parts decides whether the new integral is easier or worse: differentiating the power lowers it towards a constant, while integrating the exponential leaves it unchanged. Take e⁻ⁿ = 0.135335 and e⁻⁴ = 0.0183156.

1. The integrand is a product of two quite different functions, n and e⁻ⁿ, so use parts: ∫ u dvdn dn = uv − ∫ v dudn dn. Take u = n, because differentiating it leaves 1 and kills the product, and dvdn = e⁻ⁿ, which integrates to v = −e⁻ⁿ.
2. Then ∫ n e⁻ⁿdn = −n e⁻ⁿ − ∫(−e⁻ⁿ)dn = −n e⁻ⁿ − e⁻ⁿ + C = −(n + 1)e⁻ⁿ + C. The integral left over was an exponential alone, which is the whole point of the choice.
3. (a) ∫₀² 50n e⁻ⁿdn = 50[−(n+1)e⁻ⁿ]₀² = 50(1 − 3e⁻²) = 50(1 − 0.406006) = 29.7 kilograms.
4. Over the four hours, 50[−(n+1)e⁻ⁿ]₀⁴ = 50(1 − 5e⁻⁴) = 50(1 − 0.091578) = 45.4 kilograms.
5. (b) The second two hours therefore give 45.4 − 29.7 = 15.7 kilograms, a little over half as much as the first two. Check: differentiating −(n+1)e⁻ⁿ gives −e⁻ⁿ + (n+1)e⁻ⁿ = n e⁻ⁿ, the integrand, so the antiderivative is right.

answer(a) 50(1 − 3e⁻²) = 29.7 kilograms; (b) 15.7 kilograms, the four hours together giving 45.4 kilograms

techniqueIntegration by Parts · Integrating Exponentials and Logarithms · Differentials

examsH2

Heating depends on the square of the voltage, and a squared sine cannot be integrated as it stands. The double angle rewrites it as a constant minus a cosine, and across a whole cycle that cosine spends as long above the axis as below, so it contributes nothing at all. What is left is the constant, which is why the mean of a squared sine over a cycle is exactly one half.

1. The mean of v² over a cycle is 120∫₀²⁰ 340²sin²(π n10)dn. Replace the squared sine by its double-angle form, sin²θ = 1 − cos 2θ2, which gives sin²(π n10) = 12 − 12cos(π n5).
2. The cosine term contributes nothing. Its antiderivative is 5πsin(π n5), which is 0 at n = 0 and 0 again at n = 20, so ∫₀²⁰cos(π n5)dn = 0: over a whole number of cycles the cosine cancels itself out.
3. (a) What survives is the constant half: the mean of v² is 340²2 = 1156002 = 57800. The root mean square voltage is the square root of that, 340√2 = 240.4 volts, or 240 volts to three significant figures.
4. (b) Averaging is linear, so the mean of v²34 is the mean of v² divided by 34: the mean power is 5780034 = 1700 watts.
5. A steady 340 volts across the same element would give 340²34 = 3400 watts, exactly twice as much. Check: the mean of a squared sine is 12, so the alternating supply delivers half the peak power, and the steady voltage that would match it is smaller by a factor of √2, which is the 240.4 volts of part (a).

answer(a) the mean of v² is 57800 and the root mean square voltage is 340√2 = 240.4 volts, which is 240 volts to three significant figures; (b) the mean power is 1700 watts, against 3400 watts for a steady 340 volts, exactly twice as much

techniqueIntegrating sin² and cos² · Integrating Trigonometric Functions · The Definite Integral

examsH2

A fraction whose bottom is a product of two different brackets is not one integral but two in disguise. Partial fractions separates them, and each piece is then of the form that integrates to a logarithm. Putting the two logarithms back together as one logarithm of a quotient is what makes the answer a single clean number rather than a string of decimals.

1. Split the integrand: 2(3−x)(1−x) = A3−x + B1−x, which on clearing the denominators means 2 = A(1 − x) + B(3 − x) for every x.
2. Choose values of x that kill one bracket at a time. Putting x = 1 gives 2 = 2B, so B = 1; putting x = 3 gives 2 = −2A, so A = −1. The integrand is 11−x − 13−x.
3. Each piece has a constant on top and a bracket below whose derivative is −1, so each integrates to a logarithm with a minus sign: ∫dx1−x = −ln(1−x) and ∫dx3−x = −ln(3−x). Together, n = [ln(3−x) − ln(1−x)]₀^X = [ln3−x1−x]₀^X.
4. (a) Three quarters of the B reacted means X = 0.75. At that point ln2.250.25 = ln 9, and at x = 0 the value is ln31 = ln 3. So n = ln 9 − ln 3 = ln 3 = 1.10 minutes.
5. (b) Nine tenths reacted means X = 0.9, where ln2.10.1 = ln 21, so n = ln 21 − ln 3 = ln 7 = 1.95 minutes. Check: the last fifteen hundredths of B take 1.95 − 1.10 = 0.85 minutes, longer than everything before them took per mole, and as x climbs towards 1 the logarithm runs away, so B is never quite used up. That is what the chemistry says too.

answer(a) ln 3 = 1.10 minutes; (b) ln 7 = 1.95 minutes

techniqueIntegration by Partial Fractions · Integrals of f′ over f · Integrating Exponentials and Logarithms

examsO-Level · H2

A sum of two squares on the bottom of a fraction is the signature of an inverse tangent. The substitution that gets there sets the variable equal to the fixed length times the tangent of an angle, because then the sum of squares collapses to a single squared secant; the differential supplies a second squared secant, the two cancel, and a constant is left to integrate. The answer is the angle the line of sight turns through.

