9231 Integration

9231-2021-mj-21-q03

May/June 2021 · Paper 21 · Question 3 · 10 marks

3(a)  1 x3dx< ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3+ ( 1 )( n )3 M1 A1 Forms the sum of the areas of the n rectangles. n n n n n n n n Need last term for A1. 0 1  n r3 = n2(n+1)2 =   n+1  2 M1 A1 Applies  n r3 = 1n2(n+1)2, AG. n4 4n4  2n  4 r=1 r=1 4 3(b)  1 x3dx> ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3 M1 A1 Forms the sum of the areas of appropriate n n n n n n rectangles. 0 Need the last term for A1. 1  n−1 r3 = (n−1)2n2 =   n−1  2 M1 A1 Applies  n r3 = 1n2(n+1)2. n4 4n4  2n  4 r=1 r=1 n+1 2 1 Accept   − .  2n  n 4 3(c)   n+1  2 −   n−1  2 = 1 <10 −3 leading to n>103 M1 Simplifies their U n − L n to k n .  2n   2n  n Least value of n is 1001. A1 2 © UCLES 2021 Page 7 of 13

Official mark scheme pages: 7 · source PDF URL

9231-2021-mj-21-q08

May/June 2021 · Paper 21 · Question 8 · 13 marks

8(a) dx B1 =2sinht dt dy ( ) M1 A1 Applies 2sinh2t = cosh2t −1, AG. = 3−1cosh2t =1− 1cosh2t−1 =1−sinh2t dt 2 2 2 2 3 8(b)(i) 2 2 M1 2 2 dx dy dx dy   +   =4s2 +(s2 −1)2 =4s2 +s4 −2s2 +1=(s2 +1)2 Factorises   +   . dt  dt  dt  dt  cosh4t A1 2π 1( 3t−1sinh2t ) cosh2t dt =π 1( 3t−1sinh2t )( cosh2t+1 ) dt M1 A1 Correct formula 2 for surfa 2 ce area, AG. A0 if limits 2 4 2 4 dx dy + 0 0 missing.     does not need to be dt   dt  simplified for M1. 4 © UCLES 2021 Page 12 of 13 8(b)(ii)  1 1sinh2t ( cosh2t+1 ) dt=1 (1+cosh2t)2 1 = 1 ( 1+cosh2 )2−1 M1 A1 Integrates. 0 4 16  0 16 4 3 1 t ( cosh2t+1 ) dt=3t ( 1sinh2t+t )  1 −3 1 1sinh2t+tdt M1 A1 Integrates by parts. 2 0 2 2  0 2 0 2 3t ( 1sinh2t +t ) − 1cosh2t − 1t2 1 = 3 ( 1sinh2− 1cosh2+ 3 ) A1 2 2 4 2  0 2 2 4 4 ( ) π 3sinh2−3cosh2− 1 (1+cosh2)2 +11 A1 OE. Must be exact. 4 8 16 8 (Decimal answer is 3.980131435…) 6 © UCLES 2021 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2021-mj-22-q03

May/June 2021 · Paper 22 · Question 3 · 10 marks

3(a)  1 x3dx< ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3+ ( 1 )( n )3 M1 A1 Forms the sum of the areas of the n rectangles. n n n n n n n n Need last term for A1. 0 1  n r3 = n2(n+1)2 =   n+1  2 M1 A1 Applies  n r3 = 1n2(n+1)2, AG. n4 4n4  2n  4 r=1 r=1 4 3(b)  1 x3dx> ( 1 )( 1 )3+ ( 1 )( 2 )3++ ( 1 )( n−1 )3 M1 A1 Forms the sum of the areas of appropriate n n n n n n rectangles. 0 Need the last term for A1. 1  n−1 r3 = (n−1)2n2 =   n−1  2 M1 A1 Applies  n r3 = 1n2(n+1)2. n4 4n4  2n  4 r=1 r=1 n+1 2 1 Accept   − .  2n  n 4 3(c)   n+1  2 −   n−1  2 = 1 <10 −3 leading to n>103 M1 Simplifies their U n − L n to k n .  2n   2n  n Least value of n is 1001. A1 2 © UCLES 2021 Page 7 of 13

Official mark scheme pages: 7 · source PDF URL

9231-2021-mj-22-q08

May/June 2021 · Paper 22 · Question 8 · 13 marks

8(a) dx B1 =2sinht dt dy ( ) M1 A1 Applies 2sinh2t = cosh2t −1, AG. = 3−1cosh2t =1− 1cosh2t−1 =1−sinh2t dt 2 2 2 2 3 8(b)(i) 2 2 M1 2 2 dx dy dx dy   +   =4s2 +(s2 −1)2 =4s2 +s4 −2s2 +1=(s2 +1)2 Factorises   +   . dt  dt  dt  dt  cosh4t A1 2π 1( 3t−1sinh2t ) cosh2t dt =π 1( 3t−1sinh2t )( cosh2t+1 ) dt M1 A1 Correct formula 2 for surfa 2 ce area, AG. A0 if limits 2 4 2 4 dx dy + 0 0 missing.     does not need to be dt   dt  simplified for M1. 4 © UCLES 2021 Page 12 of 13 8(b)(ii)  1 1sinh2t ( cosh2t+1 ) dt=1 (1+cosh2t)2 1 = 1 ( 1+cosh2 )2−1 M1 A1 Integrates. 0 4 16  0 16 4 3 1 t ( cosh2t+1 ) dt=3t ( 1sinh2t+t )  1 −3 1 1sinh2t+tdt M1 A1 Integrates by parts. 2 0 2 2  0 2 0 2 3t ( 1sinh2t +t ) − 1cosh2t − 1t2 1 = 3 ( 1sinh2− 1cosh2+ 3 ) A1 2 2 4 2  0 2 2 4 4 ( ) π 3sinh2−3cosh2− 1 (1+cosh2)2 +11 A1 OE. Must be exact. 4 8 16 8 (Decimal answer is 3.980131435…) 6 © UCLES 2021 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2021-mj-23-q03

May/June 2021 · Paper 23 · Question 3 · 10 marks

3(a) N r N r M1 A1 Compares with sum of the areas of the rectangles.  =1+ 2r2 −1 2r2 −1 r=1 r=2 N x M1 Compares with integral. < 1+ dx 1 2x2 −1 N x ( ) N M1 A1 Finds integral.  dx=1ln 2x2 −1  1 2x2 −1 4  1 N r ( ) M1 A1 Inserts limits, AG.  <1+ 1ln 2N2 −1 2r2 −1 4 r=1 7 3(b) N r N−1 r N N x N M1 A1 Compares with integral.  = + > dx+ 2r2 −1 2r2 −1 2N2 −1 1 2x2 −1 2N2 −1 r=1 r=1 ( ) N A1 CWO. 1ln 2N2 −1 + 4 2N2 −1 Alternative method for question 3(b) N r N+1 x M1 A1 Compares with integral.  > dx 2r2 −1 1 2x2 −1 r=1 ( ) 1ln 2 ( N +1 )2 −1 A1 CWO. 4 3 © UCLES 2021 Page 8 of 15

Official mark scheme pages: 8 · source PDF URL

9231-2021-mj-23-q07

May/June 2021 · Paper 23 · Question 7 · 11 marks

7(a) 3( )−1 ( ) 3 M1 A1 Recognises integral or uses appropriate substitution. I =2 4+x2 2 =sinh −1 1x 2 1  2  0 0 ln2 A1 Insert limits, must simplify. 3 7(b) d  x ( 4+x2 )−1 2 n  =−nx2 ( 4+x2 )−1 2 n−1+ ( 4+x2 )−1 2 n M1 A1 Uses the product rule to differentiate. dx  −n ( 4+x2 −4 )( 4+x2 )−1 2 n−1+ ( 4+x2 )−1 2 n M1 Applies x2 =4+x2 −4.  x ( 4+x2 )−1 2 n 3 2 =−nI +4nI +I M1 Integrates both sides using the limits given.   n n+2 n   0 3 ( 2 )n =(1−n)I +4nI 4nI = 3 ( 2 )n +(n−1)I A1 Substitutes limits and rearranges, AG. 2 5 n n+2 n+2 2 5 n Alternative method for question 7(b) I = x ( 4+x2 )−1 2 n 3 2 +n 3 2x2 ( 4+x2 )−1 2 n−1 dx M1 A1 Integrates by parts. n     0 0 I = x ( 4+x2 )−1 2 n2 3 +n2 3( 4+x2 −4 )( 4+x2 )−1 2 n−1 dx M1 Applies x2 =4+x2 −4. n     0 0 I = 3 ( 2 )n +nI −4I 4nI = 3 ( 2 )n +(n−1)I M1 A1 Substitutes limits and rearranges, AG. n 2 5 n n+2 n+2 2 5 n 5 © UCLES 2021 Page 12 of 15 7(c) I = 3 B1 Applies reduction formula with n=1. 3 20 12I = 3 ( 8 ) +2I I = 33 =0.033 M1 A1 Applies reduction formula with n=3. 5 2 125 3 5 1000 3

Official mark scheme pages: 12, 13 · source PDF URL

9231-2021-on-21-q04

Oct/Nov 2021 · Paper 21 · Question 4 · 10 marks

4(a) N lnr ln2 N lnr M1 A1 Compares with sum of the areas of the rectangles.  = + r=1 r2 4 r=3 r2 M1 for writing out sum, A1 for considering N lnr . r2 r=3 ln2 Nlnx M1 Compares with integral. < + dx 4 2 x2 Nlnx  lnx+1 N M1 A1 Finds integral.  dx= − 2 x2   x   2 N lnr ln2  lnN +1 ln2+1 2+3ln2 1+lnN M1 A1 Inserts limits, AG. A1 requires second M1.  < +  − +  = − r2 4  N 2  4 N r=1 7 4(b) N lnr N−1lnr lnN Nlnx lnN M1 A1 Compares with integral. Lower limit of 1 scores M0.  = + > dx+ r=1 r2 r=2 r2 N2 2 x2 N2 Accept  N lnr > N+1lnx . r2 2 x2 r=1 ln2+1 lnN +1 lnN A1 ln2+1 ln(N +1)+1 = − + Accept − . 2 N N2 2 N +1 3 © UCLES 2021 Page 8 of 13

Official mark scheme pages: 8 · source PDF URL

9231-2021-on-21-q08

Oct/Nov 2021 · Paper 21 · Question 8 · 14 marks

8(a) coshA= 1 eA+e −A B1 Writes in exponential form 2 ( ) ( ) 2cosh2 A= 1 e2A +2+e −2A = 1 e2A +e −2A +1=cosh2A+1 M1 A1 Expands, AG. 2 2 3 8(b)(i) dx dy B1 =4sinh2t+3 =3sinh2t−4 dt dt ( 4sinh2t+3 )2 +( 3sinh2t−4 )2 =25(sinh22t+1)=25cosh22t M1 A1 Expands and applies cosh2 A=sinh2 A+1 2π 1 2 ( 2cosh2t+3t )( 5cosh2t ) dt  =π 1 2 20cosh22t+30tcosh2tdt  A1 Correct formula for surface area, AG.   −1  −1  2 2 4 8(b)(ii) 20 1 2 cosh22tdt =10 1 2 cosh4t+1dt M1 Applies cosh2 A= 1 2 ( cosh2A+1 ) . −1 −1 2 2 101sinh4t+t 1 2 =10 ( 1sinh2+1 ) M1 A1 Integrates.   4 −1 2 2 30 1 2 tcosh2tdt=30   1 [ tsinh2t ]1 2 −1 1 2 sinh2tdt   M1 A1 Integrates by parts. −1 2 2 −1 2 2 −1 2  15  tsinh2t−1cosh2t  1 2 =0 A1 Accept  1 2 tcosh2tdt =0 since tcosh2t is odd. 2 −1 2 −1 2 ( ( ) ) 10π 1 e2 −e −2 +1 A1 64.82457... 4 © UCLES 2021 Page 12 of 13 8(b)(ii) Alternative method for question 8(b)(ii) 10π 1 2 ( 2cosh2t+3t ) cosh2tdt = M1 A1 I u nt = eg 2 r c a o te s s h b 2 y t + pa 3 r t t , s . u '=4sinh2t+3 −1 2 10πcosh2tsinh2t+(3tsinh2t) 1 2 −5π 1 2 sinh2t ( 4sinh2t+3 ) dt v'=cosh2t, v= 1 2 sinh2t  2  −1 −1 2 2 10π  1sinh4t  1 2 −20π 1 2 sinh22tdt−  15π 1 2 sinh2tdt   = M1 A1 Applies 2sinh22t =cosh4t−1 2 −1 −1  −1  2 2 2 10π  1sinh4t  1 2 −10π 1 2 cosh4t−1dt−  15π 1 2 sinh2tdt   2 −1 −1  −1  2 2 2 ( ) 1 M1 A1 Integrates π5sinh4t−10sinh4t+10t− 15cosh2t 2 2 4 2  −1 2 π5sinh4t+10t− ( 15cosh2t )  1 2 =π ( 5sinh2+10 ) =10π ( 1 ( e2 −e −2 ) +1 ) A1 64.82457... 2 2  −1 4 2 7 © UCLES 2021 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2021-on-22-q03

