Integration — Paper 3 · A Level Mathematics 9709

7 The diagram shows the curve y = x2e−12x. (i) Find the x-coordinate of M, the maximum point of the curve. [4] (ii) Find the area of the shaded region enclosed by the curve, the x-axis and the line x = 1, giving your answer in terms of e. [5]

9 marks

Mark scheme: 7 (i) Use product or quotient rule M1* 1 1 − 2 x 1 2 − 2 x Obtain first derivative 2 xe − x e or equivalent A1 2 Equate derivative to zero and solve for non-zero x M1(dep*) Obtain answer x = 4 A1 4 1 1 − x − x (ii) Integrate by parts once, obtaining kx 2 e 2 + l ∫ x e 2 d x , where kl ≠ 0 M1 1 1 − x − x Obtain integral − 2 x 2 e 2 + 4 ∫ xe 2 d x , or any unsimplified equivalent A1 1 − x 2 or equivalent A1 Complete the integration, obtaining − 2(x 2 + 4 x + 8 )e Having integrated by parts twice, use limits x = 0 and x = 1 in the complete integral M1 1 − Obtain simplified answer 16 − 26 e 2 or equivalent A1 5 A Bx + C

This question in 9709/31 Oct/Nov 2004

x2 + 3x + 3 7 Let f(x) ≡ (x + 1)(x + 3). (i) Express f(x) in partial fractions. [5] 3 (ii) Hence show that f(x) dx = 3 −1 ln 2. [4] 2 0

9 marks

Mark scheme: B C 7 (i) State or imply the form A + + B1 x + 1 x + 3 State or obtain A = 1 B1 Use correct method for finding B or C M1 Obtain B = 1 A1 2 Obtain C = − 3 A1 [5] 2 (ii) Obtain integral x + 1 ln ( x + 1) − 3 ln ( x + 3 ) B2√ 2 2 [Award B1√ if only one error. The f.t. is on A, B, C.] Substitute limits correctly M1 Obtain given answer following full and exact working A1 [4] [SR: if A omitted, only M1 in part (i) is available, then in part (ii) B1√ for each correct integral and M1.] y d y

This question in 9709/31 May/June 2008

8 An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m. It is given that V = 43h3. The liquid is poured in at a rate of 20 m3 per hour, but owing to leakage, liquid is lost at a rate dh proportional to h2. When h = 1, = 4.95. dt (i) Show that h satisfies the differential equation dh 5 = −1 [4] dt h2 20. 20h2 2000 (ii) Verify that ≡−20 + [1] 100 −h2 (10 −h)(10 + h). (iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h. [5]

10 marks

Mark scheme: dV 2 dh dV 28 (i) State or obtain = 4 h , or = 4 h , or equivalent B1 dt dt d h dV 2 State or imply = 20 − kh B1 dt Use the given values to evaluate k M1 Show that k = 0.2, or equivalent, and obtain the given equation A1 [4] [The M1 is dependent on at least one B mark having been earned.] (ii) Fully justify the given identity B1 [1] (iii) Separate variables correctly and attempt integration of both sides M1 Obtain terms –20h and t, or equivalent A1  10 + h  Obtain terms aln(10 + h) + bln(10 – h), where ab ≠ 0, or k ln   M1  10 − h  Obtain correct terms, i.e. with a = 100 and b = −100, or k = 2000/20, or equivalent A1 Evaluate a constant and obtain a correct expression for t in terms of h A1 [5] 1 x 1 x ∫

This question in 9709/31 Oct/Nov 2008

10 In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When t 0, = dr r 5 and 2. = dt = (i) Show that r satisfies the differential equation dr 0.08r2. dt = [4] [The surface area A and volume V of a sphere of radius r are given by the formulae A 4πr2, 4 = V = 3πr3.] (ii) Solve this differential equation, obtaining an expression for r in terms of t. [5] (iii) Deduce from your answer to part (ii) the set of values that t can take, according to this model. [1]

10 marks

Mark scheme: dA 10 (i) State or imply = kV M1* dt dr dr 4 Obtain equation in r and , e.g. 8πr = k πr3 A1 dt dt 3 dr Use = 2, r = 5 to evaluate k M1(dep*) dt Obtain given answer A1 [4] (ii) Separate variables correctly and integrate both sides M1 1 Obtain terms – and 0.08t, or equivalent A1 + A1 r Evaluate a constant or use limits t = 0, r = 5 with a solution containing terms of the form a and bt M1 r 5 Obtain solution r = , or equivalent A1 [5] 1( − 4.0t ) (iii) State the set of values 0 Y t < 2.5, or equivalent B1 [1] [Allow t < 2.5 and 0 < t < 2.5 to earn B1.]