1. The dose is ∫₀³309 + x²dx. Substitute x = 3tanθ, since 9 + 9tan²θ = 9sec²θ turns the sum of two squares into a single square.
2. The differential is dx = 3sec²θ dθ, so the integrand becomes 309sec²θ × 3sec²θ dθ = 10 dθ. Both squared secants cancel and nothing is left but a constant.
3. The limits travel with the substitution. At x = 0, tanθ = 0, so θ = 0; at x = 3, tanθ = 1, so θ = π4.
4. (a) The dose is 10[θ]₀^π4 = 10π4 = 5π2 = 7.85 microsieverts. That working, done once and for all, is the standard result ∫dxa² + x² = 1atan⁻¹xa + C.
5. (b) With the offset 4, ∫₀⁴3016 + x²dx = 304[tan⁻¹x4]₀⁴ = 304 × π4 = 15π8 = 5.89 microsieverts. Check: each answer is 30a × π4, so moving the walkway from 3 meters out to 4 multiplies the dose by 34, and three quarters of 7.85 is 5.89.

answer(a) 5π2 = 7.85 microsieverts; (b) 15π8 = 5.89 microsieverts, three quarters of the first

techniqueIntegrals That Give tan⁻¹ and sin⁻¹ · Choosing a Trigonometric Substitution · Integrating sec²x, sec x tan x and tan x

examsH2

Each short stretch dr of the wire takes drv seconds to cover, so the time is the integral of one over the speed. Here the speed is a root of r² plus or minus a constant, and a constant over such a root is the signature of an inverse hyperbolic function: the substitution that clears the root uses cosh when the constant is subtracted and sinh when it is added, because cosh² u − 1 = sinh² u and sinh² u + 1 = cosh² u. Each inverse hyperbolic is a logarithm in disguise, which is how the exact answers are written.

1. (a) At the start v = 0 and r = 0.4, so k = 0 − 4 × 0.16 = −0.64. Then v² = 4r² − 0.64 = 4(r² − 0.16), so v = 2√r² − 0.16, and the time to go from r = 0.4 to r = 1.2 is ∫_0.4^1.2dr2√r² − 0.16 seconds.
2. A square minus a constant under the root: substitute r = 0.4cosh u. Then dr = 0.4sinh u du and √r² − 0.16 = 0.4√cosh² u − 1 = 0.4sinh u, so the integrand becomes 0.4sinh u2 × 0.4sinh udu = 12du. Done once and for all, this is ∫dr√r² − a² = arcoshra + C.
3. (a) The limits travel with the substitution: r = 0.4 gives cosh u = 1, so u = 0, and r = 1.2 gives cosh u = 3, so u = arcosh 3 = ln(3 + √8). The time is 12ln(3 + 2√2) = ln(1 + √2) = 0.881 seconds, since (1 + √2)² = 3 + 2√2. The integrand is infinite at r = 0.4, where the bead is at rest, but the area under it is finite.
4. (b) The second bead has v = 0.8 at r = 0, so k = 0.64 − 0 = 0.64 and v = 2√r² + 0.16. The sign inside the root has turned over, so substitute r = 0.4sinh u: then dr = 0.4cosh u du and √r² + 0.16 = 0.4√sinh² u + 1 = 0.4cosh u, and again the integrand is 12du. Done once and for all, this is ∫dr√r² + a² = arsinhra + C.
5. (b) The limits: r = 0 gives u = 0, and r = 1.2 gives sinh u = 3, so u = arsinh 3 = ln(3 + √10). The time is 12ln(3 + √10) = 0.909 seconds, so the bead let go from rest reaches its end first, by 0.028 seconds: the flicked bead starts faster but has 1.2 meters to cover instead of 0.8. Check: since the time is u2, the positions are r = 0.4cosh 2n and r = 0.4sinh 2n after n seconds, and 0.4cosh(2 × 0.881) = 1.20 and 0.4sinh(2 × 0.909) = 1.20.

answer(a) ln(1 + √2) = 0.881 seconds; (b) 12ln(3 + √10) = 0.909 seconds, so the bead let go from rest gets there first, by 0.028 seconds

techniqueIntegrals That Give arsinh and arcosh · Integration by Substitution · Inverse Hyperbolics as Logarithms

examsH2