Oct/Nov 2021 · Paper 22 · Question 3 · 8 marks

3(a)  1 1−x2dx< 1 + 1 ( 1− 1 ) ++ 1 ( 1−(n−1)2 ) M1 A1 Forms the sum of the areas of the rectangles. Three 0 n n n2 n n2 terms, including last for A1. Allow use of sigma. 1 n−1 (n−1)n(2n−1) M1 n−1 1− r2 =1− Applies r2 =1(n−1)n(2n−1). n3 6n3 6 r=1 r=1 6n2 −(n−1)(2n−1) 4n2 +3n−1 A1 AG. [Can get this A1 without previous A1.] = 6n2 6n2 4 © UCLES 2021 Page 7 of 14 3(b)  1 1−x2dx> 1 ( 1− 1 ) + 1 ( 1−22 ) ++ 1 1− ( n−1 )2 M1 A1 Forms the sum of the areas of appropriate rectangles. 0 n n2 n n2 n n2 Three terms, including last for A1. Allow  n 1 ( 1− r2 ) . n n2 r=1 n−1 1 n−1 n−1 (n−1)n(2n−1) M1 n−1 − r2 = − Applies r2 =1(n−1)n(2n−1). n n3 n 6n3 6 r=1 r=1 6n(n−1)−(n−1)(2n−1) 4n2 −3n−1 A1 4n2 +3n−1 1 = SC: − scores 2/4. 6n2 6n2 6n2 n 4 Question Answer Marks Guidance

Official mark scheme pages: 7, 8 · source PDF URL

9231-2021-on-22-q05

Oct/Nov 2021 · Paper 22 · Question 5 · 10 marks

5(a) x=3−2t −2 +2t2 y =4+ 3t −2 −3t2 B1 Differentiates x and y with respect to t. 2 2 4t4 +12t2 + 4 − 12 +1+ 23 + 9t4 + 12 + 9 −12t2 M1 Expands x2 + y2. Accept 25+ 25 ( t −2 −t2 )2 . 4 t4 t2 2 4 t2 4t4 25t4 + 25t −4 + 25 = 25 ( t2 +t −2 )2 M1 A1 Factorises x2 + y2. AG. 4 4 2 4 5 2 t2 +t −2 dt = 51t3 −t −1 2 = 85 M1 A1 Applies correct formula for arc length. 2 1 23  1 12 6 © UCLES 2021 Page 9 of 14 5(b) dy y 4+ 3t −2 B1 dy = = 2 Finds . dx x 3−2t −2 dx ( )( ) ( )( ) d d t     4 3 + −2 3 2 t t − − 2 2     = 3−2t −2 −3 ( t 3 − − 3 2 − t −2 4 )2 + 3 2 t −2 4t −3 B1 Differentiates d d y x with respect to t. d2y d 4+ 3t −2  dt −25t −3 M1 A1 Applies chain rule. =  2  × = =−25 when t =1. dx2 dt   3−2t −2   dx ( 3−2t −2 )3 4 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2021-on-23-q04

Oct/Nov 2021 · Paper 23 · Question 4 · 10 marks

4(a) N lnr ln2 N lnr M1 A1 Compares with sum of the areas of the rectangles.  = + r=1 r2 4 r=3 r2 M1 for writing out sum, A1 for considering N lnr . r2 r=3 ln2 Nlnx M1 Compares with integral. < + dx 4 2 x2 Nlnx  lnx+1 N M1 A1 Finds integral.  dx= − 2 x2   x   2 N lnr ln2  lnN +1 ln2+1 2+3ln2 1+lnN M1 A1 Inserts limits, AG. A1 requires second M1.  < +  − +  = − r2 4  N 2  4 N r=1 7 4(b) N lnr N−1lnr lnN Nlnx lnN M1 A1 Compares with integral. Lower limit of 1 scores M0.  = + > dx+ r=1 r2 r=2 r2 N2 2 x2 N2 Accept  N lnr > N+1lnx . r2 2 x2 r=1 ln2+1 lnN +1 lnN A1 ln2+1 ln(N +1)+1 = − + Accept − . 2 N N2 2 N +1 3 © UCLES 2021 Page 8 of 13

Official mark scheme pages: 8 · source PDF URL

9231-2021-on-23-q08

Oct/Nov 2021 · Paper 23 · Question 8 · 14 marks

8(a) coshA= 1 eA+e −A B1 Writes in exponential form 2 ( ) ( ) 2cosh2 A= 1 e2A +2+e −2A = 1 e2A +e −2A +1=cosh2A+1 M1 A1 Expands, AG. 2 2 3 8(b)(i) dx dy B1 =4sinh2t+3 =3sinh2t−4 dt dt ( 4sinh2t+3 )2 +( 3sinh2t−4 )2 =25(sinh22t+1)=25cosh22t M1 A1 Expands and applies cosh2 A=sinh2 A+1 2π 1 2 ( 2cosh2t+3t )( 5cosh2t ) dt  =π 1 2 20cosh22t+30tcosh2tdt  A1 Correct formula for surface area, AG.   −1  −1  2 2 4 8(b)(ii) 20 1 2 cosh22tdt =10 1 2 cosh4t+1dt M1 Applies cosh2 A= 1 2 ( cosh2A+1 ) . −1 −1 2 2 101sinh4t+t 1 2 =10 ( 1sinh2+1 ) M1 A1 Integrates.   4 −1 2 2 30 1 2 tcosh2tdt=30   1 [ tsinh2t ]1 2 −1 1 2 sinh2tdt   M1 A1 Integrates by parts. −1 2 2 −1 2 2 −1 2  15  tsinh2t−1cosh2t  1 2 =0 A1 Accept  1 2 tcosh2tdt =0 since tcosh2t is odd. 2 −1 2 −1 2 ( ( ) ) 10π 1 e2 −e −2 +1 A1 64.82457... 4 © UCLES 2021 Page 12 of 13 8(b)(ii) Alternative method for question 8(b)(ii) 10π 1 2 ( 2cosh2t+3t ) cosh2tdt = M1 A1 I u nt = eg 2 r c a o te s s h b 2 y t + pa 3 r t t , s . u '=4sinh2t+3 −1 2 10πcosh2tsinh2t+(3tsinh2t) 1 2 −5π 1 2 sinh2t ( 4sinh2t+3 ) dt v'=cosh2t, v= 1 2 sinh2t  2  −1 −1 2 2 10π  1sinh4t  1 2 −20π 1 2 sinh22tdt−  15π 1 2 sinh2tdt   = M1 A1 Applies 2sinh22t =cosh4t−1 2 −1 −1  −1  2 2 2 10π  1sinh4t  1 2 −10π 1 2 cosh4t−1dt−  15π 1 2 sinh2tdt   2 −1 −1  −1  2 2 2 ( ) 1 M1 A1 Integrates π5sinh4t−10sinh4t+10t− 15cosh2t 2 2 4 2  −1 2 π5sinh4t+10t− ( 15cosh2t )  1 2 =π ( 5sinh2+10 ) =10π ( 1 ( e2 −e −2 ) +1 ) A1 64.82457... 2 2  −1 4 2 7 © UCLES 2021 Page 13 of 13

Official mark scheme pages: 12, 13 · source PDF URL

9231-2022-mj-21-q01

May/June 2022 · Paper 21 · Question 1 · 5 marks

1 s    e 3 2  9 e 3 2 d 5  e 3 4 d M1 A1 Forms s correctly. 16 4 0 0 s 5e 3 4    5  e 3 4 1  M1 A1 Integrates. 3  3 0  4ln1 3s A1 3 5 5 Question Answer Marks Guidance

Official mark scheme pages: 6 · source PDF URL

9231-2022-mj-21-q04

May/June 2022 · Paper 21 · Question 4 · 10 marks

4(a)  1 2xdx 1 2N 1  1 2N 2  1 2 N N 1  1 2 N N M1 A1 Forms the sum of the areas of the rectangles. N N N N 0  1  N  2N 1 n  2N 1 M1 A1 Applies  N rn  r  rN 1  , AG. N n1 N  2N 1 1  n1 r1 4 4(b)  1 2xdx 1  1 2N 1  1 2 N N 2  1 2 N N 1 M1 A1 Forms the sum of the areas of appropriate rectangles. N N N N 0  1  N1 2N 1 n  1 M1 A1 Applies  N1 rn  rN 1 . N n0 N  2N 1 1  n0 r1 4 4(c) 1 M1 c N  2 2 N 1 N 1   N  2 1 N 1 1   N 1 104 leading to N 104 Simplifies U n L n to n . Least value of N is 10001 A1 2 © UCLES 2022 Page 8 of 13

Official mark scheme pages: 8 · source PDF URL

9231-2022-mj-22-q01

May/June 2022 · Paper 22 · Question 1 · 5 marks

1 s    e 3 2  9 e 3 2 d 5  e 3 4 d M1 A1 Forms s correctly. 16 4 0 0 s 5e 3 4    5  e 3 4 1  M1 A1 Integrates. 3  3 0  4ln1 3s A1 3 5 5 Question Answer Marks Guidance

Official mark scheme pages: 6 · source PDF URL

9231-2022-mj-22-q04

May/June 2022 · Paper 22 · Question 4 · 10 marks

4(a)  1 2xdx 1 2N 1  1 2N 2  1 2 N N 1  1 2 N N M1 A1 Forms the sum of the areas of the rectangles. N N N N 0  1  N  2N 1 n  2N 1 M1 A1 Applies  N rn  r  rN 1  , AG. N n1 N  2N 1 1  n1 r1 4 4(b)  1 2xdx 1  1 2N 1  1 2 N N 2  1 2 N N 1 M1 A1 Forms the sum of the areas of appropriate rectangles. N N N N 0  1  N1 2N 1 n  1 M1 A1 Applies  N1 rn  rN 1 . N n0 N  2N 1 1  n0 r1 4 4(c) 1 M1 c N  2 2 N 1 N 1   N  2 1 N 1 1   N 1 104 leading to N 104 Simplifies U n L n to n . Least value of N is 10001 A1 2 © UCLES 2022 Page 8 of 13

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9231-2022-mj-23-q02

May/June 2022 · Paper 23 · Question 2 · 8 marks

2(a) d B1 Finds first derivative. lncosxtanx dx d B1 Finds second derivative. tanxsec2x dx y''(0) 1 M1 A1 Evaluates second derivative at zero. Accept 1x2.  2 2! 2 4 © UCLES 2022 Page 6 of 15 2(b) 1π M1 Forms correct integral 4 1tan2 x dx 0 1π A1 4 secx dx 0 ln  tan1x 1π 1 4 π M1 Integrates (answer given in List MF19).  2 4  Accept ln(secxtanx). 0 lntan3π0.881 A1 Accept ln  21  . 8 4 Question Answer Marks Guidance

