The trapezium rule questions

Question 1

A student applies the trapezium rule to ∫₀³ f(x) dx using 7 equally spaced ordinates. State the number of strips and the strip width h.

Accepted Answer

Seven ordinates bound six strips, since n strips need n + 1 ordinates: the first and last ordinates are the ends of the interval, and each interior ordinate is shared by two neighbouring strips. B1 for 6 strips. Then h = (3 − 0)/6 = 0.5. B1 for h = 0.5. Dividing by 7 instead gives h = 0.4286 and a set of ordinates that never reaches x = 3, which is the error the ordinate count guards against.

Question 2

Use the trapezium rule with 4 strips to estimate ∫₀² √(1 + x³) dx, giving your answer to 4 decimal places.

Accepted Answer

h = 2/4 = 0.5, and the five ordinates at x = 0, 0.5, 1, 1.5, 2 are 1, √1.125 = 1.06066, √2 = 1.41421, √4.375 = 2.09165 and 3. So T = (0.5/2)[1 + 3 + 2(1.06066 + 1.41421 + 2.09165)] = 0.25 × 13.13304 = 3.2833. M1 for the five ordinates with h = 0.5, M1 for the bracket with the end ordinates once and the interior ones twice, A1 for 3.2833. This integrand has no antiderivative among the standard functions, which is the situation the rule exists for. Keep five figures in the ordinates and round once, at the end.

Question 3

For n strips of width h with ordinates y₀, y₁, …, y_n, write down the left-endpoint rectangle sum L and the right-endpoint rectangle sum R, and state the condition on f under which one of them is a lower bound and the other an upper bound for the integral.

Accepted Answer

L = h(y₀ + y₁ + … + y_n−1) and R = h(y₁ + … + y_n): the same ordinates and the same h, with the last ordinate left out of L and the first left out of R. B1 for both sums. They bound the integral when f is monotonic on every strip. Increasing throughout puts L below the integral and R above it, and decreasing throughout swaps them, because the argument uses the two end ordinates of each strip as its least and greatest values. B1 for the monotonic condition with the direction. A function that turns inside a strip takes values beyond both end ordinates there, and neither sum is then a bound.

Question 4

A car pulls away from traffic lights, and its speed v m s⁻¹ is recorded every 10 seconds. At t = 0, 10, 20, 30 and 40 s the readings are 0, 6.4, 11.3, 14.1 and 15.6. Estimate the distance travelled in the 40 seconds, giving your answer with its units, and state one assumption your estimate makes.

Accepted Answer

Distance is the area under speed against time. The five readings are the ordinates, four strips, h = 10 s, so T = (10/2)[0 + 15.6 + 2(6.4 + 11.3 + 14.1)] = 5 × 79.2 = 396. M1 for the bracket with the readings as ordinates and h = 10, A1 for 396. Metres per second multiplied by seconds gives metres, so 396 m, or about 400 m to the precision five readings support. The estimate assumes the speed changes steadily between readings, that is, each pair of consecutive readings is joined by a straight line. B1 for the units and the assumption. Nothing was measured inside any 10-second gap, so a surge or a hesitation within one leaves no trace in the arithmetic.

Question 5

Use the trapezium rule with 4 strips to estimate ∫₀¹ e^x² dx to 4 decimal places, and state with a reason whether your estimate is an overestimate or an underestimate.

Accepted Answer

h = 0.25, and the ordinates at x = 0, 0.25, 0.5, 0.75, 1 are 1, e^0.0625 = 1.06449, e^0.25 = 1.28403, e^0.5625 = 1.75505 and e = 2.71828. T = (0.25/2)[1 + 2.71828 + 2(1.06449 + 1.28403 + 1.75505)] = 0.125 × 11.92542 = 1.4907. M1 for the ordinates, A1 for 1.4907. Differentiating twice, f′(x) = 2xe^x² and f″(x) = (2 + 4x²)e^x², which is positive for every x, so on [0, 1] the curve is convex, every chord lies above it and the estimate is an overestimate. M1 for f″, A1 for convex and therefore too big, with the chord above the curve as the reason. The true value is 1.4627, so the estimate is 0.028 high. The direction comes from the sign of f″ across every strip, not from its value at a single point.