This question in 9709/31 Oct/Nov 2009

a ln x 2 5 7 (i) Given that dx 5, show that a ln [5] x2 = = 3(1 + a). ä 1 (ii) Use an iteration formula based on the equation a 5 ln to find the value of a correct to 2 decimal places. Use an initial value of 4 and give= 3(1the+ resulta) of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 7 (i) Attempt integration by parts M1 1 x ln x − x ln x 1 −1 d x , d x or equivalent A1 2 2 2 Obtain − x ln x + ∫ x x 2 + 2 ∫ x d x − 2 ∫ x Obtain − x −1 ln x − x −1 or equivalent A1 Use limits correctly, equate to 52 and attempt rearrangement to obtain a in terms of ln a M1 Obtain given answer a = 53 (1 + ln a ) correctly A1 [5] (ii) Use valid iterative formula correctly at least once M1 Obtain final answer 3.96 A1 Show sufficient iterations to > 4 dp to justify accuracy to 2 dp or show sign change in interval (3.955, 3.965) A1 [3] [4 → 3.9772 → 3.9676 → 3.9636 → 3.9619] ( 53 an −1 ) SR: Use of a n +1 = e to obtain 0.50 also earns 3/3. GCE A/AS LEVEL – October/November 2010 9709 33

This question in 9709/33 Oct/Nov 2010

9 A biologist is investigating the spread of a weed in a particular region. At time t weeks after the start of the investigation, the area covered by the weed is A m2. The biologist claims that the rate of increase of A is proportional to √(2A −5). (i) Write down a differential equation representing the biologist’s claim. [1] (ii) At the start of the investigation, the area covered by the weed was 7 m2 and, 10 weeks later, the area covered was 27 m2 . Assuming that the biologist’s claim is correct, find the area covered 20 weeks after the start of the investigation. [9]

10 marks

Mark scheme: dA 9 (i) State = k 2 A − 5 B1 [1] dt (ii) Separate variables correctly and attempt integration of each side M1 1 2 = … or equivalent A1 Obtain (2A − 5 ) Obtain = kt or equivalent A1 Use t = 0 and A = 7 to find value of arbitrary constant M1 Obtain C = 3 or equivalent A1 Use t = 10 and A = 27 to find k M1 Obtain k = 0.4 or equivalent A1 Substitute t = 20 and values for C and k to find value of A M1 Obtain 63 cwo A1 [9]

This question in 9709/33 Oct/Nov 2010

10 y M P x O 3 The diagram shows the curve y = x2e−x. (i) Show that the area of the shaded region bounded by the curve, the x-axis and the line x = 3 is equal to 2 −17 . [5] e3 (ii) Find the x-coordinate of the maximum point M on the curve. [4] (iii) Find the x-coordinate of the point P at which the tangent to the curve passes through the origin. [2]

11 marks

Mark scheme: 10 (i) Attempt integration by parts and reach ± x 2 e −±x ∫ 2 xe − x dx M1* Obtain − x 2 e −+x ∫ 2 xe − x d x , or equivalent A1 Integrate and obtain –x2e–x – 2xe–x – 2e–x, or equivalent A1 Use limits x = 0 and x = 3, having integrated by parts twice M1(dep*) Obtain the given answer correctly A1 [5] (ii) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for non-zero x M1 Obtain x = 2 with no errors send A1 [4] (iii) Carry out a complete method for finding the x-coordinate of P M1 Obtain answer x =1 A1 [2]

This question in 9709/32 May/June 2011

4 During an experiment, the number of organisms present at time t days is denoted by N, where N is treated as a continuous variable. It is given that dN dt = 1.2e−0.02tN0.5. When t 0, the number of organisms present is 100. = (i) Find an expression for N in terms of t. [6] (ii) State what happens to the number of organisms present after a long time. [1]

7 marks

Mark scheme: 4 (i) Separate variables and attempt integration on both sides M1* Obtain 2N0.5 on left-hand side or equivalent A1 Obtain –60e–0.02t on right-hand side or equivalent A1 Use 0 and 100 to evaluate a constant or as limits in a solution containing terms DM1* aNo.5 and be –0.02t Obtain 2N0.5 = –60e–0.02t + 80 or equivalent A1 Conclude with N = (40 – 30e–0.02t)2 or equivalent A1 [6] (ii) State number approaches 1600 or equivalent, following expression of form B1√ [1] (c + de–0.02t)n (i) Eith

This question in 9709/33 Oct/Nov 2011

7 y x O The diagram shows part of the curve y sin32x cos32x. The shaded region shown is bounded by the = curve and the x-axis and its exact area is denoted by A. (i) Use the substitution u sin 2x in a suitable integral to find the value of A. [6] = kπ (ii) Given that dx 40A, find the value of the constant k. [2] ã 0 |sin32x cos32x| = [Questions 8, 9 and 10 are printed on the next page.]

