2023 May/June · Paper 22 · Q7 (a, b, c)11 marksRectangles, bounds & limits of sums · Substitution & standard integrals
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2022 Oct/Nov · Paper 22 · Q3 (a)6 marksSurface area of revolution
2022 Oct/Nov · Paper 22 · Q610 marksRectangles, bounds & limits of sums
2022 May/June · Paper 22 · Q15 marksArc length
2022 May/June · Paper 22 · Q4 (a, b, c)10 marksRectangles, bounds & limits of sums
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Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Rectangles, bounds & limits of sums · Core · Paper 2 style · 5 marks
(a)The region R is bounded by the curve y = 1/x, the x-axis and the lines x = 1 and x = 2. The interval [1, 2] is divided into 5 equal strips of width 0.2.Using 5 rectangles each time, find an upper bound and a lower bound for the area of R. Give your answers correct to 4 decimal places.[4]
(b)Hence write down an inequality satisfied by ln 2.[1]
Show answer
(a) upper 0.7456, lower 0.6456
B1: 1/x is decreasing, so the upper bound uses left ends (1, 1.2, 1.4, 1.6, 1.8) and the lower bound uses right ends (1.2, …, 2)M1: 0.2(1 + 1/1.2 + 1/1.4 + 1/1.6 + 1/1.8)A1: 0.7456A1: lower 0.2(1/1.2 + … + 1/2) = 0.6456
(b) 0.6456 < ln 2 < 0.7456
B1: area of R = ∫₁² (1/x) dx = ln 2
Q2Rectangles, bounds & limits of sums · Core · Paper 2 style · 5 marks
(a)Using 4 rectangles of width 1, find a lower bound and an upper bound for ∫₀⁴ √x dx. Give your answers correct to 3 significant figures.[3]
(b)Find the exact value of the integral and check that it lies between your bounds.[2]
Q3Rectangles, bounds & limits of sums · Core · Paper 2 style · 6 marks
(a)Show that ∫₀¹ 1/(1 + x)³ dx = 3/8.[1]
(b)Using four rectangles of equal width, find a lower bound and an upper bound for this integral. Give your answers correct to 4 decimal places.[4]
(c)Without further calculation of heights, show that the difference between your two bounds is exactly 7/32.[1]
Show answer
(a) [−1/(2(1 + x)²)]₀¹ = −1/8 + 1/2 = 3/8
B1: correct integration and limits
(b) lower 0.2800, upper 0.4987
B1: 1/(1 + x)³ is decreasing on [0, 1], so the lower bound uses right ends x = 1/4, 1/2, 3/4, 1 and the upper bound uses left ends x = 0, 1/4, 1/2, 3/4M1: (1/4)(64/125 + 8/27 + 64/343 + 1/8)A1: lower 0.2800A1: upper (1/4)(1 + 64/125 + 8/27 + 64/343) = 0.4987
(c) (1/4)(f(0) − f(1)) = (1/4)(1 − 1/8) = 7/32
B1: all other heights cancel, leaving (1/4)(1 − 1/8) = 7/32
Q4Rectangles, bounds & limits of sums · Core · Paper 2 style · 6 marks
(a)The region under the curve y = 4 − x² from x = 0 to x = 2 is divided into 4 strips of width 0.5. Find the total area of the 4 upper rectangles and the total area of the 4 lower rectangles.[3]
(b)Find the exact area of the region and the percentage error if the mean of the two bounds is used as an estimate.[3]
Show answer
(a) upper 6.25, lower 4.25
B1: decreasing on [0, 2]; heights at x = 0, 0.5, 1, 1.5, 2 are 4, 3.75, 3, 1.75, 0M1: upper 0.5(4 + 3.75 + 3 + 1.75)A1: upper 6.25, lower 0.5(3.75 + 3 + 1.75 + 0) = 4.25
(b) 16/3; mean 5.25, percentage error 1.56% (under-estimate)
Q5Rectangles, bounds & limits of sums · Standard · Paper 2 style · 7 marks
(a)The integral I = ∫₀¹ e^(−x²) dx cannot be found exactly using elementary functions.Explain why, for 0 ≤ x ≤ 1, a set of rectangles using the values of e^(−x²) at the left-hand end of each strip gives an upper bound for I.[1]
(b)Using 5 strips of equal width, find an upper bound and a lower bound for I, giving each correct to 3 decimal places.[4]
(c)Use your answers to give an estimate of I, and state the greatest possible error in this estimate.[2]
Show answer
(a) e^(−x²) is decreasing on [0, 1], so each left-end height is the largest value on its strip.
