Tiêu đề Integration Practice Sheet Solution HSC FRB 25.pdf ✅

†hvMRxKiY  Final Revision Batch '25 1 Board Questions Analysis m„Rbkxj cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 1 1 1 1 2 2 2 1 1 2022 1 1 1 1 2 2 1 1 1 eûwbe©vPwb cÖkœ †evW© mvj XvKv gqgbwmsn ivRkvnx Kzwgjøv h‡kvi PÆMÖvg ewikvj wm‡jU w`bvRcyi 2023 3 6 4 3 3 5 2 3 4 2022 4 4 5 3 4 4 4 4 3 weMZ mv‡j †ev‡W© Avmv m„Rbkxj cÖkœ 1| `„k ̈Kí-1: x 2 + y2 = 4 `„k ̈Kí-2: f(x) = (sin–1 x) [XvKv †evW©- Õ23] (K)  1 a 2 + 4x2 dx wbY©q Ki| (L) `„k ̈Kí-2 Gi Av‡jv‡K  x f(x) dx wbY©q Ki| (M) mgvKjb c×wZ‡Z `„k ̈Kí-1 I x + y = 2 †iLv Øviv Ave× ÿz`aZi As‡ki †ÿÎdj wbY©q Ki| mgvavb: (K)  1 a 2 + 4x2 dx =  1 4     a 2 4 + x2 dx = 1 4  1     a 2 2 + x2 dx = 1 4 . 1 a 2 tan–1x a 2 + c = 1 2a tan–12x a + c (Ans.) (L) †`Iqv Av‡Q, f(x) = sin–1 x   xf(x) dx =  x sin–1 x dx = sin–1 x  xdx –      d  dx sin–1 x  xdx dx = x 2 2 sin–1 x –  1 1 – x 2 . x 2 2 dx awi, x = sin  dx = cos d = x 2 2 sin–1 x –  sin2  2 1 – sin2  cos d = x 2 2 sin–1 x –  sin2  2cos cos d = x 2 2 sin–1 x – 1 4  (1 – cos2) d = x 2 2 sin–1 x – 1 4      – sin2 2 + c = x 2 2 sin–1 x – 1 4  + 1 4  2sin cos 2 + c = x 2 2 sin–1 x – 1 4  + 1 4  sin  1 – sin2  + c = x 2 2 sin–1 x – 1 4 sin–1 x + 1 4 x . 1 – x 2 + c [⸪ x = sin   = sin–1 x] = x 2 2 sin–1 x – 1 4 sin–1 x + 1 4 x 1 – x 2 + c (Ans.) (M) `„k ̈Kí-1 n‡Z, x 2 + y2 = 4......(i) `„k ̈Kí-2 n‡Z, x + y = 2 .......(ii) (i) bs n‡Z, y 2 = 4 – x 2  y =  4 – x 2 1g PZzf©v‡M y abvZ¥K A_©vr, y = 4 – x 2 (i) bs e„ËØviv 1g PZzf©v‡M MwVZ As‡ki †ÿÎdj A1 =  2 0 y dx =  2 0 4 – x 2 dx =   2 0 4 – 4 sin2  2 cos d =   2 0 2 4(1 – sin2 ) cos d =   2 0 2  2 cos2  .cos d awi, x = 2sin  dx = 2cos d x 0 2  0  2

†hvMRxKiY  Final Revision Batch '25 3 (M) GLv‡b, mij‡iLv, y = 4x ......(i) eμ‡iLv, y = x – x 2 .....(2) (i) bs I (ii) bs n‡Z cvB,  x 2 + 4x – x = 0  x 2 + 3x = 0  x(x + 3) = 0  x = 0, – 3 y = 4x X C X Y Y O y = x– x 2 x = 0 n‡j, (i) bs n‡Z cvB, y = 4  0  y = 0 x = – 3 n‡j, (i) bs n‡Z cvB, y = 4  (– 3)  y = – 12  (i) bs I (ii) bs ci ̄úi (0, 0) Ges (– 3, – 12) we›`y‡Z †Q` K‡i| GLb, (i) bs n‡Z, y = 4x = y1 (awi) (ii) bs n‡Z, y = x – x 2 = y2 (awi) wP‡Îi QvqvK...Z As‡ki †ÿÎdj, =  –3 0 (y1 – y2) dx =  –3 0 (4x – x + x2 )dx =  –3 0 (3x + x2 ) dx = 3     x 2 2 –3 0 +     x 3 3 –3 0 = 3 2  9 + 1 3 (– 27) = 27 2 – 9 = 27 – 18 2 = 9 2 eM© GKK (Ans.) 3| `„k ̈Kí-1: f(x) = 1 (4 + x2 ) 3 2 `„k ̈Kí-2: x 2 + y2 = 64 ; y = 5 [h‡kvi †evW©- Õ23] (K)  sin9x sin11x dx wbY©q Ki| (L) `„k ̈Kí-1 e ̈envi K‡i  4 0 f(x) dx wbY©q Ki| (M) `„k ̈Kí-2 Gi e„Ë I mij‡iLv Øviv Ave× ÿz`aZi As‡ki †ÿÎdj wbY©q Ki| mgvavb: (K)  sin9x sin11x dx = 1 2  2 sin11x sin9x dx = 1 2  {cos(11x – 9x) – cos(11x + 9x)} dx = 1 2  (cos2x – cos20x) dx = 1 2     sin2x 2 – sin20x 20 + c = sin2x 4 – sin20x 40 + c (Ans.) (L) †`Iqv Av‡Q, f(x)= 1 (4 + x2 ) 3 2   4 0 f(x) dx =  4 0 1 (4 + x2 ) 3 2 dx =  tan–1 2 0 2 sec2  (4 + 4tan2 ) 3 2 d awi, x = 2 tan  dx = 2 sec2  d x 0 4  0 tan–1 2 =  tan–1 2 0 2 sec2 d (4sec2 ) 3 2 =  tan–1 2 0 2 sec2  8 sec3  d =  tan–1 2 0 1 4sec d = 1 4  tan–1 2 0 cos d = 1 4 [sin] tan–1 2 0 = 1 4 [sin(tan–1 2) – 0] [⸪ x = 2 tan   = tan–1 2] = 1 4 sin    sin–1 2 5 = 1 4  2 5 = 1 2 5 (Ans.) 1 2 2 + 12 1 = 5 tan–1 2 2 (M) cÖ`Ë e„Ë, x 2 + y2 = 64  x 2 + y2 = 82 .....(i) Ges cÖ`Ë mij‡iLv, y = 5 ......(ii) (0, – 8) X X Y Y (– 8, 0) O (8, 0) (0, 5) (0, 8) y = 5 (i) bs I (ii) bs Øviv Ave× ÿz`aZi As‡ki †ÿÎdj = 2  8 5 xdy = 2  8 5 64 – y 2 dy = 2   2 sin–15 8 64 – 64sin2  8cos d awi, y = 8sin  dy = 8 cos d y 5 8  sin–15 8  2 = 2  8  8   2 sin–1 5 8 1 – sin2  . cos d = 128  2 sin–1 5 8 cos2  .cos d = 128   2 sin–1 5 8 cos2  d