8 marks

Mark scheme: 7 (i) State or imply du = 2cos2x dx or equivalent B1 Express integrand in terms of u and du M1 1 3 2 Obtain u (1 − u ) du or equivalent A1 ∫ 2 Integration to obtain an integral of the form k 1 u 4 + k 2 u 6 , k 1 , k 2 ≠ 0 M1 1 Use limits 0 and 1 or (if reverting to x) 0 and π correctly DM1 4 1 Obtain , or equivalent A1 [6] 24 (ii) Use 40 and upper limit from part (i) in appropriate calculation M1 Obtain k = 10 with no errors seen A1 [2]

This question in 9709/33 Oct/Nov 2012

4 ln x 3 Find the exact value of dx. [5] x Ô1

5 marks

Mark scheme: 1 3 EITHER: Integrate by parts and reach kx 2 ln x − m ∫ x 2 . x dx M1* 1 1 Obtain 2 x 2 ln x − 2 ∫ 1 d x , or equivalent A1 x 2 1 1 Integrate again and obtain 2 x 2 ln x − 4 x 2 , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4(ln 4 − )1 , or exact equivalent A1 1 1 u u OR1: Using u = ln x, or equivalent, integrate by parts and reach kue 2 − m ∫ e 2 d u M1* 1 1 u u Obtain 2ue 2 − 2 ∫ e 2 du , or equivalent A1 1 1 u u Integrate again and obtain 2u e 2 − 4 e 2 , or equivalent A1 Substitute limits u = 0 and u = ln4, having integrated twice M1(dep*) Obtain answer 4 ln 4 − 4 , or exact equivalent A1 1 OR2: Using u = x , or equivalent, integrate and obtain ku ln u − m ∫ u . u d u M1* Obtain 4u ln u − 4 ∫ 1du , or equivalent A1 Integrate again and obtain 4u ln u − 4u , or equivalent A1 Substitute limits u = 1 and u = 2, having integrated twice or quoted ∫ ln u du as u ln u ± u M1(dep*) Obtain answer 8 ln 2 − 4 , or exact equivalent A1 x ln x ± x x ln x ± x OR3: Integrate by parts and reach I = + k ∫ dx M1* x x x x ln x − x 1 1 1 I − Obtain I = + 2 2 ∫ dx A1 x x Integrate and obtain I = 2 x ln x − 4 x , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4 ln 4 − 4 , or exact equivalent A1 [5] GCE A LEVEL – October/November 2013 9709 31

This question in 9709/31 Oct/Nov 2013

3x 2 Use the substitution u 3x 1 to find dx. [4] 3x 1 = + Ô +

4 marks

Mark scheme: du 2 Carry out complete substitution including the use of = 3 M1 dx  1 1  − du A1  Obtain ∫  3 3u   Integrate to obtain form k1u + k 2 ln u or k1u + k 2 ln 3u where k1 k 2 ≠ 0 M1 1 1 Obtain (3 x + 1) − ln (3 x + )1 or equivalent, condoning absence of modulus signs and + c A1 [4] 3 3

This question in 9709/33 Oct/Nov 2013

8 y x 0 O The diagram shows the curve y = x cos 12x for 0 ≤x ≤0. dy 1 (i) Find and show that 4d2y + y + 4 sin = 0. [5] 2x dx dx2 (ii) Find the exact value of the area of the region enclosed by this part of the curve and the x-axis. [5]

10 marks

Mark scheme: 8 (i) Use product rule M1 Obtain derivative in any correct form A1 Differentiate first derivative using the product rule M1 Obtain second derivative in any correct form, e.g. − 12 sin 12 x − 14 x cos 12 x − 12 sin 12 x A1 Verify the given statement A1 5 2 x d x M1* (ii) Integrate and reach kx sin 12 x + l ∫ sin 1 2 x d x , or equivalent A1 Obtain 2 x sin 12 x − 2 ∫ sin 1 Obtain indefinite integral 2 x sin 12 x + 4 cos 12 x A1 Use correct limits x = 0, x = π correctly M1(dep*) Obtain answer 2π − 4 , or exact equivalent A1 5 dN

This question in 9709/32 May/June 2014

1 x 6 Let I dx. 2 x = Ô 0 − 2 2 2 2 (i) Using the substitution u 2 x, show that I du. [4] −u u = − = Ô 1 (ii) Hence show that I 8 ln 2 [4] = −5.

8 marks

Mark scheme: 1 6 (i) State or imply d u = − d x , or equivalent B1 2 x Substitute for x and dx throughout M1 ± 2 ( 2 − u 2) Obtain integrand , or equivalent A1 u Show correct working to justify the change in limits and obtain the given answer with no errors seen A1 [4] (ii) Integrate and obtain at least two terms of the form a ln u , bu , and cu 2 M1* Obtain indefinite integral 8 ln u − 8u + u 2 , or equivalent A1 Substitute limits correctly M1(dep*) Obtain the given answer correctly having shown sufficient working A1 [4]

This question in 9709/32 May/June 2015

1 x5 7 Let I dx. 3 = Ô0 1 x2 + 2 u 2 (i) Using the substitution u 1 x2, show that I du. [3] −1 2u3 = + = Ô1 (ii) Hence find the exact value of I. [5]

8 marks

Mark scheme: 7 (i) State or imply du = 2x dx , or equivalent B1 Substitute for x and dx throughout M1 Reduce to the given form and justify the change in limits A1 [3] (ii) Convert integrand to a sum of integrable terms and attempt integration M1 1 1 1 Obtain integral 2 ln u + − 2 , or equivalent A1 + A1 u 4u (deduct A1 for each error or omission) Substitute limits in an integral containing two terms of the form a ln u and bu− 2 M1 Obtain answer 12 ln2 − 165 , exact simplified equivalent A1 [5]

This question in 9709/33 May/June 2016

4 x 6 Let I dx. −1 2 x x = Ô1 + 2 u (i) Using the substitution u x, show that I du. [3] −1 u 1 = = Ô1 + (ii) Hence show that I 1 ln 49. [6] = +

9 marks

Mark scheme: 1 6 (i) State or imply d u = d x B1 2 x Substitute for x and dx throughout M1 Justify the change in limits and obtain the given answer A1 [3] B (ii) Convert integrand into the form A + M1* u + 1 Obtain integrand A =1, B = −2 A1 Integrate and obtain u − 2ln(u + 1) A1 + A1 Substitute limits correctly in an integral containing terms au and bln(u + 1), where ab ≠ 0 DM1 Obtain the given answer following full and correct working A1 [6] [The f.t. is on A and B.]