Question 6

The trapezium rule with 4 strips gives 3.2833 for ∫₀² √(1 + x³) dx, and the true value is 3.2413 to 4 decimal places. State how the estimate can be improved, find the estimate with 8 strips to 4 decimal places, and describe what has happened to the error.

Accepted Answer

Halve the strip width, that is, double the number of strips: shorter chords lie closer to the arcs they replace. B1 for more strips. Eight strips need nine ordinates, at x = 0, 0.25, …, 2 with h = 0.25: 1, 1.00778, 1.06066, 1.19242, 1.41421, 1.71847, 2.09165, 2.52178 and 3. Every other one is an ordinate already used; the four between them are new. T = (0.25/2)[1 + 3 + 2(1.00778 + 1.06066 + 1.19242 + 1.41421 + 1.71847 + 2.09165 + 2.52178)] = 0.125 × 26.01394 = 3.2517. M1 for the nine ordinates with h = 0.25, A1 for 3.2517. The error has fallen from 0.0420 to 0.0104, about a quarter of what it was, and it has kept its sign: f″(x) = (3x + ¾x⁴)(1 + x³)^−3/2 is positive on [0, 2], so every estimate is an overestimate and the refined value approaches the true one from above. B1 for the error being smaller and of the same sign. Halving h roughly quarters the trapezium rule's error, which is what makes the extra ordinates worth computing.

Question 7

A student estimating ∫₀³ f(x) dx with 6 strips writes down the ordinates at x = 0, 0.5, 1, 1.5, 2 and 2.5 and begins the bracket. Explain the mistake, state how many ordinates the rule needs here, and state how the ordinates and the strip width change if the number of strips is doubled.

Accepted Answer

The list stops one ordinate short: the sixth strip runs from 2.5 to 3, and its right-hand side, f(3), is missing. B1 for the missing ordinate at x = 3. Six strips have seven ordinates, since the ends of the interval count once and each of the five interior points is shared by two strips; the bracket takes f(0) and f(3) once and f(0.5), …, f(2.5) twice. B1 for seven ordinates. With 12 strips, h = 3/12 = 0.25 and 13 ordinates are needed, at x = 0, 0.25, …, 3, of which the seven correct ones are every other one. B1 for 13 ordinates and h = 0.25. The strip width is (b − a)/n with n the number of strips, so h was right; what goes wrong with six ordinates is the last term of the bracket, and with it the whole estimate.

Question 8

Using 4 strips, find the left-endpoint and right-endpoint rectangle sums L and R for ∫₀¹ e^−x² dx, giving each to 4 decimal places. State which is the upper bound and which the lower, with the condition that makes them bounds, and show that the trapezium estimate from the same ordinates is the mean of the two.

Accepted Answer

h = 0.25, and the ordinates at x = 0, 0.25, 0.5, 0.75, 1 are 1, 0.93941, 0.77880, 0.56978 and 0.36788. L = 0.25(1 + 0.93941 + 0.77880 + 0.56978) = 0.25 × 3.28799 = 0.8220 and R = 0.25(0.93941 + 0.77880 + 0.56978 + 0.36788) = 0.25 × 2.65587 = 0.6640. M1 for the ordinates, A1 for L, A1 for R. On [0, 1] the function decreases throughout, so on every strip the left-hand ordinate is the greater and the right-hand ordinate the lesser: L is the upper bound and R the lower, and decreasing on every strip is the condition that makes them so. B1 for the assignment with the condition. Adding the two sums counts each interior ordinate twice and each end ordinate once, which is the trapezium bracket, so L + R = 2T: (0.8220 + 0.6640)/2 = 0.7430, and T = (0.25/2)[1 + 0.36788 + 2(0.93941 + 0.77880 + 0.56978)] = 0.7430 confirms it. B1 for L + R = 2T. The true value is 0.7468, inside the bounds and within half their width, 0.079, of T. Here f″(x) = (4x² − 2)e^−x² changes sign at x = 1/√2, so the sign of f″ gives no direction, and it is the bounds that say anything certain.