B1: derivative −2x e^(−x²) ≤ 0, so the function is decreasing
M1: mean of bounds (0.808 + 0.681)/2A1: 0.744 ± 0.063 (half the gap; exact gap is 0.2(1 − e⁻¹))
Q6Rectangles, bounds & limits of sums · Standard · Paper 2 style · 7 marks
(a)By considering the area under the curve y = 1/x, with rectangles of width 1, show thatln(n + 1) < 1 + 1/2 + 1/3 + … + 1/n < 1 + ln n, for n ≥ 2.[5]
(b)Hence find, correct to 2 decimal places, a lower bound and an upper bound for 1 + 1/2 + 1/3 + … + 1/1000.[2]
Show answer
(a) Shown (see working).
M1: rectangles of width 1 with heights 1/r on [r, r + 1] lie above the curve (1/x decreasing)A1: Σ_{r=1}^{n} 1/r > ∫₁^(n+1) dx/x = ln(n + 1)M1: rectangles with heights 1/r on [r − 1, r] for r = 2, …, n lie below the curveA1: Σ_{r=2}^{n} 1/r < ∫₁ⁿ dx/x = ln nA1: add 1 to complete the result
Q10Rectangles, bounds & limits of sums · Standard · Paper 2 style · 6 marks
(a)Use rectangles of width 1 under or above the curve y = 1/√x to show that2√(n + 1) − 2 < Σ_{r=1}^{n} 1/√r < 2√n − 1.[4]
(b)Hence show that 18.0 < Σ_{r=1}^{100} 1/√r < 19.[2]
Show answer
(a) Shown (see working).
M1: 1/√x decreasing; rectangles of height 1/√r on [r, r + 1] lie above the curve: Σ > ∫₁^(n+1) x^(−1/2) dxA1: = 2√(n + 1) − 2M1: for r ≥ 2 rectangles on [r − 1, r] lie below: Σ_{r=2}^{n} < ∫₁ⁿ x^(−1/2) dx = 2√n − 2A1: add the first term 1: Σ < 2√n − 1
(b) lower bound 18.100 > 18.0; upper bound 19
M1: 2√101 − 2 = 18.100A1: 2√100 − 1 = 19
Q11Rectangles, bounds & limits of sums · Challenge · Paper 2 style · 8 marks
(a)The interval [0, 1] is divided into n equal strips. Show that the upper rectangle sum Uₙ for the area under y = x² from x = 0 to x = 1 isUₙ = (n + 1)(2n + 1)/(6n²).[3]
(b)Find a similar expression for the lower sum Lₙ, and show that Uₙ − Lₙ = 1/n.[3]
(c)Find the least value of n for which the difference between the bounds is less than 0.01, and state the limit of Uₙ as n → ∞.[2]
B1: 1/n < 0.01 ⇒ n > 100, so n = 101B1: Uₙ → 2n²/(6n²) = 1/3
Q12Rectangles, bounds & limits of sums · Challenge · Paper 2 style · 8 marks
(a)The region under y = eˣ from x = 0 to x = 1 is divided into n strips of equal width. Show that the sum of the areas of the lower rectangles isLₙ = (e − 1) / [n(e^(1/n) − 1)].[4]
(b)Find L₁₀ correct to 4 decimal places.[1]
(c)By writing h = 1/n, find the limit of Lₙ as n → ∞, and explain why this is the value you should expect.[3]
Show answer
(a) Shown (see working).