This question in 9709/33 Oct/Nov 2016

4 x 5 Let I dx. 1 = Ô 1 4 −x 1 30 (i) Using the substitution x show that I 2 [4] 1 = cos21, = Ó cos21 d1. 60 … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the exact value of I. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) State or imply dx = −2cosθsinθdθ, or equivalent B1 Substitute for x and dx, and use Pythagoras M1 Obtain integrand ± 2cos 2θ A1 Justify change of limits and obtain given answer correctly A1 4 5(ii) Obtain indefinite integral of the form aθ+ b sin 2θ M1* 1 A1 Obtain θ+ sin2θ 2 Use correct limits correctly M1(dep*) 1 A1 Obtain answer 6π with no errors seen 4

This question in 9709/31 May/June 2018

7 y M R x O 120 The diagram shows the curve y 5 sin2x cos3x for 0 and its maximum point M. The shaded = ≤x ≤120, region R is bounded by the curve and the x-axis. (i) Find the x-coordinate of M, giving your answer correct to 3 decimal places. [5] … … … … … … … … … … … … … … … … … … … (ii) Using the substitution u sin x and showing all necessary working, find the exact area of R. [4] = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Use product rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and obtain an equation in a single trig function depM1* Obtain a correct equation, e.g. 2 3 tan 2 x = A1 Obtain answer x = 0.685 A1 5 Question Answer Marks Guidance 7(ii) Use the given substitution and reach ( ) 2 4 d a u u u ∫ − M1 Obtain correct integral with a = 5 and limits 0 and 1 A1 Use correct limits in an integral of the form 3 5 1 1 3 5 a u u   −     M1 Obtain answer 2 3 A1 4

This question in 9709/33 Oct/Nov 2018

4 4 Show that 2 4. [5] 2 ln x dx = −ln Ó 1 x−3 … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Integrate by parts and reach 1 1 2 2 1 ln . d − − + ∫ ax x b x x x Obtain 1 1 2 2 1 2 ln 2 . d − − − + ∫ x x x x x , or equivalent A1 Complete the integration, obtaining 1 1 2 2 2 ln 4 − − − − x x x , or equivalent A1 Substitute limits correctly, having integrated twice M1(dep*) Obtain the given answer following full and correct working A1 5

This question in 9709/32 Feb/March 2019

3 Find the exact value of 4 3 x 2 ln x dx. [5] Ó1 … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Commence integration and reach 5 5 2 2 1 ln . d +  ax x b x x x Obtain 5 5 2 2 2 2 1 ln . d 5 5 − x x x x x A1 Complete the integration and obtain 5 5 2 2 2 4 ln 5 25 − x x x , or equivalent A1 Use limits correctly, having integrated twice e.g 2 4 2 4 32ln 4 32 0 5 25 5 25 × − × − × +       DM1 Obtain answer 128 124 ln2 5 25 − , or exact equivalent A1 5

This question in 9709/32 May/June 2020

2 4 Using integration by parts, find the exact value of tan−1 1 dx. [5] 2x Ó0 … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Commence integration and reach 1 2 1 1 tan d 2 − +  + ax x b x x c x *M1 OE. Denominator might be 2 2 1 or 2 4 2 + + x x . Obtain 1 1 tan . 2 − −     x x x 2 2 4 + x dx A1 OE Complete integration and obtain ( ) 1 2 1 tan ln 4 2 − − +     x x x A1 OE e.g. with 2 1 ln 4 x +       Substitute limits correctly in an expression of the form ( ) 1 2 tan ln − + + px x q c x DM1 1 2tan 1 ln8 ln 4 − − + OE Obtain final answer 1 π ln 2 2 − A1 OE exact answer. Needs a value for 1 tan 1 − and a single log term Alternative method for Question 4 Use the substitution 1 tan 2 θ − = x to obtain 2 2 sec d λ θ θ θ  and reach tan tan d θ θ θ θ +  p q *M1 Question Answer Marks Guidance 4 Obtain 2 tan 2 tan d θ θ θ θ − A1 OE Complete integration and obtain ( ) 2 tan 2ln cos θ θ θ + A1 OE Substitute correct limits correctly in an expression of the form ( ) tan ln cos θ θ θ + r s DM1 Limits should be 4 π and 0. Limits must be in radians. Obtain final answer 1 π ln 2 2 − A1 OE exact answer. Need values for trig. functions and a single log term. 5