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9231-2022-mj-23-q06

May/June 2022 · Paper 23 · Question 6 · 10 marks

6(a)  1 ln1xdx1ln1 11ln1 21ln1 n M1 A1 Forms the sum of the areas of the rectangles. n n n n n n 0  1nln1 lnn1lnn2ln2n 1ln 2n! lnn M1 A1 Applies laws of logarithms, AG. n n n n! 4 6(b)  1 ln1xdx1ln1 11ln1 21ln1 n1 M1 A1 Forms the sum of the areas of appropriate n n n n n n rectangles. 0  1(n1)ln1 lnn1lnn2ln2n1 M1 Applies laws of logarithms. n n 1ln 2n1! n1lnn A1 Equivalent answers: 1ln(2n)!lnn1ln2; n n! n n n! n 1ln(2n1)!lnn. n (n1)! 4 6(c) ln2 M1 c U L  Simplifies their U L to . n n n n n n ln2 A1 AG 0 as n n 2 © UCLES 2022 Page 11 of 15

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9231-2022-mj-23-q08

May/June 2022 · Paper 23 · Question 8 · 16 marks

8(a) cosn1 M1 A1 Allow with "C"missing. sincosnd (C) n1 2 8(b) 1π M1 A1 Integrates by parts using part (a). 1π  sinm1cosn12 m1 1π 2 sinm1sincosnd   2 cosn2sinm2d Allow with limits missing. 0  n1  n1 0 0  m1  1 2 π cosnsinm2(1sin2)d M1 Applies cos21sin2. n1 0 m1 m1 M1 Combines with LHS. I  I  I m,n n1 m2,n n1 m,n nm m1 m1 A1 AG. I  I leading to I  I . n1 m,n n1 m2,n m,n mn m2,n © UCLES 2022 Page 13 of 15 8(b) Alternative method for question 8(b)  1 2 π sinm2sin2cosnd 1 2 π sinm2  1cos2  cosnd M1 Applies sin21cos2 0 0 1π 1π 2 sinm2cosnd2 sinm2cosn2d 0 0 1π I 2 sinm2coscosn1d m2,n 0  1  1 2 π n1 1π M1 A1 ucosn1u'(n1)cosnsin I  sinm1cosn1  2 cosnsinmd m2,n  m1   0 m1 0 v'sinm2cosv 1 sinm1 m1 n1  n1 M1 A1 Combines with LHS. I m,n I m2,n  m1 I m,n    1 m1  I m,n I m2,n AG 5 8(c) zz1 2cos B1  zz15   z5 z5 5  z3 z3 10  zz1 M1 A1 Expands and groups. A0 if no grouping shown. 25cos52cos552cos3102cos M1 Substitutes zn zn 2cosn. cos5 1 cos5 5 cos3 5cos A1 16 16 8 5 © UCLES 2022 Page 14 of 15 8(d) I  1I M1 Applies reduction formula in part (b). 2,5 7 0,5 1π 1π M1 Applies part (c). 1I  12 cos5d 12 1 cos5 5 cos35cosd 7 0,5 7 7 16 16 8 0 0 1π A1  11 sin5 5 sin3 5sin2 780 48 8  0  8 A1 105 4 © UCLES 2022 Page 15 of 15

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9231-2022-on-21-q03

Oct/Nov 2022 · Paper 21 · Question 3 · 6 marks

3 x=et −t2 B1 y=2te 1 2 t B1 ( et −t2)2 +4t2et = e2t +2t2et +t4 =et +t2 M1 A1 Finds x2 + y2. 2 2 M1 A1  et +t2dt =et +1t3 =e2 + 5 0  3  0 3 6 Question Answer Marks Guidance

Official mark scheme pages: 7 · source PDF URL

9231-2022-on-22-q03

Oct/Nov 2022 · Paper 22 · Question 3 · 8 marks

3(a) M1 1+ ( ex −1e−x)2 = 1+e2x −1+ 1 e−2x = e2x +1+ 1 e−2x dy 2 4 2 16 2 16 Finds 1+   dx A1 ( ex +1e−x)2 =ex +1e−x 4 4 2π 1( ex + 1e−x)( ex + 1e−x) dx M1 Correct formula for surface area with limits. 4 4 0 1 A1 Expanded integrand. 2π e2x + 1 + 1 e−2x dx 2 16 0 1 M1 Integrates and substitutes limits. Integrand must be 21e2x + 1x− 1 e−2x 2 2 32  0 of the form ae2x +be−2x +c, with a and b nonzero. 2π ( 1e2 − 1 e−2 + 1 ) =π ( e2 − 1 e−2 + 1 ) A1 2 32 32 16 16 6 3(b)  x2  1 (−x)2  M1 Uses expansion of ex or finds first and second 1+x+ + + 1−x+ +  2! 4 2! derivatives ex −1e−x and ex +1e−x.     4 4 5+3x+5x2 A1 4 4 8 2 © UCLES 2022 Page 8 of 15

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9231-2022-on-22-q06

Oct/Nov 2022 · Paper 22 · Question 6 · 10 marks

6 1 1 1 n 1 M1A1 Forms sum of the areas of the rectangles. At least Sum= + + + = three terms, including first and last for A1.    ( n2 +2n ) r2 +2r r=2 SC Areas of these three rectangles written down but sum not clearly formed scores B1. n M1 Compares with integral.  1     ( x2 +2x ) dx 1 x2+2x=(x+1)2 −1 B1 Completes the square. n M1 A1 Uses formula. Correct limits for A1.  1 n  dx=  cosh−1(x+1)   (x+1)2 −1 1 1  ln ( x+1+ x2 +2x ) n M1 A1 Uses logarithmic form of cosh−1.   1 ln ( n+1+ n2 +2n ) −ln ( 2+ 3 ) M1 Inserts limits.  n 1 ln ( n+1+ n2 +2n ) −ln ( 2+ 3 ) +1 3 A1 Adds term from r=1. AG r2 +2r 3 r=1 10 © UCLES 2022 Page 11 of 15

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9231-2022-on-23-q03

Oct/Nov 2022 · Paper 23 · Question 3 · 6 marks

3 x=et −t2 B1 y=2te 1 2 t B1 ( et −t2)2 +4t2et = e2t +2t2et +t4 =et +t2 M1 A1 Finds x2 + y2. 2 2 M1 A1  et +t2dt =et +1t3 =e2 + 5 0  3  0 3 6 Question Answer Marks Guidance

Official mark scheme pages: 7 · source PDF URL

9231-2023-mj-21-q04

May/June 2023 · Paper 21 · Question 4 · 9 marks

4(a) d  x  1x5n  5nx5 1x5n1   1x5n M1 A1 Uses the product rule to differentiate. dx 5n  1x5 1  1x5n1   1x5n M1* Uses x5 x5 11. 1 DM1 Integrates both sides using the limits given. Requires  x  1x5n 5nI 5nI I   n n1 n previous method mark. 0 2n (5n1)I 5nI (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG. n n1 n n1 Alternative method for question 4(a) I  1 1x5n dx  x  1x5n 1 5n 1 x5 1x5n1 dx M1 A1 Integrates by parts. n 0   0 0 I   x  1x5n 1 5n 1 1x5n dx5n 1 1x5n1 dx M1* Uses x5 x5 11. n   0 0 0 1 DM1 Forms recursive formula. Requires previous method I   x  1x5n 5nI 5nI n   n n1 mark. 0 (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG. n n1 5 © UCLES 2023 Page 9 of 15 4(b) 1 B1 I x1x6  7 or I 1 1  6  0 6 0 11I 22 10I I  47 M1 A1 Applies reduction formula. 2 1 2 33 16I 2315I I 323 A1 3 2 3 176 4 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2023-mj-21-q07

May/June 2023 · Paper 21 · Question 7 · 11 marks

7(a)  x  1 M1 A1 Applies given substitution.  dx 1 du  x2 1 2 u x21C A1 Allow with “C ” missing. Answer must be in terms of x. 3 7(b)   B1 cosh1r ln r r2 1 cosh12cosh13cosh1N M1 Forms sum of the areas of the rectangles. N M1 Compares with integral with correct limits.  cosh1xdx 1 N A1 Evaluates integral. N N  x  cosh1xdxxcosh1x  dx   1 1  x2 1 1 N     A1 AG. ln r r2 1 Nln N  N2 1  N2 1 r2 5 © UCLES 2023 Page 13 of 15 7(c) (cosh11)cosh12cosh1(N1)  N cosh1xdx(or N cosh1xdx) M1 A1 Compares with integral with correct limits. 1 2  N ln  r r2 1  (N 1)ln  N  N2 1   N2 1 A1 Adds ln  N  N2 1  to both sides. r2 or  N ln  r r2 1  (N 1)ln  N  N2 1   N2 12ln  2 3   3 r2 Alternative method for question 7(c) (cosh11)cosh12cosh1(N1)cosh1N M1 A1 Compares with integral with correct limits.  N1 cosh1xdx  xln  x x2 1   x2 1  N1 (or  N1 cosh1xdx) 1   1 2 N     A1 ln r r2 1 N 1ln N 1 N2 2N  N2 2N r2 or N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 Second alternative method for question 7(c) cosh1(11)cosh1(21)cosh1(N11) M1 A1 Compares with integral with correct limits.  N cosh1x1dx x1ln  x1 x2 2x   x2 2x  N 1   1 N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 A1 3 © UCLES 2023 Page 14 of 15

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9231-2023-mj-22-q04

May/June 2023 · Paper 22 · Question 4 · 9 marks

4(a) d  x  1x5n  5nx5 1x5n1   1x5n M1 A1 Uses the product rule to differentiate. dx 5n  1x5 1  1x5n1   1x5n M1* Uses x5 x5 11. 1 DM1 Integrates both sides using the limits given. Requires  x  1x5n 5nI 5nI I   n n1 n previous method mark. 0 2n (5n1)I 5nI (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG. n n1 n n1 Alternative method for question 4(a) I  1 1x5n dx  x  1x5n 1 5n 1 x5 1x5n1 dx M1 A1 Integrates by parts. n 0   0 0 I   x  1x5n 1 5n 1 1x5n dx5n 1 1x5n1 dx M1* Uses x5 x5 11. n   0 0 0 1 DM1 Forms recursive formula. Requires previous method I   x  1x5n 5nI 5nI n   n n1 mark. 0 (5n1)I 2n 5nI A1 Substitutes limits and rearranges. AG. n n1 5 © UCLES 2023 Page 9 of 15 4(b) 1 B1 I x1x6  7 or I 1 1  6  0 6 0 11I 22 10I I  47 M1 A1 Applies reduction formula. 2 1 2 33 16I 2315I I 323 A1 3 2 3 176 4 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2023-mj-22-q07

May/June 2023 · Paper 22 · Question 7 · 11 marks

7(a)  x  1 M1 A1 Applies given substitution.  dx 1 du  x2 1 2 u x21C A1 Allow with “C ” missing. Answer must be in terms of x. 3 7(b)   B1 cosh1r ln r r2 1 cosh12cosh13cosh1N M1 Forms sum of the areas of the rectangles. N M1 Compares with integral with correct limits.  cosh1xdx 1 N A1 Evaluates integral. N N  x  cosh1xdxxcosh1x  dx   1 1  x2 1 1 N     A1 AG. ln r r2 1 Nln N  N2 1  N2 1 r2 5 © UCLES 2023 Page 13 of 15 7(c) (cosh11)cosh12cosh1(N1)  N cosh1xdx(or N cosh1xdx) M1 A1 Compares with integral with correct limits. 1 2  N ln  r r2 1  (N 1)ln  N  N2 1   N2 1 A1 Adds ln  N  N2 1  to both sides. r2 or  N ln  r r2 1  (N 1)ln  N  N2 1   N2 12ln  2 3   3 r2 Alternative method for question 7(c) (cosh11)cosh12cosh1(N1)cosh1N M1 A1 Compares with integral with correct limits.  N1 cosh1xdx  xln  x x2 1   x2 1  N1 (or  N1 cosh1xdx) 1   1 2 N     A1 ln r r2 1 N 1ln N 1 N2 2N  N2 2N r2 or N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 Second alternative method for question 7(c) cosh1(11)cosh1(21)cosh1(N11) M1 A1 Compares with integral with correct limits.  N cosh1x1dx x1ln  x1 x2 2x   x2 2x  N 1   1 N 1ln  N 1 N2 2N   N2 2N 2ln  2 3   3 A1 3 © UCLES 2023 Page 14 of 15