M1: eˣ increasing: Lₙ = (1/n)[1 + e^(1/n) + e^(2/n) + … + e^((n−1)/n)]M1: geometric series with ratio e^(1/n), n termsA1: sum = (e − 1)/(e^(1/n) − 1)A1: Lₙ = (e − 1)/[n(e^(1/n) − 1)] (AG)
(b) 1.6338
B1: 1.6338
(c) e − 1; it equals ∫₀¹ eˣ dx
M1: Lₙ = (e − 1) × h/(eʰ − 1)M1: h/(eʰ − 1) → 1 as h → 0 (eʰ − 1 ≈ h, or l'Hôpital / series)A1: limit e − 1, which is the exact area ∫₀¹ eˣ dx
Q13Rectangles, bounds & limits of sums · Challenge · Paper 2 style · 8 marks
(a)The area A under the curve y = 1/x² between x = 1 and x = 2 is estimated using N rectangles of equal width, giving an upper bound U and a lower bound L.Show that U − L = 3/(4N).[3]
(b)Find the least value of N for which U − L < 0.01.[2]
(c)Find the exact value of A and verify that U − A < 3/(4N) for N = 2.[3]
Show answer
(a) Shown (see working).
M1: decreasing function; U uses left ends x = 1, 1 + h, …, 2 − h and L uses right ends 1 + h, …, 2, with h = 1/NM1: all terms cancel except h[f(1) − f(2)]A1: (1/N)(1 − 1/4) = 3/(4N) (AG)
(b) N = 76
M1: 3/(4N) < 0.01 ⇒ N > 75A1: N = 76
(c) A = 1/2; for N = 2, U = 13/18 ≈ 0.7222, so U − A = 2/9 < 3/8
B1: A = [−1/x]₁² = 1/2M1: U = 0.5(1 + 1/1.5²) = 13/18A1: U − A = 2/9 ≈ 0.222 < 0.375
Q14Rectangles, bounds & limits of sums · Challenge · Paper 2 style · 8 marks
(a)Let Sₙ = Σ_{r=1}^{n} n/(n² + r²). Show that Sₙ is the total area of n rectangles of width 1/n under the curve y = 1/(1 + x²) for 0 ≤ x ≤ 1, and explain why Sₙ < π/4.[4]
(b)Find the corresponding upper sum Tₙ in terms of n, and hence use n = 4 to show that 2.88 < π < 3.39.[3]
(c)State the limit of Sₙ as n → ∞.[1]
Show answer
(a) Sₙ = (1/n)Σ 1/(1 + (r/n)²), a right-end (lower) rectangle sum for a decreasing function, so Sₙ < ∫₀¹ dx/(1 + x²) = π/4.
M1: n/(n² + r²) = (1/n) × 1/(1 + (r/n)²)A1: heights at x = r/n, width 1/nB1: 1/(1 + x²) is decreasing on [0, 1], so right ends give rectangles below the curveB1: ∫₀¹ dx/(1 + x²) = tan⁻¹1 = π/4
Q16Reduction formulae · Standard · Paper 2 style · 6 marks
(a)Let Iₙ = ∫₀¹ xⁿ eˣ dx for n ≥ 0. Show that Iₙ = e − nIₙ₋₁ for n ≥ 1.[3]
(b)Find I₄ in the form ae + b, where a and b are integers.[3]
Show answer
(a) Shown (see working).
M1: by parts with u = xⁿ, dv = eˣ dxA1: [xⁿeˣ]₀¹ − n∫₀¹ xⁿ⁻¹eˣ dxA1: = e − nIₙ₋₁ (AG)
(b) I₄ = 9e − 24
B1: I₀ = e − 1M1: I₁ = 1, I₂ = e − 2, I₃ = 6 − 2eA1: I₄ = e − 4(6 − 2e) = 9e − 24
Q17Reduction formulae · Standard · Paper 2 style · 8 marks
(a)Let Iₙ = ∫₀^(π/2) sinⁿx dx. By writing sinⁿx = sinⁿ⁻¹x sin x, show that nIₙ = (n − 1)Iₙ₋₂ for n ≥ 2.[4]
(b)Find the exact value of I₆.[2]
(c)Hence find the exact value of ∫₀^(π/2) sin⁴x cos²x dx.[2]
Show answer
(a) Shown (see working).