This question in 9709/32 May/June 2021

7 The variables x and satisfy the differential equation 1 dx x cot sin21 = tan21 −2 1, d1 for 0 1 and x 0. It is given that x 2 when 1 < 1 < 2π > = 1 = 4π. d cot (a) Show that 1 . cot21 = −2 d1 sin21 (You may assume without proof that the derivative of cot with respect to is [1] 1 1 −cosec21.) … … … … … … … … (b) Solve the differential equation and find the value of x when 1 [7] 1 = 6π. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Show sufficient working to justify the given statement B1 e.g. see 2cot − cosec 2 in the working or express in terms of sinand cos and use quotient rule to obtain the given result. Solution must have θ present throughout and must reach the given answer. 1 7(b) Separate variables correctly B1 tan 2  2cot  xdx = 2 − 2 d  sin  sin  Check for relevant working in (a) Condone incorrect notation e.g. missing dx. Need either the integral sign or the dx, dθ. 1 2 B1 Obtain term x 2 Obtain terms tan+ cot 2 B1 + B1  2cot   2cos 1 d = − C ) Alternative:  2 d =  3 2 ( +  sin   sin  sin  1 M1 Need to have 3 terms. Constant of correct form. Form an equation for the constant of integration, or use limits x = 2, = π , in 4 a solution with at least two correctly obtained terms of the form ax 2 , btan and ccot 2, where abc ≠ 0 A1 1 or 12 x 2 = tan+ cosec 2− 1 x 2 = tan+ cot 2 State correct solution in any form, e.g. 2 If everything else is correct, allow a correct final answer to imply this A1. 1 A1 18 + 2 3 Substitute = π and obtain answer x = 2.67 2.6748… 3 If see a correctly rounded value ISW. 6 7

This question in 9709/32 Oct/Nov 2022

7 The variables x and y satisfy the differential equation dy 4 tan 2x cos 2x , dx = sin2 3y where 0 1 It is given that y 0 when x 1 ≤x < 4π. = = 6π. Solve the differential equation to obtain the value of x when y 1 Give your answer correct to = 6π. 3 decimal places. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7 Correct separation of variables B1 2 sin 3 d 4sec2 tan 2 d y y x x x    or equivalent. Condone missing integral signs or dx and dy. Integrate to obtain sec2 k x M1 Obtain 2sec2x A1 Use double angle formula and integrate to obtain sin6 py q y  M1 Or two cycles of integration by parts. Obtain 1 1 sin6 2 12 y y  A1 Use 0 y  , π 6 x  in a solution containing terms sec2x  and sin6y  to find the constant of integration M1 Obtain 1 1 sin6 2sec2 4 2 12 y y x    A1 Or equivalent seen or implied by π 1 sin π 2sec2 4 2 12 x          . Obtain 0.541 x  A1 From correct working (not by using the calculator to integrate). 8

This question in 9709/31 May/June 2023

7 (a) Use the substitution u cosx to show that = 1 π cos x sin 2x e2 dx 2ue2u du. [4] Ó 0 = Ó −1 … … … … … … … … … … … … … … … … … … … … … … … π cos x(b) Hence find the exact value of sin 2x e2 dx. [4] Ó 0 … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) d sin d  u x x B1 SOI Use double angle formula and substitute for x and dx throughout the integral M1 All x’s must be removed, can be coefficient errors provided 2 seen in working. Obtain 2 2 e d  u u u A1 Limits may be omitted, or left as 0 and , during the change of variable stage. Justify new limits and obtain 1 2 12 e d  u u u from correct working A1 AG Must see x = 0, u = 1 and x = , u = −1. Inequalities alone e.g. 0 ⩽ x ⩽ π and 1 ⩽ u ⩽ −1 or –1 ⩽ u ⩽ 1 for limits are insufficient A0 If sign in expression and order of limits incorrect then A0. If negative sign is present in the integrand then this can be removed and limits introduced in correct order in a single step. 4 Question Answer Marks Guidance 7(b) Commence integration and reach 2 2 e e d ,  u u au b u where 0,  ab 0  b M1* Condone dx. Complete integration and obtain 2 2 1 2 e e  u u u A1 OE Allow (2 2 2 1 1 e ) e 2 2  u u u . Use correct limits correctly in c 2 e  u u d e2u having integrated twice or in c cos x e2 cos x + d e2cos x DM1 1 and −1 for u, 0 and π for x e.g. ce2 + de2 − (− ce−2 + de−2). Not decimals. Allow one sign error at most in going from c 2 e  u u d e2u or c cos x e2 cos x + d e2 cos x to ce2 + de2 − (− ce−2 + de−2). [e2 − ½ e2 − (− e−2 – ½ e−2)] Complete reversal of sign by converting back to cos x and not making x = 0 upper limit is DM0 A0. Obtain 2 2 1 3 e e 2 2   A1 ISW Or equivalent 2-term expression e.g. 4 2 e 3 2e  or 2 2 1 3 e 2 e        . 4