Question 9

The current I amperes through a component is recorded every 5 seconds. At t = 0, 5, 10, 15, 20, 25 and 30 s the readings are 2.0, 2.6, 3.1, 3.4, 3.5, 3.5 and 3.4. The charge that passes is the area under current against time. Estimate the charge that passes in the 30 seconds, with its units, and explain why the two rectangle sums from this table cannot be quoted as bounds for it.

Accepted Answer

Seven readings, six strips, h = 5 s. T = (5/2)[2.0 + 3.4 + 2(2.6 + 3.1 + 3.4 + 3.5 + 3.5)] = 2.5 × 37.6 = 94. M1 for the bracket with the readings as ordinates and h = 5, A1 for 94. Amperes multiplied by seconds are coulombs, so about 94 C passes, on the assumption that the current changes steadily between readings. B1 for the units with the assumption. The rectangle sums, L = 5 × 18.1 = 90.5 C and R = 5 × 19.5 = 97.5 C, bound the integral only where the current is monotonic on every strip. The readings themselves rise to 3.5 and then fall to 3.4, so the current turns somewhere in the last two intervals: before the turn the right-hand ordinate of a strip is the larger, after it the left-hand one is, and neither sum is on the same side of the area on every strip. B1 for the reason. Even a table that rose throughout would not show what happens inside its gaps. What can be reported is the estimate, about 94 C, and that readings taken more often would tighten it.

Question 10

A function f is increasing on [0, 3], with f(0) = 2 and f(3) = 8, and ∫₀³ f(x) dx is to be bounded by rectangle sums with n strips. Show that the two bounds differ by h(y_n − y₀), find the difference when n = 6, and find the least n for which the bounds are no more than 0.1 apart. State how close to the integral the trapezium estimate with that n is then guaranteed to be.

Accepted Answer

R − L = h(y₁ + … + y_n) − h(y₀ + … + y_n−1): every interior ordinate appears in both sums and cancels, leaving h(y_n − y₀), the strip width times the total rise. M1 for the cancellation, A1 for the printed difference. Here y_n − y₀ = 6, so with n = 6 and h = 0.5 the bounds are 3 apart. B1 for 3. In general h = 3/n, so R − L = 18/n, and 18/n ≤ 0.1 requires n ≥ 180. M1 for the inequality, A1 for n = 180. The trapezium estimate is the mean of the two bounds and the integral lies between them, so with 180 strips T is within 0.05 of the integral, half the width of the bounds. B1 for 0.05. The number of strips is settled before a single interior ordinate is computed, which is the use of the formula. It is a guarantee about the bounds: the trapezium estimate itself is usually far closer than 0.05.

Question 11

Use the trapezium rule with 4 strips to estimate ∫₀² √(4 − x) dx to 4 decimal places, and state with a reason whether the estimate is above or below the true value. Then find the exact value by integration and confirm your answer.

Accepted Answer

h = 0.5, and the ordinates at x = 0, 0.5, 1, 1.5, 2 are 2, √3.5 = 1.870829, √3 = 1.732051, √2.5 = 1.581139 and √2 = 1.414214. T = (0.5/2)[2 + 1.414214 + 2(1.870829 + 1.732051 + 1.581139)] = 0.25 × 13.782252 = 3.4456. M1 for the ordinates, A1 for 3.4456. With f(x) = (4 − x)^1/2, f′(x) = −½(4 − x)^−1/2 and f″(x) = −¼(4 − x)^−3/2, negative on [0, 2], so the curve is concave, each chord lies below it and the estimate is below the true value. M1 for f″, A1 for concave and therefore too small. Exactly, ∫(4 − x)^1/2 dx = −(2/3)(4 − x)^3/2, so the integral is (2/3)(8 − 2√2) = 3.4477 to 4 decimal places, about 0.002 above the estimate, as predicted. A1 for the exact value confirming the direction. Where an integral can be done exactly the trapezium rule is a check on the method rather than a route to the answer, and this one shows both the direction and the size of a four-strip error on a gently curved function.