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9231-2023-mj-23-q06

May/June 2023 · Paper 23 · Question 6 · 11 marks

6(a)  1 1x2 dx 1  1 02  1  1 1 2  1  1 n12 M1 A1 Forms the sum of the areas of the rectangles. Must have first two terms and last term for A1. Accept 0 n n n n n n 1  n2 1 for the last term. Accept written in n n summation form with correct limits. 1  1  n1 nr2  1  1  n1  n2 2nrr2 M1 A1 Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as n n3 n n3 r1 r1 1 n  2 n1 1 n1  r2 1 r r2  n3  n2 r1 n3 r1  r1 1 n2n1 n2n1 n(n1)(2n1) 1 2n2 3n1 2n2 3n1 A1 AG. Must have gained previous accuracy mark.       n n3 n3 6n3 n 6n2 6n2 5 © UCLES 2023 Page 11 of 15 6(b)  1 1x2 dx 1 112  1 122  1 1n12 M1 A1 Forms the sum of the areas of appropriate n n n n n n rectangles. Must have first two and last terms for 0 A1. Accept 11 n2 for the last term. n n Accept written in summation form with correct limits. 1  n1  n2 2nrr2  n2n1  n2n1  n(n1)(2n1) M1 S M e F tt 1 in 9 g c u a p n c b o e r a re p c p t l i s e e d r . i e R s e s c o o g th n a i t s i f n o g rm su u m la e a s f rom n3 n3 n3 6n3 r1 1 n1 1 n 1  2 n 1 n  r2  r2  without wrong working   1 n2  r1 r n3  r1 r2   s n c 3 or r e  s 1 M1 A n 1 3 M r1 1. n 2n2 3n1 A1 6n2 4 6(c) 1 M1 c U L  Simplifies U L to . n n n n n n l A1 AG. 0 as n n 2 © UCLES 2023 Page 12 of 15

Official mark scheme pages: 11, 12 · source PDF URL

9231-2023-mj-23-q07

May/June 2023 · Paper 23 · Question 7 · 11 marks

7(a) I  4 3 1x21 2dxsinh1x 4 3 ln3 M1 A1 Recognises integral. 1   0 0 2 7(b) d  x  1x21 2 n nx2 1x21 2 n1   1x21 2 n M1 A1 Uses the product rule to differentiate. dx  n  1x2 1  1x21 2 n1   1x21 2 n M1* Uses x2 1x2 1. 4 DM1 Integrates both sides using the limits given. Requires  x  1x21 2 n3 nI nI I previous method mark.   n n2 n   0 45n n1I nI n1I nI  45n A1 Substitutes limits and rearranges. AG. 3 3 n n2 n n2 3 3 5 7(c) 2 M1 Forms correct integral with correct limits. s3 14x2 dx 0 1  4 3 1u2 du 1 I A1 AG. 2 2 1 0 45 M1 Applies reduction formula with n1. 2I I  1 1 3 3 1 5 A1 s ln3 4 9 4 © UCLES 2023 Page 13 of 15

Official mark scheme pages: 13 · source PDF URL

9231-2023-on-21-q05

Oct/Nov 2023 · Paper 21 · Question 5 · 10 marks

5(a) 1 1 B1 Differentiates x and y with respect to t. − x=t2 −t 2, y=2  1 1  2  1 1 2 M1 A1 Factorises x2 + y2. − − x2 + y2 =t2 −t 2 +4=t+2+t−1 =t2 +t 2         3 M1 A1 Applies correct formula for arc length. 3 1 − 1 2 3 1  0 t2 +t 2dt =  3 t2 +2t2  =4 3 M1 for their  3 x2 + y2 dt. 0 0 Answer must be simplified to 4 3 for A1. 5 © UCLES 2023 Page 8 of 14 5(b) dy y 2 B1 Finds first derivative. = = dx x 1 − 1 t2 −t 2 1 − 1 1 − 3 B1 Differentiates dy with respect to t. −2 t 2 + t 2 d   2     2 2   dx = dt 1 − 1   1 1  2 t2 −t 2  − t2 −t 2     1 − 1 1 − 3 M1 A1 Applies chain rule. OE. Does not have to be simplified −2 t 2 + t 2 for A1. d2y d   2   dt   2 2   t+1 =  = =− dx2 dt 1 − 1  dx  1 1  3 (t−1)3 t2 −t 2  − t2 −t 2     0t1 A1 Accept −1t1. CWO. 5 Question Answer Marks Guidance

Official mark scheme pages: 8, 9 · source PDF URL

9231-2023-on-22-q07

Oct/Nov 2023 · Paper 22 · Question 7 · 12 marks

7(a) cosh2A= 1 ( e2A+e−2A) sinhA= 1 ( eA−e−A) B1 2 2 2sinh2 A= 1 ( eA −e−A)2 = 1 ( e2A −2+e−2A) =cosh2A−1 M1 A1 Expands, AG. A0 for mixing 2 2 variables e.g. sinhA=1 ( ex −e−x) . 2 3 © UCLES 2023 Page 12 of 15 7(b) 2 M1 A1 Correct formula with correct limits. 3 dy 2 2 S = 2y 1+ dx=23x2 1+4x2dx Correct limits for M1. (Limits may     dx 0 be recovered.) 0 sinh−14 M1 Applies given substitution to their S =π 31sinh2u 1+sinh2ucoshudu 4 expression with correct limits. 0 (Limits may be recovered.) sinh−14 A1 Must be simplified. =1π 3sinh2ucosh2udu 4 0 sinh−14 M1 Applies sinh2u=2sinhucoshu. = 1 π 3sinh22udu 16 May use double angle formulae for 0 cosh and sinh instead. sinh−14 M1 Applies sinh2 A=1(cosh2A−1). S = 1 π 3cosh4u−1du 2 32 0 must have the form asinh2ucosh2udu. sinh−14 A1 = 1 π1sinh4u−u 3   32 4 0 sinh−14 =ln3 B1 3 S = 1 π ( 1 ( eln81−e−ln81) −ln3 ) = 1 π ( 1( 81− 1 ) −ln3 ) = 1 π (820−ln3 ) A1 AG. 32 8 32 8 81 32 81 9 © UCLES 2023 Page 13 of 15

Official mark scheme pages: 12, 13 · source PDF URL

9231-2023-on-23-q05

Oct/Nov 2023 · Paper 23 · Question 5 · 10 marks

5(a) 1 1 B1 Differentiates x and y with respect to t. − x=t2 −t 2, y=2  1 1  2  1 1 2 M1 A1 Factorises x2 + y2. − − x2 + y2 =t2 −t 2 +4=t+2+t−1 =t2 +t 2         3 M1 A1 Applies correct formula for arc length. 3 1 − 1 2 3 1  0 t2 +t 2dt =  3 t2 +2t2  =4 3 M1 for their  3 x2 + y2 dt. 0 0 Answer must be simplified to 4 3 for A1. 5 © UCLES 2023 Page 8 of 14 5(b) dy y 2 B1 Finds first derivative. = = dx x 1 − 1 t2 −t 2 1 − 1 1 − 3 B1 Differentiates dy with respect to t. −2 t 2 + t 2 d   2     2 2   dx = dt 1 − 1   1 1  2 t2 −t 2  − t2 −t 2     1 − 1 1 − 3 M1 A1 Applies chain rule. OE. Does not have to be simplified −2 t 2 + t 2 for A1. d2y d   2   dt   2 2   t+1 =  = =− dx2 dt 1 − 1  dx  1 1  3 (t−1)3 t2 −t 2  − t2 −t 2     0t1 A1 Accept −1t1. CWO. 5 Question Answer Marks Guidance

Official mark scheme pages: 8, 9 · source PDF URL

9231-2024-mj-21-q04

May/June 2024 · Paper 21 · Question 4 · 8 marks

4(a) ln3 B1 Separates into correct structure. I  sechn2 xsech2 xdx n 0 ln3 ln3 ln3 *M1 Uses integration by parts correctly. I  sechn2xsech2 xdxsechn2xtanhx (n2) sechn2xtanh2xdx n   0 0 0 3n24(n2)I I  DM1 A1 Uses 1sech2 xtanh2 x. 5 5 n2 n 1n2I 3n24(n2)I (n1)I 3n24(n2)I A1 AG n 5 5 n2 n 5 5 n2 5 © Cambridge University Press & Assessment 2024 Page 8 of 14 4(b) I  4 B1 2 5 3I  324 2I I  236 0.629 M1 A1 Applies reduction formula with n4. 4 5 5 2 4 375 3 Question Answer Marks Guidance

Official mark scheme pages: 8, 9 · source PDF URL

9231-2024-mj-21-q05

May/June 2024 · Paper 21 · Question 5 · 11 marks

5(a)   1 2xx2dx 1  21  12  1  22  22  1  2n  n2 M1 A1 Forms the sum of the areas of the rectangles,   0   n n n n n n n n n may be in summation form. M1 is for correct number of rectangles. n n M1 A1 Applies formulae from MF19. Summations must  2 r 1 r2  1 n(n1) 1 n(n1)(2n1) have (correct) limits for A1. Substituted in n2 n3 n2 6n3 r1 r1 correctly once for M1. 1 1 11 12 1  1 12 1   2 1  1 A1 AG. This A1 requires the previous A1. n 6 n n n 3 6n 3 2n 6n2 5 5(b)  1 2xx2dx1  2112 1  2222 1  2n1n12 M1 A1 Forms the sum of the areas of appropriate n n n n n n n n n rectangles. M1 is for seeing correct height of last 0 rectangle. n1 n1 M1 Applies formulae from MF19, substitutes correct 2 r 1 r2  1 (n1)(n) 1 (n1)(n)(2n1) limit. Substituted in correctly once for M1. n2 n3 n2 6n3 r1 r1 112 1   2 1  1 A1  112 1  1 or fully expanded or fully n 3 6n 3 2n 6n2 n 3 6n n factorised. 4 © Cambridge University Press & Assessment 2024 Page 9 of 14 5(c) U L   1 12 1    112 1  ( 1) M1 Expresses their U L in terms of 1. n n n 3 6n n 3 6n n n n n For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.) 2 2 A1 AG, CWO. 1 0 as n or  0 n 3 3 Their L n must be correct. 2 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2024-mj-22-q04

May/June 2024 · Paper 22 · Question 4 · 8 marks

4(a) ln3 B1 Separates into correct structure. I  sechn2 xsech2 xdx n 0 ln3 ln3 ln3 *M1 Uses integration by parts correctly. I  sechn2xsech2 xdxsechn2xtanhx (n2) sechn2xtanh2xdx n   0 0 0 3n24(n2)I I  DM1 A1 Uses 1sech2 xtanh2 x. 5 5 n2 n 1n2I 3n24(n2)I (n1)I 3n24(n2)I A1 AG n 5 5 n2 n 5 5 n2 5 © Cambridge University Press & Assessment 2024 Page 8 of 14 4(b) I  4 B1 2 5 3I  324 2I I  236 0.629 M1 A1 Applies reduction formula with n4. 4 5 5 2 4 375 3 Question Answer Marks Guidance

Official mark scheme pages: 8, 9 · source PDF URL

9231-2024-mj-22-q05

May/June 2024 · Paper 22 · Question 5 · 11 marks

5(a)   1 2xx2dx 1  21  12  1  22  22  1  2n  n2 M1 A1 Forms the sum of the areas of the rectangles,   0   n n n n n n n n n may be in summation form. M1 is for correct number of rectangles. n n M1 A1 Applies formulae from MF19. Summations must  2 r 1 r2  1 n(n1) 1 n(n1)(2n1) have (correct) limits for A1. Substituted in n2 n3 n2 6n3 r1 r1 correctly once for M1. 1 1 11 12 1  1 12 1   2 1  1 A1 AG. This A1 requires the previous A1. n 6 n n n 3 6n 3 2n 6n2 5 5(b)  1 2xx2dx1  2112 1  2222 1  2n1n12 M1 A1 Forms the sum of the areas of appropriate n n n n n n n n n rectangles. M1 is for seeing correct height of last 0 rectangle. n1 n1 M1 Applies formulae from MF19, substitutes correct 2 r 1 r2  1 (n1)(n) 1 (n1)(n)(2n1) limit. Substituted in correctly once for M1. n2 n3 n2 6n3 r1 r1 112 1   2 1  1 A1  112 1  1 or fully expanded or fully n 3 6n 3 2n 6n2 n 3 6n n factorised. 4 © Cambridge University Press & Assessment 2024 Page 9 of 14 5(c) U L   1 12 1    112 1  ( 1) M1 Expresses their U L in terms of 1. n n n 3 6n n 3 6n n n n n For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.) 2 2 A1 AG, CWO. 1 0 as n or  0 n 3 3 Their L n must be correct. 2 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2024-mj-23-q01