This question in 9709/33 May/June 2023

36a 210 Let f ( x) = , where a is a positive constant. ( 2a + x)( 2a - x)( 5a - 2x) (a) Express f ( x) in partial fractions. [5] … … … … … … … … … … … … … … … … … … … … … … … … … a (b) Hence find the exact value of f ( x) d x , giving your answer in the form p ln q + r ln s where p and -ya r are integers and q and s are prime numbers. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) A B C B1 Allow if seen prior to assigning a value for a. State or imply the form + + 2 a + x 2 a − x 5 a − 2 x Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = 9, C = −16 A1 Obtain a second value A1 Obtain the third value A1 5 Dx + E C SC + B0 M1 and C = −16 4 a ^ 2 − x ^ 2 5 a − 2 x A1 Max 2/5. SC Allow M1 only for other incorrect partial fraction. 10(b) Integrate and obtain one of the terms ln 2 a + x − 9ln 2 a − x + 8ln 5 a − 2 x B1 FT Condone missing modulus signs. Use their A, B and C. Obtain a second correct term B1 FT Obtain the third correct term B1 FT Max 3/5 if value is assigned for a (award M0 A0). Substitute limits correctly in an integral of the form M1 Either (i) collect terms with same coeeficient and p ln 2 a + x + q ln 2 a − x + + r ln 5 a − 2 x and remove all a’s remove all a’s e.g. pln 3a −pln a + qln a − qln 3a + rln 3a – rln7a hence pln 3 − qln 3 + rln 3 − rln7 or (ii) collect same ln terms and remove all a’s e.g. (p – q + r) ln 3a – ( p – q) ln a – rln7a and − (p − q) ln a = (−p + q − r ) ln a + r ln a hence p ln 3 – q ln3 + rln 3 – r ln 7. Obtain 18ln3 − 8ln7 from correct working A1 A0 if the solution involves logarithms of negative numbers. 5

This question in 9709/32 Feb/March 2024

10 (a) Given that 2x = tan y , show that = 2 . [3] dx 1 + 4x … … … … … … … … … … … … … 3 (b) Hence find the exact value of 2 x tan -1 ( 2x) dx . [7] y1 2 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Obtain 2 = sec2y d d y x or equivalent d x y = sec2y by differentiation with respect to y. Use sec2y = 1 + tan2 y M1 Replace tan y with 2x and rearrange to obtain given answer 2 d 2 d 1 4   y x x A1 3 10(b) Integrate by parts and reach 2 2 1 2 tan 2 d 1 4     x ax x b x x *M1 Obtain 2 2 1 1 2 2 tan 2 d 1 4 x x x x x     A1 OE Reduce integral to expression of the form 2 d 1 4    n m x x M1 Complete integration and reach 2 1 1 tan 2 tan 2     px x qx r x M1 Obtain 2 1 1 1 1 1 2 4 8 tan 2 tan 2 x x x x     A1 OE Use limits of 1 2 x  and 1 2 3 x  in the correct order, having integrated twice DM1 Obtain answer 5 1 1 48 8 8 π 3   or exact equivalent A1 7

This question in 9709/31 May/June 2024

9 10- x x x A container in the shape of a cuboid has a square base of side x and a height of ( 10- )x . It is given that x varies with time, t, where t 2 0 . The container decreases in volume at a rate which is inversely proportional to t. When t = 1 , x = 1 and the rate of decrease of x is 20. 10 2 37 (a) Show that x and t satisfy the differential equation dx - 1 = . [5] td 2t `20 x - 3x 2j … … … … … … … … … … … … … … … … … (b) Solve the differential equation, obtaining an expression for t in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 9(a) Obtain  d d V k t t  or  d 1 d V t kt  B1 Obtain 2 d 20 3 d V x x x   B1 Correct use of chain rule involving k M1 Use d d V t = d d . d d V x x t  Expressions for d d V t and d d V x must be seen to get M1. Obtain   2 d d 20 3 x k t t x x   or equivalent, A1 If this expression is first seen with numerical values, allow A1 when their value of k is substituted back into the general expression. Use 1 10 t  , 1 2 x  and d 20 d 37 x t  to obtain given answer which must be stated d 20 d 37 x t  needed to score final A1 A1   2 d 1 d 2 20 3 x t t x x    AG Need to at least see 20 37  = 1 3 10 10 4 k        if k t or 20 37  = 1 3 10 10 4 k         if k t  in working for correct k. d 20 d 37 x t  seen anywhere, then A0. 5 Question Answer Marks Guidance 9(b) Separate variables correctly & integrate at least one side correctly B1 Obtain terms 10x2 – x3 B1 May see –10x2 + x3 if negative sign moved across or e.g. 20x2 – 2x3 if 2 moved across. Allow 2 3 20 3 . 2 3 x x  Obtain term ln t with ‘correct’ coefficient from their separation of variables, for example a ln t for a t . B1FT FT sign and position of 2 from their separation but B0 if error from later manipulation. Use 1 10 t  , 1 2 x  to evaluate a constant or as limits in a solution containing terms of the form x2, x3 and ln t (or ln 2t) M1 Allow numerical and sign errors and decimals. Allow if exponentiate before substitution, even if exponentiation done incorrectly, allow for c or ec. Obtain correct answer in any form, for example 2 3 ln 19 ln0.1 10 2 8 2 t x x     A1 2 3 ln2 19 ln0.2 10 2 8 2 t x x     or 2 3 ln 10 2.5 0.125 1.15 2 t x x       Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45 Obtain answer 3 2 19 2 20 1 4 10 e x x t    or equivalent A1 ISW Need t =……… E.g. 3 3 2 2 2 3 19 2 19 4 2 20 4 19 20 20 2 4 0 . .1 e 1 , , e e 10 10e e x x x x x x     Allow decimals, allow 2.44 to 2.46 for 2.45, e.g. 3 2 2 20 2.45 e . x x   A0 if 1 ln10 e present in final answer. 6