Question 12

The table gives five values of y for a curve y = f(x), each correct to 1 decimal place. No equation for f is known. Use the trapezium rule with four strips to estimate ∫₀² y dx. Use the second differences of the tabulated values to decide whether the curve is convex or concave on this interval, and hence state, with a reason, whether your estimate is an overestimate or an underestimate.

Accepted Answer

The five ordinates are 0.5 apart, so h = 0.5 and there are four strips. The rule puts the two end ordinates in once and the three interior ones twice: T = (0.5/2)[5.0 + 3.0 + 2(4.2 + 3.6 + 3.2)] = 0.25[8.0 + 2 × 11.0] = 0.25 × 30.0. M1 for the bracket built that way with h = 0.5. So the estimate is 7.5. A1 for 7.5. The first differences of the y values are −0.8, −0.6, −0.4 and −0.2, so the second differences are +0.2, +0.2 and +0.2. M1 for the two rounds of differences. They are positive, so the gradient is increasing, f″ is positive and the curve is convex across the whole interval. A1 for convex, with the positive second differences as the evidence. A chord joining two points of a convex curve lies above the curve between them, so each trapezium covers more than the strip beneath it and the estimate is an overestimate. B1 for the conclusion with the chord-above-the-curve reason. The values here fall while the curve stays convex, so a falling curve is no evidence either way: it is the second difference, not the first, that decides the direction of the error.

Question 13

A pond is surveyed by measuring its width at right angles to a base line at 10 m intervals. At distances 0, 10, 20, 30, 40, 50 and 60 m along the line the widths are 0, 24, 37, 42, 40, 31 and 0 metres. The pond has a roughly uniform depth of 2.5 m. Estimate the surface area of the pond and the volume of water it holds, giving each with its units, state the assumptions the estimates rest on, and explain why the survey cannot settle whether the area estimate is too large or too small.

Accepted Answer

Seven readings, six strips of width h = 10 m, and the two end readings are 0 because the base line starts and ends at the water's edge. Area ≈ (10/2)[0 + 0 + 2(24 + 37 + 42 + 40 + 31)] = 5 × 348 = 1740 m²: a width in metres times a distance in metres. B1 for h = 10 with the seven widths as ordinates, M1 for the bracket, A1 for 1740 m². Volume ≈ 1740 × 2.5 = 4350 m³, about 4400 m³ to the precision the survey supports. A1 for the volume with its units. The area assumes the edge of the pond runs straight between consecutive measured widths, and the volume assumes the depth really is 2.5 m everywhere. B1 for both assumptions. Nothing was measured between the survey lines, so the table does not fix the curve of the shoreline there: a bay between two lines would make the estimate too large and a bulge too small, and the pattern of a finite table, whatever it suggests, is not the sign of f″ on the strips. Nor can the rectangle sums be offered as bounds, because the widths rise and then fall. B1 for the reason. More closely spaced survey lines would improve the estimate; nothing in this table can say in which direction.

Question 14

Use the trapezium rule with 4 strips to estimate ∫₀² 1/(1 + x²) dx to 4 decimal places. A student writes: f″(2) is positive, so the curve is convex and the estimate is an overestimate. Explain why the argument fails, and, given that the exact value is 1.1071 to 4 decimal places, state whether the estimate is in fact too large or too small.

Accepted Answer

h = 0.5, and the ordinates at x = 0, 0.5, 1, 1.5, 2 are 1, 0.8, 0.5, 0.30769 and 0.2. T = (0.5/2)[1 + 0.2 + 2(0.8 + 0.5 + 0.30769)] = 0.25 × 4.41538 = 1.1038. M1 for the ordinates, A1 for 1.1038. Differentiating twice, f′(x) = −2x(1 + x²)⁻² and f″(x) = (6x² − 2)/(1 + x²)³, which is negative for x below 1/√3 = 0.577 and positive beyond it. M1 for f″ with its sign change. The sign at one point says nothing about the other strips: on [0, 0.5] the curve is concave and the chord lies below it, on [1, 1.5] and [1.5, 2] it is convex and the chords lie above, and the strip [0.5, 1] contains the change. The errors have opposite signs, so no direction for the total follows from f″. A1 for the explanation. Against the exact 1.1071, the estimate 1.1038 is in fact too small: the first strip, where the curve bends most sharply, loses more than the gently convex later strips gain. B1 for the comparison. A direction can be claimed only when f″ keeps one sign on every strip, and this integral is the case the lesson warns about.