M1: by parts: u = sinⁿ⁻¹x, dv = sin x dxA1: [−sinⁿ⁻¹x cos x]₀^(π/2) + (n − 1)∫ sinⁿ⁻²x cos²x dxM1: cos²x = 1 − sin²xA1: Iₙ = (n − 1)(Iₙ₋₂ − Iₙ) ⇒ nIₙ = (n − 1)Iₙ₋₂ (AG)
(b) 5π/32
M1: I₀ = π/2; I₆ = (5/6)(3/4)(1/2)(π/2)A1: 5π/32
(c) π/32
M1: sin⁴x cos²x = sin⁴x − sin⁶x, so integral = I₄ − I₆A1: 3π/16 − 5π/32 = π/32
Q18Reduction formulae · Standard · Paper 2 style · 7 marks
(a)Let Iₙ = ∫₀^(π/2) xⁿ cos x dx for n ≥ 0. Show that Iₙ = (π/2)ⁿ − n(n − 1)Iₙ₋₂ for n ≥ 2.[4]
(b)Find the exact value of I₄, and give it correct to 3 significant figures.[3]
Show answer
(a) Shown (see working).
M1: by parts: [xⁿ sin x]₀^(π/2) − n∫ xⁿ⁻¹ sin x dx = (π/2)ⁿ − n∫ xⁿ⁻¹ sin x dxM1: by parts again: ∫ xⁿ⁻¹ sin x dx = [−xⁿ⁻¹cos x] + (n − 1)∫ xⁿ⁻² cos x dxA1: boundary terms vanish (cos(π/2) = 0, and x^(n−1) = 0 at 0)A1: Iₙ = (π/2)ⁿ − n(n − 1)Iₙ₋₂ (AG)
Q24Substitution & standard integrals · Challenge · Paper 2 style · 6 marks
(a)Use the substitution u = eˣ to find the exact value of ∫₀^(ln 2) 1/(eˣ + 4e⁻ˣ) dx.[5]
(b)Give your answer correct to 4 significant figures.[1]
Show answer
(a) (1/2)(π/4 − tan⁻¹(1/2)) = (1/2)tan⁻¹(1/3)
M1: multiply top and bottom by eˣ: eˣ/(e²ˣ + 4)M1: u = eˣ, du = eˣ dx, limits 1 to 2: ∫₁² du/(u² + 4)A1: [(1/2)tan⁻¹(u/2)]₁²A1: (1/2)(π/4 − tan⁻¹(1/2))A1: (or (1/2)tan⁻¹(1/3), using the tan subtraction formula)
(b) 0.1609
B1: 0.1609
Q25Arc length · Core · Paper 2 style · 4 marks
(a)Find the exact length of the arc of the curve y = cosh x from x = 0 to x = ln 3.[4]
Q27Arc length · Challenge · Paper 2 style · 8 marks
(a)A curve is defined parametrically by x = t², y = (2/3)t³, for 0 ≤ t ≤ √3. Show that (dx/dt)² + (dy/dt)² = 4t²(1 + t²).[2]
(b)Hence find the exact length of the curve.[3]
(c)A wire is bent into the shape of this curve, with lengths in centimetres. A second wire of the same shape is made from the curve for 0 ≤ t ≤ T. Find T if the second wire is exactly half as long as the first.[3]
Q29Surface area of revolution · Standard · Paper 2 style · 6 marks
(a)A lampshade is modelled by rotating the arc of y = cosh x between x = 0 and x = 1 through 2π radians about the x-axis (units are decimetres). Show that the curved surface area is π(1 + (1/2)sinh 2).[5]
(b)Find this area correct to 3 significant figures.[1]