May/June 2024 · Paper 23 · Question 1 · 5 marks

1 x2 4x13(x2)2 B1 Completes the square. 7 7 M1 A1 Applies formula. 2 1  x22  dxsin1    x2 7 2  x2 7 2  2 3(x2)2   3  2 sin1   cos1     3    3  2 2 sin1 1 3  sin10 1π M1 A1 Uses limits. Answer must be in radians. 2 3 5 Question Answer Marks Guidance

Official mark scheme pages: 6 · source PDF URL

9231-2024-mj-23-q04

May/June 2024 · Paper 23 · Question 4 · 10 marks

4(a) N1 1 B1 Forms sum of the areas of the rectangles. Need 22 (N 1)2  to see this sum written explicitly, may be in r2 r2 summation form, cannot be embedded in working. N N 1 1 M1 A1 Evaluates integral, correct limits for M1.  x2dxx1     2 2 2 N N N 1 M1 Compares with integral correctly (using >). r2  x2dx1 2 N2 r1 N 3 1 1 A1 AG r2    2 N N2 r1 5 4(b) 32 42 N2 M1* Forms the sum of the areas of appropriate rectangles. Allow addition of one extra rectangle or shift one left or right if limits on the integral are adjusted accordingly. N N 1 1 DM1 Compares with integral, correct limits and  x2dxx1     correct inequality symbol (<). 2 2 2 N N 7 1 A1 r2   4 N r1 3 © Cambridge University Press & Assessment 2024 Page 9 of 15 4(c)  3 B1 Needs to clear (using > or words) is that value is r2  lower bound. 2 r1  7 B1 FT Needs to clear (using < or words) is that value is r2  upper bound. Must have scored final A mark in 4 r1 part (b). FT on fully correct alternative to part 4(b). 2 Question Answer Marks Guidance

Official mark scheme pages: 9, 10 · source PDF URL

9231-2024-on-21-q03

Oct/Nov 2024 · Paper 21 · Question 3 · 8 marks

3(a) A=2π ln1 5 2 ex 1+e2x dx=2π 1 5 2 u 1+u2(1) du M1A1 Correct formula and applies substitution, AG. ln4 4 u 3 3 2 3(b) u=sinhvdu=coshvdv B1 A=2π ln5 1+sinh2vcoshvdv A=2π ln5 cosh2vdv M1A1 Applies substitution andcosh2x=1+sinh2x. ln3 ln3 Need limits for A1. π ln5 (cosh2v+1)dv M1 Applies 2cosh2x=cosh2x+1. ln3 ln5 A1 Correct integration. =π1sinh2v+v   2 ln3 π (312+ln5−40−ln3 ) =π (904+ln5) A1 AG. 50 18 225 3 6 Question Answer Marks Guidance

Official mark scheme pages: 7 · source PDF URL

9231-2024-on-21-q06

Oct/Nov 2024 · Paper 21 · Question 6 · 14 marks

6(a)   1(1)x dx (1)(1) N 1 + (1)(1) N 2 + + (1)(1)N N −1 + (1)(1) N N M1A1 Forms the sum of the areas of the rectangles. M1   0 2   N 2 N 2 N 2 N 2 for correct number of rectangles. = 1  N (1) N n = (1 2 )N 1 (1 2 −1) = (1 2 )(1 2 )N 1 = 1 . M1A1 Applies  N rn = r ( rN −1 ) , AG. N n=1 2 N(1)N 1 −N N −N(1)N 1 2N ( 2N 1 −1 ) n=1 r−1 2 2 4 6(b)  1(1)x dx (1) + (1)(1) N 1 + + (1)(1)N N −2 + (1)(1)N N −1 M1A1 Forms the sum of the areas of appropriate 2 N N 2 N 2 N 2 rectangles. 0 1  N−1 (1) N n = 1 2 −1 = 1 = 2N 1 M1A1 Applies  N−1 rn = rN −1 . N n=0 2 N(1)N 1 −N 2N ( 1−(1) N 1 ) 2N ( 2N 1 −1 ) n=0 r−1 2 2 4 6(c) 1 M1 c 2N − 1 = 1 10−32N 103 Simplifies U N −L N to . ( 1 ) ( 1 ) 2N N 2N 2N −1 2N 2N −1 Least value of N is 500 A1 CWO. 2 6(d) L  1 U  1 ln2 1 M1A1FT Forms inequality. FT on their U . N 2ln2 N 2U 2L N N N 0.693= 1 ln2 1 =0.694 A1A1 CWO. Must have used N =500. 2U 2L N N 4 © Cambridge University Press & Assessment 2024 Page 11 of 14

Official mark scheme pages: 11 · source PDF URL

9231-2024-on-21-q08

Oct/Nov 2024 · Paper 21 · Question 8 · 14 marks

8(a) z+z−1 =2cos B1 For LHS. ( z+z−1)7 = ( z7 +z−7) +7 ( z5 +z−5) +21 ( z3+z−3) +35 ( z+z−1) M1A1 Expands and groups. A0 if not grouped clearly. 27cos7=2cos7+7(2cos5)+21(2cos3)+35(2cos) M1 Substitutes zn +z−n =2cosn. cos7= 1 cos7+ 7 cos5+21cos3+35cos A1 64 64 64 64 5 8(b) 1π 1π 1π M1A1 Applies integration by parts. I =4 cosn−1cosd=cosn−1sin4 +(n−1)4 cosn−2sin2d n   0 0 0 I =cosn−1sin 1 4 π +(n−1) 1 4 π cosn−2 ( 1−cos2 ) d M1 Applies sin2=1−cos2 n   0 0 I =2−n 2 +(n−1)I −(n−1)I nI =2−n 2 +(n−1)I A1 AG. n n−2 n n n−2 4 © Cambridge University Press & Assessment 2024 Page 13 of 14 8(c) 9I =2−9 2 +8I = 1 ( 2−1 2 ) +8I M1A1 Applies reduction formula from (b). 9 7 16 7 1π M1 Applies identity from (a). Allow I =4 1 cos7+ 7 cos5+ 21cos3+35cosd 7 64 64 64 64 missing/incorrect limits. 0 1π A1 I = 1 1sin7+7sin5+ 21sin3+35sin4 7 647 5 3  0 9I = 1 ( 2−1 2 ) +1 ( 1 ( −2−1 2 ) +7 ( −2−1 2 ) +21 ( 2−1 2 ) +35 ( 2−1 2 )) A1 Check their exact answer when I 9 is the subject, 9 16 8 7 5 3 like terms collected. ISW. 0.402237 I = 2867 ( 2−1 2 ) = 2867 2 9 5040 10080 5 © Cambridge University Press & Assessment 2024 Page 14 of 14

Official mark scheme pages: 13, 14 · source PDF URL

9231-2024-on-22-q01

Oct/Nov 2024 · Paper 22 · Question 1 · 4 marks

1 7 M1A1 Applies formula.  1 7  dx=  cosh−1(x−5)   (x−5)2 −12 6 6 =cosh−1(2)−cosh−1(1) M1 Uses limits. =ln ( 2+ 3 ) A1 4 Question Answer Marks Guidance

Official mark scheme pages: 6 · source PDF URL

9231-2024-on-22-q03

Oct/Nov 2024 · Paper 22 · Question 3 · 7 marks

3 x=e2t −t2, y=2et +2(t−1)et =2tet B1B1 Differentiates x and y with respect to t. x2 + y2 = ( e2t −t2)2 +4t2e2t =e4t +2t2e2t +t4 = ( e2t +t2)2 M1A1 Factorises x2 + y2. 1 1 M1A1 Applies correct formula for arc length.  e2t +t2dt =1e2t +1t3 0 2 3  0 = 1e2 −1 A1 2 6 7 © Cambridge University Press & Assessment 2024 Page 7 of 13

Official mark scheme pages: 7 · source PDF URL

9231-2024-on-22-q06

Oct/Nov 2024 · Paper 22 · Question 6 · 13 marks

6(a)   1 e1−xdx (1) e1−0 n + (1) e1−1 n + + (1) e1−n n −1 M1A1 Forms the sum of the areas of the rectangles.   0   n n n M1 for correct number of rectangles. e n−1 e1−e−1  e−1 M1A1 Applies sum of geometric progression, AG. = e− n r =  = n r=0 n 1−e−1 n  n ( 1−e−1 n ) 4 6(b)  1 e1−x dx(1)e1−1 n +(1)e1− n 2 +(1)e1−n n −1 +(1)e1− n n M1A1 Forms the sum of the areas of appropriate n n n n rectangles. M1 for correct number rectangles. 0 e−1 e 1 e−1 M1A1 Applies sum of geometric progression. = − + = n ( 1−e−1 n ) n n n ( e 1 n −1 ) e−1 n (e−1) e1−1 n ( 1−e−1) OE. E.g. or . n ( 1−e−1 n ) n ( 1−e−1 n ) 4 6(c) e−1 M1 c U −L = Simplifies U −L to , where c is a n n n n n n nonzero constant. e−1 A1 AG. CWO. →0 as n→ n 2 © Cambridge University Press & Assessment 2024 Page 10 of 13 6(d) z ( 1−e−1 z ) =z ( 1− ( 1−1+ 1 − 1 + )) =1− 1 + 1 + M1A1 Substitutes power series for e−1 z. A1 for z 2z2 6z3 2z 6z2 enough terms. lim(U )=e−1 B1 n n→ 3 Question Answer Marks Guidance

Official mark scheme pages: 10, 11 · source PDF URL

9231-2024-on-23-q03

Oct/Nov 2024 · Paper 23 · Question 3 · 8 marks

3(a) A=2π ln1 5 2 ex 1+e2x dx=2π 1 5 2 u 1+u2(1) du M1A1 Correct formula and applies substitution, AG. ln4 4 u 3 3 2 3(b) u=sinhvdu=coshvdv B1 A=2π ln5 1+sinh2vcoshvdv A=2π ln5 cosh2vdv M1A1 Applies substitution andcosh2x=1+sinh2x. ln3 ln3 Need limits for A1. π ln5 (cosh2v+1)dv M1 Applies 2cosh2x=cosh2x+1. ln3 ln5 A1 Correct integration. =π1sinh2v+v   2 ln3 π (312+ln5−40−ln3 ) =π (904+ln5) A1 AG. 50 18 225 3 6 Question Answer Marks Guidance

Official mark scheme pages: 7 · source PDF URL

9231-2024-on-23-q06

Oct/Nov 2024 · Paper 23 · Question 6 · 14 marks

6(a)   1(1)x dx (1)(1) N 1 + (1)(1) N 2 + + (1)(1)N N −1 + (1)(1) N N M1A1 Forms the sum of the areas of the rectangles. M1   0 2   N 2 N 2 N 2 N 2 for correct number of rectangles. = 1  N (1) N n = (1 2 ) N 1 (1 2 −1) = (1 2 )(1 2 ) N 1 = 1 . M1A1 Applies  N rn = r ( rN −1 ) , AG. N n=1 2 N(1) N 1 −N N −N(1) N 1 2N ( 2N 1 −1 ) n=1 r−1 2 2 4 6(b)  1(1)x dx(1)+(1)(1) N 1 + +(1)(1)N N −2 +(1)(1)N N −1 M1A1 Forms the sum of the areas of appropriate 2 N N 2 N 2 N 2 rectangles. 0 1  N−1 (1) N n = 1 2 −1 = 1 = 2N 1 M1A1 Applies  N−1 rn = rN −1 . N n=0 2 N(1) N 1 −N 2N ( 1−(1) N 1 ) 2N ( 2N 1 −1 ) n=0 r−1 2 2 4 6(c) ( 2N 1 1 ) − ( 1 1 ) = 2 1 N 10−32N 103 M1 Simplifies U N −L N to N c . 2N 2N −1 2N 2N −1 Least value of N is 500 A1 CWO. 2 6(d) L  1 U  1 ln2 1 M1A1FT Forms inequality. FT on their U . N 2ln2 N 2U 2L N N N 0.693= 1 ln2 1 =0.694 A1A1 CWO. Must have used N =500. 2U 2L N N 4 © Cambridge University Press & Assessment 2024 Page 11 of 14