Question 15

f(x) = xe^−x, and ∫₀³ f(x) dx = 1 − 4e⁻³ = 0.8009 to 4 decimal places. Using 3 strips, find the left-endpoint and right-endpoint rectangle sums, explain why neither is a bound for the integral, and use the same ordinates to find two values between which the integral must lie.

Accepted Answer

h = 1, and the ordinates at x = 0, 1, 2, 3 are 0, e⁻¹ = 0.36788, 2e⁻² = 0.27067 and 3e⁻³ = 0.14936. L = 1(0 + 0.36788 + 0.27067) = 0.6386 and R = 1(0.36788 + 0.27067 + 0.14936) = 0.7879. M1 for the ordinates and the two sums, A1 for both values. Both lie below 0.8009, so neither is an upper bound. The reason is that f′(x) = (1 − x)e^−x changes sign at x = 1: f increases on the first strip and decreases on the other two, so the right-hand ordinate is the larger on [0, 1] but the smaller on [1, 2] and [2, 3], and R mixes one overcount with two undercounts. B1 for the turning point at x = 1 and what it does to the sums. Take, strip by strip, the larger ordinate for an upper bound and the smaller for a lower bound: upper 1(0.36788 + 0.36788 + 0.27067) = 1.0064, lower 1(0 + 0.27067 + 0.14936) = 0.4200. M1 for choosing the extreme ordinate on each strip, A1 for both bounds. The integral, 0.8009, lies between them, as it must. The bounds are wide because h = 1 is coarse and the function's whole rise and fall sits inside three strips. Splitting at the turning point is what makes each piece monotonic, and it needs no extra ordinate here because x = 1 is already a strip boundary.

Question 16

Oil leaks from a damaged tank, and the rate of leakage R litres per minute is recorded every 15 minutes. At t = 0, 15, 30, 45, 60, 75 and 90 minutes the readings are 3.2, 2.7, 2.3, 2.0, 1.8, 1.7 and 1.6. Estimate the volume of oil lost in the 90 minutes, with its units and the assumption it rests on. An engineer says that, because the readings decrease throughout, the two rectangle sums bound the volume: state the extra assumption that claim needs, and find the two sums. State how the estimate could be improved, and how far apart the two sums would then be if readings were taken every 5 minutes and the first and last readings were unchanged.

Accepted Answer

Seven readings, six strips, h = 15 min. T = (15/2)[3.2 + 1.6 + 2(2.7 + 2.3 + 2.0 + 1.8 + 1.7)] = 7.5 × 25.8 = 193.5, and litres per minute multiplied by minutes are litres, so about 190 litres to the precision the readings support, assuming the rate changes steadily between readings. M1 for the bracket with h = 15, A1 for 193.5 with the units, B1 for the assumption. The engineer's claim needs the rate to be decreasing throughout every 15-minute interval, not only at the seven instants recorded: a table cannot show what happens inside its gaps. B1 for the monotonic assumption stated about the intervals. With that assumption, L = 15(3.2 + 2.7 + 2.3 + 2.0 + 1.8 + 1.7) = 15 × 13.7 = 205.5 litres is the upper bound and R = 15 × 12.1 = 181.5 litres the lower, 24 litres apart, which is h(y₀ − y_n) = 15 × 1.6. A1 for the two sums, correctly assigned. The estimate is improved by reading the rate more often: with h = 5 the bounds would be 5 × 1.6 = 8 litres apart, a third of the present gap, and the trapezium estimate would then be within 4 litres of the true volume rather than 12. B1 for more frequent readings with the gap of 8 litres.

x	y
0	5.0
0.5	4.2
1.0	3.6
1.5	3.2
2.0	3.0