This question in 9709/33 May/June 2024

3 2 Find the exact value of x 2 ln 3x dx . Give your answer in the form a ln b + c , where a and c are rational y 1 and b is an integer. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 3 2 M1* x dx Integrate to obtain px ln3 x + q  1 3 1 2 A1 Or unsimplified equivalent. Obtain x ln3 x x dx 3 −3 1 3 1 3 A1 Complete integration to obtain x ln3 x − x 3 9 Use correct limits correctly in an expression of the form rx 3 ln3 x + sx 3 DM1 9ln9 − 3 − 13 ln3 + 19 An exact expression for their integral. 53 26 A1 Or 2-term equivalent. Obtain ln3 − 3 9 5

This question in 9709/31 Oct/Nov 2024

2 e 2 x 11 Let f ( )x = . e 2 x - 3 e x + 2 (a) Find f l ( x) and hence find the exact coordinates of the stationary point of the curve with equation y = f ( )x . [5] … … … … … … … … … … … … … … … … … … … … … … … … … ln 5 (b) Use the substitution u = ex and partial fractions to find the exact value of f ( )x d x . y ln 3 Give your answer in the form lna, where a is a rational number in its simplest form. [9] … … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 11(a) Use correct quotient rule NB the question asks for f x( ) so need complete form M1 Or correct product rule. 4e 2 x (e 2 x − 3e x + 2) − 2e 2 x (2e 2 x − 3e x ) A1 Obtain correct derivative in any form, e.g. 2 e 2 x − 3e x + 2 ( ) Equate their derivative to zero *M1 Can be implied by numerator equated to zero for quotient rule. 8 = 6ex ( ) Solve for x to obtain x = lna DM1 a positive. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept equivalent exact forms, e.g. x = ln . 6 11(a) Alternative Method for Question 11(a) Complete method to express f ( x ) in partial fractions M1 As far as p + q + r with values for p, e x − 2 e x − 1  8 2  q and r  2 + x − x  .  e − 2 e − 1  u = e x Allow in u ( ) . s e x t e x *M1 Note: the question requires f’(x) so if they have Differentiate to obtain f  ( x ) = 2 + 2 substituted for ex, they will also need chain rule. e x − 2 e x − 1 ( ) ( ) − 8e x 2e x A1 From correct work. Obtain f  ( x ) = 2 + 2 e x − 2 e x − 1 ( ) ( ) Equate derivative to zero and solve for x to obtain x = lna DM1 Must follow correctly to give a positive value of a. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept x = ln , or equivalent. 6 5 11(b) du x B1 State or imply = e dx 2u B1 Correct expression in u. Obtain  u du or equivalent 2 − 3u + 2 Condone missing du or missing integral but not both.  2  8 2 + Or1   − du Allow FT if using their partial fractions from   u  u ( u − 2 ) u ( u − 1) (a). A B B1 FT Complete reduction to partial fractions. State or imply partial fractions of the form + Correct form for their integrand. u − 1 u − 2 C D E 2u − 3 3 2u − 3 F G Or1 + + Or2 + = + + u u − 2 u − 1 u 2 − 3u + 2 u 2 − 3u + 2 u 2 − 3u + 2 u − 2 u − 1 Use a correct method for finding a constant M1 Available if they have incorrect form. −2 4 A1 Obtain correct + u − 1 u − 2 2u − 3 3 3 Or2 + − u 2 − 3u + 2 u − 2 u − 1 Integrate to obtain a ln (u – 1) + b ln (u – 2) or equivalent *M1 M0 if they have additional terms that do not cancel out. Obtain correct –2 ln (u – 1) + 4 ln (u – 2) or equivalent A1FT FT values of their partial fraction coefficients. Correctly use limits u = 5 and 3 in an expression of the form a ln (u – 1) + b ln (u – 2) DM1 or x = ln 5 and ln 3 in an expression of the form a ln (ex – 1) + b ln (ex – 2) 81 A1 Accept ln 20.25. Obtain ln 4 9