Official mark scheme pages: 11 · source PDF URL

9231-2024-on-23-q08

Oct/Nov 2024 · Paper 23 · Question 8 · 14 marks

8(a) z+z−1 =2cos B1 For LHS. ( z+z−1)7 = ( z7 +z−7) +7 ( z5 +z−5) +21 ( z3+z−3) +35 ( z+z−1) M1A1 Expands and groups. A0 if not grouped clearly. 27cos7=2cos7+7(2cos5)+21(2cos3)+35(2cos) M1 Substitutes zn +z−n =2cosn. cos7= 1 cos7+ 7 cos5+21cos3+35cos A1 64 64 64 64 5 8(b) 1π 1π 1π M1A1 Applies integration by parts. I =4 cosn−1cosd=cosn−1sin4 +(n−1)4 cosn−2sin2d n   0 0 0 I =cosn−1sin 1 4 π +(n−1) 1 4 π cosn−2 ( 1−cos2 ) d M1 Applies sin2=1−cos2 n   0 0 I =2−n 2 +(n−1)I −(n−1)I nI =2−n 2 +(n−1)I A1 AG. n n−2 n n n−2 4 © Cambridge University Press & Assessment 2024 Page 13 of 14 8(c) 9I =2−9 2 +8I = 1 ( 2−1 2 ) +8I M1A1 Applies reduction formula from (b). 9 7 16 7 1π M1 Applies identity from (a). Allow I =4 1 cos7+ 7 cos5+ 21cos3+35cosd 7 64 64 64 64 missing/incorrect limits. 0 1π A1 I = 1 1sin7+7sin5+ 21sin3+35sin4 7 647 5 3  0 9I = 1 ( 2−1 2 ) +1 ( 1 ( −2−1 2 ) +7 ( −2−1 2 ) +21 ( 2−1 2 ) +35 ( 2−1 2 )) A1 Check their exact answer when I 9 is the subject, 9 16 8 7 5 3 like terms collected. ISW. 0.402237 I = 2867 ( 2−1 2 ) = 2867 2 9 5040 10080 5 © Cambridge University Press & Assessment 2024 Page 14 of 14

Official mark scheme pages: 13, 14 · source PDF URL

9231-2025-mj-21-q02

May/June 2025 · Paper 21 · Question 2 · 7 marks

2(a) I = 1 (1−x)n sinhxdx=(1−x)ncoshx 1 + 1 n(1−x)n−1coshxdx M1 A1 Applies integration by parts. Appropriate choice of n   parts for M1. 0 0 0 I =−1+n(1−x)n−1sinhx 1 +n(n−1) 1 (1−x)n−2 sinhxdx M1 Applies integration by parts again. Appropriate choice n   of parts for M1. 0 0 I =−1+n(n−1)I A1 AG, withhold for no evidence of substitution or n n−2 incorrect use of limits. 4 2(b) I = 1 sinhxdx=coshx1 =cosh1−1. B1 Finds I 0 . 0 0 0 I =−1+2I M1 Uses reduction formula from (a). 2 0 I =2cosh1−3 A1 Accept e+e−1−3. 2 3 © Cambridge University Press & Assessment 2025 Page 9 of 18

Official mark scheme pages: 9 · source PDF URL

9231-2025-mj-21-q04

May/June 2025 · Paper 21 · Question 4 · 9 marks

4(a) 1 e 1 + 1 e 2 + + 1 e n−1 M1 Forms sum of the areas of the rectangles. 1 2 n−1 n A1 Compares with integral. Wrong comparison  1 e x dx 1 x (missing/wrong inequality or wrong limits) is A0. n n M1 A1 Correct integration. Attempt at integration by parts is  1 e xdx=2e x 1 x   1 M0 A0. n n ( ) A1 AG, must have obtained all previous marks.  1 e r 2e x + 1 e n = 2+ 1 e n −2e r   1 n n r=1 5 © Cambridge University Press & Assessment 2025 Page 10 of 18 4(b) 1 e 2 + 1 e 3 + + 1 e n M1 Forms the sum of the areas of appropriate rectangles. 2 3 n n n A1 Compares with integral.  1 e xdx=2e x 1 x   1 n n M1 A1 Adds term from r=1.  1 e r 2e x +e=2e n −e r   1 r=1 4 © Cambridge University Press & Assessment 2025 Page 11 of 18

Official mark scheme pages: 10, 11 · source PDF URL

9231-2025-mj-22-q02

May/June 2025 · Paper 22 · Question 2 · 7 marks

2(a) I = 1 (1−x)n sinhxdx=(1−x)ncoshx 1 + 1 n(1−x)n−1coshxdx M1 A1 Applies integration by parts. Appropriate choice of n   parts for M1. 0 0 0 I =−1+n(1−x)n−1sinhx 1 +n(n−1) 1 (1−x)n−2 sinhxdx M1 Applies integration by parts again. Appropriate choice n   of parts for M1. 0 0 I =−1+n(n−1)I A1 AG, withhold for no evidence of substitution or n n−2 incorrect use of limits. 4 2(b) I = 1 sinhxdx=coshx1 =cosh1−1. B1 Finds I 0 . 0 0 0 I =−1+2I M1 Uses reduction formula from (a). 2 0 I =2cosh1−3 A1 Accept e+e−1−3. 2 3 © Cambridge University Press & Assessment 2025 Page 9 of 18

Official mark scheme pages: 9 · source PDF URL

9231-2025-mj-22-q04

May/June 2025 · Paper 22 · Question 4 · 9 marks

4(a) 1 e 1 + 1 e 2 + + 1 e n−1 M1 Forms sum of the areas of the rectangles. 1 2 n−1 n A1 Compares with integral. Wrong comparison  1 e x dx 1 x (missing/wrong inequality or wrong limits) is A0. n n M1 A1 Correct integration. Attempt at integration by parts is  1 e xdx=2e x 1 x   1 M0 A0. n n ( ) A1 AG, must have obtained all previous marks.  1 e r 2e x + 1 e n = 2+ 1 e n −2e r   1 n n r=1 5 © Cambridge University Press & Assessment 2025 Page 10 of 18 4(b) 1 e 2 + 1 e 3 + + 1 e n M1 Forms the sum of the areas of appropriate rectangles. 2 3 n n n A1 Compares with integral.  1 e xdx=2e x 1 x   1 n n M1 A1 Adds term from r=1.  1 e r 2e x +e=2e n −e r   1 r=1 4 © Cambridge University Press & Assessment 2025 Page 11 of 18

Official mark scheme pages: 10, 11 · source PDF URL

9231-2025-mj-23-q06

May/June 2025 · Paper 23 · Question 6 · 10 marks

6(a) 1  1 1 1  1 n2 n2 n2  M1 A1 Forms sum of the areas of the rectangles.  + + + =  + + +  n  1 n 2 2 +1 n 22 2 +1 n n 2 2 +1  n 12 +n2 22 +n2 n2 +n2    1 1 dx=tan−1x 1 = 1π M1 Evaluates integral.  x2 +1   0 4 0 A1 CAO 1 n n2 n n A1 AG. Clear and convincing comparison with  =  1π n n2 +r2 n2 +r2 4 the integral. r=1 r=1 5 6(b)   *M1 A1 Forms the sum of the areas of appropriate 1 1 1 1 1 n2 n2 n2   + + + =  + + +  rectangles. M0 for more than n rectangles. n02 +1 12 +1 (n−1)2 +1 n 02 +n2 12 +n2 (n−1)2 +n2   n2 n2  1 n n 1 n n 1 DM1 A1 Compares with integral. Correctly adjust for n + n2 +r2 − 2n  1 4 π n2 +r2 1 4 π− 2n 0th and last rectangles. (Allow sign errors r=1 r=1 only for M1). Limits must match area covered by chosen rectangles. 4 6(c) 1π B1 4 1 © Cambridge University Press & Assessment 2025 Page 14 of 17

Official mark scheme pages: 14 · source PDF URL

9231-2025-mj-24-q02

May/June 2025 · Paper 24 · Question 2 · 6 marks

2 x2 −2x+5=4+(x−1)2 B1 Completes the square. OE. 5 5 M1 A1 Applies formula. 2 1  x−12  dx=  sinh−1    4+(x−1)2   2  1 1 =sinh−1(3) −sinh−1(0) M1 Uses limits. 4 ( ) M1 A1 Uses logarithmic form. =ln 3+ 9 +1 =ln2 4 16 6 © Cambridge University Press & Assessment 2025 Page 10 of 17

Official mark scheme pages: 10 · source PDF URL

9231-2025-mj-24-q08

May/June 2025 · Paper 24 · Question 8 · 15 marks

8(a) B1 Correct shape and position. Ignore x0 on their sketch. x y=1 B1 States the equation of the asymptote. Ignore y=−1. 2 8(b) N M1 A1 Compares sum with integral. Limits on integral N tanhr tanhxdx must be present and correct for M1. 0 r=1  N tanhx=ln(coshx) N =ln(coshN) A1 Evaluates integral, AG. Must see ln(coshx).   0 0 3 © Cambridge University Press & Assessment 2025 Page 16 of 17 8(c)(i)  1 2 ln3 dy 2 1ln3 M1 A1 Correct formula for surface area. Missing limits is S =2π  y 1+  dx=2π2 tanh(x) 1+sech4(x)dx M1 A0. y outside of integral is M0.  dx 0 0 u= 1+sech4x  du =−2 ( 1+sech4x )−1 2sech4xtanhx M1 A1 Finds du . Sign error gets M1 A0. dx dx x=1ln3coshx= 2 3 sechx= 1 3u= 1+ 9 M1 A1 Finds u when x= 1ln3. Need to see working to  2 3  2 16 2 justify given limit. Working without evidence of evaluating sechx is M1 A0.  5 4 u  2 u2 A1 AG, must have gained all previous marks. S =−π u du=π du  u2 −1  u2 −1 2 5 4 7 8(c)(ii) = 1π   2 2+ 1 − 1 du=πu+ 1ln(u−1)−1ln(u+1) 2 M1 A1 Uses partial fractions or u2 =1+ 1 . 2  u−1 u+1  2 2  5 u2 −1 u2 −1 5 4 4   2−1 1  ( ( )) A1 OE. =π   2+1 2 ln   2+1    −5 4 −1 2 ln  9      =π 2−5 4 +ln3+ 1 2 ln 3−2 2 3 © Cambridge University Press & Assessment 2025 Page 17 of 17

Official mark scheme pages: 16, 17 · source PDF URL

9231-2025-on-21-q03

Oct/Nov 2025 · Paper 21 · Question 3 · 9 marks

3(a) d ( x ( 1+x2)n ) =2nx2( 1+x2)n−1 + ( 1+x2)n M1 A1 Uses the product rule to differentiate. dx =2n ( 1+x2 −1 )( 1+x2)n−1 + ( 1+x2)n *M1 Uses x2 =x2 +1−1. 1 DM1 Integrates both sides using the limits given.  x ( 1+x2)n =2nI −2nI +I   n n−1 n 0 2n =(2n+1)I −2nI (2n+1)I =2n +2nI A1 Substitutes limits and rearranges. AG. n n−1 n n−1 5 3(b) I =tan−1x 1 = 1π B1 −1   0 4 −I = 1 −2I M1 A1 Applies reduction formula with n=−1. −1 2 −2 −1π= 1 −2I I = 1 +1π A1 4 2 −2 −2 4 8 4 © Cambridge University Press & Assessment 2025 Page 10 of 19

Official mark scheme pages: 10 · source PDF URL

9231-2025-on-21-q05

Oct/Nov 2025 · Paper 21 · Question 5 · 11 marks

5(a)   1( 1x3+x ) dx (1) ( 1(1)3 +(1) ) +(1) ( 1(2)3 +(2) ) + +(1) ( 1(n)3 +(n) ) M1 A1 Forms the sum of the areas of the rectangles. M1 for   0 3   n 3 n n n 3 n n n 3 n n correct width and at least three complete terms including first and last. n n n2(n+1)2 n(n+1) M1 A1 Applies formulae from MF19. = 1 r3+ 1 r= + 3n4 n2 12n4 2n2 r=1 r=1 = 1 (1+1)2 +1(1+1)= 1 (1+1)(7+ 1) A1 AG. 12 n 2 n 12 n n 5 5(b)  1( 1x3+x ) dx (1) ( 1(1)3 + (1) ) + (1) ( 1(2)3 + (2) ) + + (1) ( 1(n−1)3 + (n−1) ) M1 A1 Forms the sum of the areas of appropriate rectangles. 3 n 3 n n n 3 n n n 3 n n 0 n−1 n−1 (n−1)2n2 (n−1)n M1 4 = 3 1 n4 r3+ n 1 2 r= 12n4 + 2n2 Applies formulae from MF19. Accept L n =U n − 3n . r=1 r=1 = 1 (1−1)2 +1(1−1)= 1 (1−1)(7−1) A1 At least one line of simplification after subbing in 12 n 2 n 12 n n formula (powers of n simplified). 4 5(c) 1 16 4 M1 c U n −L n = 12   n   = 3n Simplifies U n −L n to n . 4 A1 AG. →0 as n→ 3n 2 © Cambridge University Press & Assessment 2025 Page 12 of 19

Official mark scheme pages: 12 · source PDF URL

9231-2025-on-22-q01

Oct/Nov 2025 · Paper 22 · Question 1 · 4 marks

1 s= π e 8 3 +16e 8 3 d= 5 π e3 4d M1 A1 Forms length of C correctly. 9 3 0 0 s= 5e 4 3  π M1 Integrates an expression of the form aeb. One 4  error in the a or b. 0 = 5 ( e 4 3 π −1 ) A1 Must be exact. Allow equivalent form but must 4 have the e0evaluated to be one. 4 Question Answer Marks Guidance

Official mark scheme pages: 9 · source PDF URL

9231-2025-on-22-q06

Oct/Nov 2025 · Paper 22 · Question 6 · 13 marks

6(a) dx  1  coshu M1 A1 Applies substitution and sinh2u+1=cosh2u. = 1 2coshu  dx= 1 2 du=1 21du du 2  2x2 +1 2  sinh2u+1 2 = 1 2sinh−1 ( x 2 )+C A1 sinh−1 ( x 2 ) =ln ( x 2+ 2x2 +1 ) Condone 2 C missing. 3 6(b) *M1 A1 Forms sum of the areas of the rectangles. Just   1  1 + 1 + + 1  required the first line of working for the M1A1. n 212 +1 222 +1 2n2 +1   n2 n2 n2  1 n n n  =  + + +  n  212 +n2 222 +n2 2n2 +n2  1 DM1 Compares with integral.  1  dx  2x2 +1 0 1 B1 Evaluates integral.   1 dx=1 2sinh−1 ( x 2 ) 1 = 1 2ln ( 2+ 3 ) 2  2  2x2 +1 0 0 1 n n n 1 ( ) A1 AG. Must see  =  1 2ln 2+ 3 n r=1 2r2 +n2 r=1 2r2 +n2 2 1  n n = n 1 or equivalent n 2r2 +n2 2r2 +n2 r=1 r=1 and limits on integral correct. 5 © Cambridge University Press & Assessment 2025 Page 13 of 17 6(c)   *M1 A1 Forms the sum of the areas of appropriate 1 1 1 1  rectangles. + + +   n  202 +1 212 +1 2 (n−1)2 +1  n2 n2    1 n n = 1+ + +  n 212 +n2 2(n−1)2 +n2    1 DM1 Compares with integral.    1 dx=1 2sinh−1 ( x 2 ) 1 = 1 2ln ( 2+ 3 ) 2  2  2x2 +1 0 0 1 n 1 1 ( ) A1 + −  1 2ln 2+ 3 n 2r2 +n2 3n2 2 r=1 n 1 ( ) 1( )   1 2ln 2+ 3 + 1 3−1 2r2 +n2 2 n 3 r=1 4 6(d) ( ) B1 1 2ln 2+ 3 2 1 © Cambridge University Press & Assessment 2025 Page 14 of 17

Official mark scheme pages: 13, 14 · source PDF URL

9231-2025-on-23-q03

Oct/Nov 2025 · Paper 23 · Question 3 · 9 marks

3(a) d ( x ( 1+x2)n ) =2nx2( 1+x2)n−1 + ( 1+x2)n M1 A1 Uses the product rule to differentiate. dx =2n ( 1+x2 −1 )( 1+x2)n−1 + ( 1+x2)n *M1 Uses x2 =x2 +1−1. 1 DM1 Integrates both sides using the limits given.  x ( 1+x2)n =2nI −2nI +I   n n−1 n 0 2n =(2n+1)I −2nI (2n+1)I =2n +2nI A1 Substitutes limits and rearranges. AG. n n−1 n n−1 5 3(b) I =tan−1x 1 = 1π B1 −1   0 4 −I = 1 −2I M1 A1 Applies reduction formula with n=−1. −1 2 −2 −1π= 1 −2I I = 1 +1π A1 4 2 −2 −2 4 8 4 © Cambridge University Press & Assessment 2025 Page 10 of 19

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9231-2025-on-23-q05

Oct/Nov 2025 · Paper 23 · Question 5 · 11 marks

5(a)   1( 1x3+x ) dx (1) ( 1(1)3 +(1) ) +(1) ( 1(2)3 +(2) ) + +(1) ( 1(n)3 +(n) ) M1 A1 Forms the sum of the areas of the rectangles. M1 for   0 3   n 3 n n n 3 n n n 3 n n correct width and at least three complete terms including first and last. n n n2(n+1)2 n(n+1) M1 A1 Applies formulae from MF19. = 1 r3+ 1 r= + 3n4 n2 12n4 2n2 r=1 r=1 = 1 (1+1)2 +1(1+1)= 1 (1+1)(7+ 1) A1 AG. 12 n 2 n 12 n n 5 5(b)  1( 1x3+x ) dx (1) ( 1(1)3 + (1) ) + (1) ( 1(2)3 + (2) ) + + (1) ( 1(n−1)3 + (n−1) ) M1 A1 Forms the sum of the areas of appropriate rectangles. 3 n 3 n n n 3 n n n 3 n n 0 n−1 n−1 (n−1)2n2 (n−1)n M1 4 = 3 1 n4 r3+ n 1 2 r= 12n4 + 2n2 Applies formulae from MF19. Accept L n =U n − 3n . r=1 r=1 = 1 (1−1)2 +1(1−1)= 1 (1−1)(7−1) A1 At least one line of simplification after subbing in 12 n 2 n 12 n n formula (powers of n simplified). 4 5(c) 1 16 4 M1 c U n −L n = 12   n   = 3n Simplifies U n −L n to n . 4 A1 AG. →0 as n→ 3n 2 © Cambridge University Press & Assessment 2025 Page 12 of 19

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9231-2025-on-24-q01

Oct/Nov 2025 · Paper 24 · Question 1 · 5 marks

1(a) 1 1 M1 Integrates by parts. Must see first term I =xncoshx −n xn−1coshxdx n   0 0 xncoshx. =xncoshx 1 −n ( xn−1sinhx 1 −(n−1) 1 xn−2sinhxdx ) M1 Integrates by parts again.     0 0 0 I =cosh1−n ( sinh1−(n−1)I ) =cosh1−nsinh1+n(n−1)I A1 Substitutes limits, AG. Omission of n n−2 n−2 brackets or limits in their working is A0. 3 1(b) I =cosh1−2sinh1+2I M1 Applies reduction formula. 2 0 I =3cosh1−2sinh1−2 A1 Accept 1 ( e+5e−1) −2. 2 2 2 © Cambridge University Press & Assessment 2025 Page 9 of 19

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9231-2025-on-24-q04

Oct/Nov 2025 · Paper 24 · Question 4 · 10 marks

4(a)  1 1 ( 3x) dx 1 ( (1)3N 1 +(1)3N 2 + +(1)3 N N −1 +(1)3N N) M1 A1 Forms the sum of the areas of the 2 2 N N N N rectangles. 0 M1 for at least three terms including first 1 and last, and common ratio 3N . Missing “+” is A0. = 1  N ( 3N 1 )n = 3N 1 M1 A1 Applies  N rn = r ( rN −1 ) , AG. 2N n=1 N ( 3N 1 −1 ) n=1 r−1 4 4(b)  1 1 ( 3x) dx 1 ( (1)+(1)3N 1 + +(1)3 N N −2 +(1)3 N N −1) M1 A1 Forms the sum of the areas of appropriate 2 2 N N N N rectangles. 0 M1 for at least three terms including first and last. = 1  N−1( 3N 1 )n = 1 M1 A1 Applies  N−1 rn = rN −1 . ( ) 2N n=0 N 3N 1 −1 n=0 r−1 4 4(c) 1 M1 c 3N 1 1 Simplifies their U −L to . N ( 3N 1 −1 ) − N ( 3N 1 −1 ) = N N N N 1 A1 →0 as N →. N 2 © Cambridge University Press & Assessment 2025 Page 13 of 19

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9231-2026-mj-21-q02

May/June 2026 · Paper 21 · Question 2 · 7 marks

2(a) I = 1 (1+t)n sinπtdt=−1(1+t)ncosπt 1 + n 1 (1+t)n−1cosπtdt *M1 A1 Applies integration by parts. Allow limits missing. n 0  π  0 π 0 M1 for correct trigonometric functions and powers of (1+t). © Cambridge University Press & Assessment 2026 Page 10 of 18 I n = 1π ( 2 n + 1 ) + nπ  1π (1 + t ) n − 1 s i n π t  1 0 − nπ  1 0 1π ( n − 1 ) ( 1 + t ) n − 2 s i n π t d t DM1 Applies integration by parts again. M1 for correct trigonometric functions and powers of (1 + t ) . I n = 1π ( 2 n + 1 ) − 12 π n ( n − 1 ) I n − 2  π 2 I n = π ( 2 n + 1 ) − n ( n − 1 ) I n − 2 A1 AG. Must have seen limits and correct use of brackets. CWO. 4 2(b) I 0 =  1 0 s i n π t d t =  − 1π c o s π t  1 0 = 2π B1 Finds . I 0 . π 2 I 2 = π ( 2 2 + 1 ) − 2 ( 2 − 1 ) I 0  = 5 π − 4π  M1 Uses reduction formula from (a). I 2 = 5π − 43 π A1 3 Question Answer Marks Guidance

Official mark scheme pages: 10 · source PDF URL

9231-2026-mj-21-q05

May/June 2026 · Paper 21 · Question 5 · 10 marks

5(a) © Cambridge University Press & Assessment 2026 Page 14 of 18  1 0 ( 11 10 x 3 + 11 0 ) d x  ( 1n ) ( 11 10 ( 1n ) 3 + 11 0 ) + ( 1n ) ( 11 10 ( 2n ) 3 + 11 0 ) +  + ( 1n ) ( 11 10 ( nn ) 3 + 11 0 ) M1 A1 Forms the sum of the areas of the rectangles. M1 for sum of n complete terms. A1 for at least 3 correct terms including first and last. May be written in summation form. = 1 10 1n 4 r n = 1 r 3 + 11 0 = 1 1 n 2 4 ( 0 n n + 4 1 ) 2 + 1 1 0 M1 Applies formula for r3 from MF19. = 14 10 ( 1 + 1n ) 2 + 11 0 A1 AG. CWO. 4 5(b)  1 0 ( 11 10 x 3 + 11 0 ) d x  1 10 n + ( 1n ) ( 11 10 ( 1n ) 3 + 11 0 ) + ( 1n ) ( 11 10 ( 2n ) 3 + 11 0 ) +  + ( 1n ) ( 11 10 ( n −n 1 ) 3 + 11 0 ) M1 A1 Forms the sum of the areas of appropriate rectangles. M1 for sum of n complete terms. A1 for at least 3 correct terms including first and last. = 1 10 n + 1 10 1n 4 nr − = 1 1 r 3 + n − 1 1 0 n = 1 1 ( n 4 − 0 1 n ) 4 2 n 2 + 11 0 M1 Applies formulae from MF19. Allow U n + 1 10 n − 1 2 1 0 n = 14 10 ( 1 − 1n ) 2 + 11 0 A1 OE. e.g. 14 10 ( 1 + 1n ) 2 + 11 0 − 1 1 1 0 n . 4 5(c) U n − L n = 14 10 ( 1 + 1n ) 2 − 14 10 ( 1 − 1n ) 2 M1 Their U n − L n c simplifies to . n Accept 14 10 ( 1 ) 2 + 11 0 − 14 10 ( 1 ) 2 − 11 0 . 11 A1 AG. Their →0 as n→ 10n L n must be correct. 2 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 14 · source PDF URL