This question in 9709/32 Oct/Nov 2024

9 The constant a is such that ; 6x ln x dx = 4 . 1 1 5 (a) Show that a = exp f e 2 + 3op, where exp(x) denotes ex. [5] 6 a … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2 and 2.1. [2] … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) M1* x 2 x dx  d x. Allow M1 with q Commence integration and reach px 2 ln x + q   x A1 x 2 3 x dx must be simplified. Obtain 3 x 2 ln x − x Complete integration and obtain 3 x 2 ln x − 32 x 2 A1 Use limits correctly in an expression of the form ax 2 ln x − bx 2 and equate to 4, having DM1 3a 2 ln a − 32 a 2 − 0 + 32 = 4 integrated twice  1  5   + 3 Obtain a = exp   correctly A1 AG   2  6  a   5 9(b) Calculate the values of a relevant expression or pair of expressions at M1 Not from using calculator to evaluate the original a = 2 and a = 2.1 integral. Justify the given statement with correct calculated values A1 E.g., using f( x ) = 6 x 2 ln x − 3 x 2 f(2) = 4.636  5 and f(2.1) = 6.402  5, or using the exponential, 2  2.0306 and 2.1  1.9917 or 0.0306 > 0 and −0.1083  0. . 2 9(c)  1  5   M1 Use the iterative process a n +1 = exp   2 + 3   correctly at least once.  6  a n   Obtain final answer 2.02 A1 Show sufficient iterations to at least 4 d.p. to justify 2.02 to 2 d.p. A1 E.g. 2 → 2.0306 → 2.0180 → 2.0231 … or show that there is a sign change in ( 2.015, 2.025 ) 2.05 → 2.0103 → 2.0263 → 2.0197 → 2.0224.. 2.1 → 1.9917 → 2.0342 → 2.0166 → 2.0237.. 3

This question in 9709/31 May/June 2025

1 Find the exact value of ln 3x dx . Give your answer in the form a + ln b , where a and b are integers. 2y 1 [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Commence integration by parts and reach Ax ln3 x +  Bdx *M1 1 Accept Ax ln3 x − B x  dx. Allow unsimplified.  x Obtain x ln3x – x A1 Substitute limits correctly in an expression of the form Ax ln3x + Bx DM1 2ln6 −−2 ln3 + 1 Obtain –1 + ln12 A1 If no working seen, no marks are available. 4

This question in 9709/31 Oct/Nov 2025

x 2 + 4 ax + 6a 2 9 Let f ( x) = , where a is a positive constant. ( x + 2 a)( x + 3 a) (a) Express f ( x) in partial fractions. [5] … … … … … … … … … … … … … … … … … … … … … … … … … a (b) Hence find the exact value of f ( x) d x . Give your answer in the form a ( p + ln q) , where p and q -ya are rational. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) B C B1 State or imply the form A + + x + 2 a x + 3a Use a correct method for finding a coefficient M1 Obtain one of A = 1, B = 2a and C = −3a A1 SC: If B0 is scored because of an omission of ‘A’, then maximum M1A1 is available for one constant correct. SC: If they substitute a value for a, or if their working implies the use of a = 1, they can score maximum B1M1. Obtain a second value A1 Obtain the third value A1 Alternative Method for Question 9(a) linear expression B1 − ax State or imply 1 + 1 + ( x + 2 a )( x + 3a ) ( x + 2 a )( x + 3a ) B C B1 State or imply the form 1 + + x + 2a x + 3a Use a correct method for finding B or C M1 Obtain one of B = 2a and C = −3a A1 Obtain the second value A1 5 9(b) Integrate and obtain terms Ax + B ln ( x + 2 a ) + C ln ( x + 3a ) B2FT B1 for any two terms correct, B2 for all three terms correct. The FT is on A, B and C: x + 2 a ln ( x + 2 a ) − 3a ln ( x + 3a ) . Allow for FT on a split completed in (b). Substitute limits correctly in an integral containing at least 2 terms from the M1 3 4 E.g. 2 a + 2 a ln − 3a ln ( 1 ) ( 2 ) form rx + s ln ( x + 2 a ) + t ln ( x + 3a ) 9 A1 Accept equivalent fractions with integers. 2 + ln Obtain a from correct working ( ( 8 ) ) 9 2 + ln A0 XP if a is following an error in (a). ( 8 ( ) ) 4

This question in 9709/32 Oct/Nov 2025

7 (a) Use the substitution u = x 2 - 3 to show that 12 4x 3 b 2 ( u + 3) dx = du , y 2 y 7 x - 3 a u where a and b are values to be found. [4] … … … … … … … … … … … … … … … … … … … … … … … … 12 4 x 3 (b) Hence, find the exact value of dx. [4] y 2 7 x - 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) du B1 State or imply = 2 x dx B1 Allow if just written down. Obtain x = 7  u = 4 ( = a ) and x = 12  u = 9 ( = b ) 12 2 du M1 Use = 2 x and substitute u for x in the integral  2 x  2 x E.g., using d x. dx 2  7 x − 3 Condone missing or incorrect limits. 12 3 9 A1 AG  4 x 2 ( u + 3 ) du from full and correct working with Allow from working in reverse if fully correct. Obtain  2 dx =   7 x − 3  4 u sufficient detail 4 7(b) 32 12 *M1 May integrate by parts. Integrate to obtain cu + du Must be in terms of u as ‘Hence’. 4 32 12 A1 Obtain 3 u + 12u Use their limits correctly DM1 For example: 4 104 152 ( 27 − 8 ) + 12 ( 3 − 2 ) or 72 − or 88 − . 9 3 3 Obtain 1123 or 37 13 A1 Exact answer only. Answer only is 0/4. 4

This question in 9709/35 Oct/Nov 2025