9231-2026-mj-22-q02

May/June 2026 · Paper 22 · Question 2 · 7 marks

2(a) I = 1 (1+t)n sinπtdt=−1(1+t)ncosπt 1 + n 1 (1+t)n−1cosπtdt *M1 A1 Applies integration by parts. Allow limits missing. n 0  π  0 π 0 M1 for correct trigonometric functions and powers of (1+t). © Cambridge University Press & Assessment 2026 Page 10 of 18 I n = 1π ( 2 n + 1 ) + nπ  1π (1 + t ) n − 1 s i n π t  1 0 − nπ  1 0 1π ( n − 1 ) ( 1 + t ) n − 2 s i n π t d t DM1 Applies integration by parts again. M1 for correct trigonometric functions and powers of (1 + t ) . I n = 1π ( 2 n + 1 ) − 12 π n ( n − 1 ) I n − 2  π 2 I n = π ( 2 n + 1 ) − n ( n − 1 ) I n − 2 A1 AG. Must have seen limits and correct use of brackets. CWO. 4 2(b) I 0 =  1 0 s i n π t d t =  − 1π c o s π t  1 0 = 2π B1 Finds . I 0 . π 2 I 2 = π ( 2 2 + 1 ) − 2 ( 2 − 1 ) I 0  = 5 π − 4π  M1 Uses reduction formula from (a). I 2 = 5π − 43 π A1 3 Question Answer Marks Guidance

Official mark scheme pages: 10 · source PDF URL

9231-2026-mj-22-q05

May/June 2026 · Paper 22 · Question 5 · 10 marks

5(a) © Cambridge University Press & Assessment 2026 Page 14 of 18  1 0 ( 11 10 x 3 + 11 0 ) d x  ( 1n ) ( 11 10 ( 1n ) 3 + 11 0 ) + ( 1n ) ( 11 10 ( 2n ) 3 + 11 0 ) +  + ( 1n ) ( 11 10 ( nn ) 3 + 11 0 ) M1 A1 Forms the sum of the areas of the rectangles. M1 for sum of n complete terms. A1 for at least 3 correct terms including first and last. May be written in summation form. = 1 10 1n 4 r n = 1 r 3 + 11 0 = 1 1 n 2 4 ( 0 n n + 4 1 ) 2 + 1 1 0 M1 Applies formula for r3 from MF19. = 14 10 ( 1 + 1n ) 2 + 11 0 A1 AG. CWO. 4 5(b)  1 0 ( 11 10 x 3 + 11 0 ) d x  1 10 n + ( 1n ) ( 11 10 ( 1n ) 3 + 11 0 ) + ( 1n ) ( 11 10 ( 2n ) 3 + 11 0 ) +  + ( 1n ) ( 11 10 ( n −n 1 ) 3 + 11 0 ) M1 A1 Forms the sum of the areas of appropriate rectangles. M1 for sum of n complete terms. A1 for at least 3 correct terms including first and last. = 1 10 n + 1 10 1n 4 nr − = 1 1 r 3 + n − 1 1 0 n = 1 1 ( n 4 − 0 1 n ) 4 2 n 2 + 11 0 M1 Applies formulae from MF19. Allow U n + 1 10 n − 1 2 1 0 n = 14 10 ( 1 − 1n ) 2 + 11 0 A1 OE. e.g. 14 10 ( 1 + 1n ) 2 + 11 0 − 1 1 1 0 n . 4 5(c) U n − L n = 14 10 ( 1 + 1n ) 2 − 14 10 ( 1 − 1n ) 2 M1 Their U n − L n c simplifies to . n Accept 14 10 ( 1 ) 2 + 11 0 − 14 10 ( 1 ) 2 − 11 0 . 11 A1 AG. Their →0 as n→ 10n L n must be correct. 2 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 14 · source PDF URL

9231-2026-mj-23-q02

May/June 2026 · Paper 23 · Question 2 · 5 marks

2 1 + ( 12 x − 12 ) 2 = 1 + 14 x − 1 B1 2 dy Finds 1+   . dx 2 π  2 0 x 1 + 1 4 x − 1 d x M1 Correct formula for surface area with limits. Integral must be in terms of x. 2 2 A1 Integrand simplified correctly. =2π x+ 1 dx or π 4x+1dx 4 0 0 = 2 π  2 3  x + 1 4  32  2 0 M1 Integrates and substitutes limits. Integrand must 2 π 3 or  (4x+1) 2 be of the form 6  0 ( p x + a ) n with a and p nonzero and consistent with previous working and n  1 . = 1 33 π A1 AEF. 5 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 9 · source PDF URL

9231-2026-mj-23-q03

May/June 2026 · Paper 23 · Question 3 · 6 marks

3(a) EITHER: © Cambridge University Press & Assessment 2026 Page 10 of 19 d d x ( ( x + 1 ) ln n ( x + 1 ) ) = n ( x + 1 ) ln n − 1 ( x ( + x 1 + ) 1 ) + ln n ( x + 1 ) (B1 CAO (AEF).  ( x + 1 ) ln n ( x + 1 )  a 0 = n I n − 1 + I n M1 Integrates both sides using the limits given. I n = ( a + 1 ) ln n ( a + 1 ) − n I n − 1 A1) Substitutes limits and rearranges. AG. OR: Alternative method for question 3(a) I n =  a 0 ln n ( x + 1 ) d x =  x ln n ( x + 1 )  a 0 − n  a 0 x ln n ( − x ( 1 + x 1 + ) 1 ) d x (B1 CAO (Condone missing limits).  = a l n n ( a + 1 ) − n  a 0 ln n − 1 ( x + 1 ) d x + n  a 0 ln n − 1 ( x ( + x 1 + ) 1 )  = a l n n ( a + 1 ) − n I n − 1 +  ln n ( x + 1 )  a 0 M1 Expression needs to be fully integrated and in terms of a and I n − 1 . I n = ( a + 1 ) ln n ( a + 1 ) − n I n − 1 A1) Substitutes limits, convincing work to AG. Available marks 3 3(b) I =(a+1)ln(a+1)−a B1 CAO. 1 I 2 = ( a + 1 ) ln 2 ( a + 1 ) − 2 I 1 M1 Applies reduction formula to find I 2 . I 2 = ( a + 1 ) ln 2 ( a + 1 ) − 2 ( a + 1 ) ln ( a + 1 ) + 2 a A1 3 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 10 · source PDF URL

9231-2026-mj-23-q07

May/June 2026 · Paper 23 · Question 7 · 12 marks

7(a) © Cambridge University Press & Assessment 2026 Page 15 of 19  1 x − x 2 d x = − 1 2  1 u d u M1 A1 Applies given substitution. Allow  a u d u for M1. = − 1 − x 2 [ + C ] A1 Accept without ‘ + C ‘. Must be in terms of x . 3 7(b)  1 n s i n − 1 1 n + 1 n s i n − 1 2 n +  + 1 n s i n − 1 n n  = r n = 1 1 n s i n − 1  r n    1 s0 i n − 1 x d x M1 M1 correct comparison (no limits required) A1 A1 correct limits on both sides  1 0 s i n − 1 x d x =  x s i n − 1 x  1 0 −  1 0 1 x − x 2 d x  x s i n − 1 x + 1 − x 2  1 0 M1 Integrates by parts. A1 CAO with correct signs = 1 2 π − 1 A1 Clear substitution of limits to given result 5 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 16 of 19 1 n s i n − 1 1 n + 1 n s i n − 1 2 n +  + 1 n s i n − 1 n − n 1 = n r − = 1 1 1 n s i n − 1  r n  M1 Forms the sum of the areas of appropriate rectangles. nr − = 1 1 1 n s i n − 1  r n  = n r = 1 1 n s i n − 1  r n  −  1 n   π 2  A1 Adjusts summation with reference to missing term.  nr − = 1 1 1 n s i n − 1  r n    1 s0 i n − 1 x d x  A1 n 1 r 1 π  sin−1   π−1+ n n 2 2n r=1 CAO 3 7(d) 12 π − 1 B1 Dependent on obtaining full marks in 7(c). 1 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 15, 16 · source PDF URL

9231-2026-mj-24-q05

May/June 2026 · Paper 24 · Question 5 · 9 marks

5 Sum © Cambridge University Press & Assessment 2026 Page 14 of 19 =  1   +  1  7 +  +  ( 4 n 2 1 + 4 n + 3 )   n 0  ( 4 x 2 1 + 4 x + 3 ) d x M1A1 Forms sum of the areas of the rectangles. M1 for correct comparison (inequality). A1 for limits. 4x2+4x+3=(2x+1)2 +2 B1 Completes the square  n 0 ( 2 x + 1 1 ) 2 + 2 d x =  12 s i n h − 1  2 x  +  1   n 0 M1A1 Uses formula =  12 ln  2 x  +  1 + ( 2 x + 1 2 ) 2 + 2   n 0 =  12 ln  2 x + 1 +  ( 4  x  2 + 4 x + 3 )   n 0 M1A1 Uses logarithmic form of s i n h − 1 . 2n+1+ ( 4n2 +4n+3 ) 1+ M1 Inserts limits. = 1ln −1ln  2    2      1+2n+ ( 4n2 +4n+3 ) A1 AG. CWO. = 1ln  2  +    9 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 14 · source PDF URL

9231-2026-mj-24-q06

May/June 2026 · Paper 24 · Question 6 · 10 marks

6(a) 1π 1π 1π *M1A1 Applies integration by parts. For M1, parts I =2 excosn xdx=excosn x2 +n2 excosn−1xsinxdx n −1π   −1π −1π chosen must lead to the given answer. Correct 2 2 2 exponents. © Cambridge University Press & Assessment 2026 Page 15 of 19 I n = n   e x c o s n − 1 x s i n x  12 − π 12 π −  1 π 2 1 − π 2 e x ( c o s n x − ( n − 1 ) c o s n − 2 x s i n 2 x ) d x  DM1 Applies integration by parts again. Correct exponents. I n = − n I n + n ( n − 1 )  1 π 2 1 − π 2 e x c o s n − 2 x s i n 2 x d x A1 AG. Withhold for missing x in trig functions; for example,cosn−2sin2x. No limits seen is A0. I n = − n I n + n ( n − 1 )  1 π 2 1 − π 2 e x c o s n − 2 x d x − n ( n − 1 )  1 π 2 1 − π 2 e x c o s n x d x M1 Applies sin2x=1−cos2x to given answer. I =−nI +n(n−1)I −n(n−1)I  ( n2 +1 ) I =n(n−1)I A1 AG. n n n−2 n n n−2 6 6(b) I = 1 2 π exdx=e 1 2 π −e−1 2 π. B1 Finds 0 −1π 2 I 0 . 5 I 2 = 2 I 0 or 17I =12I . M1 Uses reduction formula from (a). 4 2 I = 2 ( e 1 2 π −e−1 2 π ) A1 2 5 17I =12I I = 24 ( e 1 2 π −e−1 2 π ) A1 4 2 4 85 4 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

Official mark scheme pages: 15 · source PDF URL