Title Integration Engineering Question Bank Solution (HSC 26).pdf ✅

2  Higher Math 1st Paper Chapter-10 5. y = 2 – |x – 2| mgxKi‡Yi MÖvd‡K y = k †iLvwU x A‡ÿi mv‡_ mgvb `yB fv‡M fvM K‡i| k Gi gvb wbY©q Ki| [BUET 21-22] mgvavb: †`Iqv Av‡Q, y = 2 – |x – 2| y =   x hLb x – 2 < 0 4 – x hLb x – 2 > 0 2 – k k B(2, 2) O(0, 0) y = k C(4 – k, k) A(4, 0) X Y y = x I y = k Gi †Q`we›`y D(k, k) y = 4 – x I y = k Gi †Q`we›`y C(4 – k, k) kZ©vbymv‡i, BDC Gi †ÿÎdj = OACD UavwcwRqv‡gi †ÿÎdj  1 2  (4 – 2k) (2 – k) = 1 2  (4 – 2k + 4)  k  (4 – 2k)(2 – k) = (8 – 2k)k  (2 – k 2 ) = (4 – k)k  k 2 – 4k + 4 = 4k – k 2  k 2 – 4k + 2 = 0  k = 2 – 2 (Ans.) [GLv‡b, k = 2 + 2 MÖnY‡hvM ̈ bq KviY k Gi gvb 2 A‡cÿv †QvU n‡e] A_ev, OACD Gi †ÿÎdj =  k 0 (4 – y) dy –  k 0 y dy =  k 0 (4 – 2y) dy = [4y – y 2 ] k 0 = 4k – k 2 y = x I y = 4 – x Gi †Q`we›`y B(2, 2)  BCD =  2 k (4 – 2y) dy = [4y – y 2 ] 2 k = 8 – 4 – 4k + k2 kZ©g‡Z, 4k – k 2 = 4 – 4k + k2  2k2 – 8k + 4 = 0  k = 2  2  k = 2 – 2 (Ans.) [ k = 2 + 2 MÖnY‡hvM ̈ bq] 6. y 2 = 4ax Ges y = 2x Gi Ave× †ÿ‡Îi †ÿÎdj 3 eM© GKK n‡j, a Gi gvb KZ? [BUET 20-21] mgvavb: GLv‡b, y 2 = 4ax  4x2 = 4ax  4x2 – 4ax = 0  4x(x – a) = 0  x = 0, x = a y 2 = 4ax O y = 2x (a,2a) Y X  †ÿÎdj =  a 0 (y1 – y2) dx =  a 0 ( 4ax – 2x) dx =         2 a 2 3 x 3 2 – 2  x 2 2 a 0 = 4 3 a 2 – a 2 = 1 3 a 2 eM© GKK cÖkœg‡Z, 1 3 a 2 = 3  a 2 = 9  a =  3 (Ans.) 7.  1 – 1 x 2 4 – x 2 dx Gi gvb wbY©q Ki| [BUET 19-20; MIST 19-20] mgvavb:  1 – 1 x 2 4 – x 2 dx =   6 –  6 (2sin) 2 4 – 4sin2   2cos d =   6 –  6 4sin2  × 2cos × 2cos d awi, x = 2sin  dx = 2cos d x – 1 1  –  6  6 =   6 –  6 16 sin2 cos2  d = 4   6 –  6 (2sincos) 2 d = 4   6 –  6 sin2 2 d = 2   6 –  6 (1 – cos4) d = 2      – sin 4 4  6 –  6 = 2            6 –    –   6 – 1 4      3  2 –     – 3 2 = 2      3 – 1 4      3  2 –     – 3 2 = 2 3 – 3 2 (Ans.)

4  Higher Math 1st Paper Chapter-10 12. †hvMRxKiY Ki :  x 2 – 1 x 4 + x2 + 1 dx [BUET 15-16] mgvavb:  x 2 – 1 x 4 + x2 + 1 dx =  1 – 1 x 2 x 2 + 1 + 1 x 2 dx =  d     x + 1 x     x + 1 x 2 – 1 = 1 2 ln x + 1 x – 1 x + 1 x + 1 + c = 1 2 ln x 2 – x + 1 x 2 + x + 1 + c (Ans.) 13. †hvMR wbY©q Ki:  x 2 + 1 x 4 + 1 dx [BUET 14-15] mgvavb:  x 2 + 1 x 4 + 1 dx =  1 + 1 x 2 x 2 + 1 x 2 dx =  1 + 1 x 2     x – 1 x 2 + ( 2) 2 dx =  d     x – 1 x     x – 1 x 2 + ( 2) 2 = 1 2 tan–1       x – 1 x 2 + c (Ans.) 14. y 2 = x – 1 cive„Ë Ges 2y = x – 1 mij‡iLv w`‡q Ave× †ÿ‡Îi †ÿÎdj wbY©q Ki| [BUET 14-15] mgvavb: y 2 = x – 1; 2y = x – 1  y 2 = 2y  y(y – 2) = 0  y = 0, 2  x = 1, 5  †Q`we›`yØq (1, 0) I (5, 2)  A =  5 1 (y1 – y2) dx =  5 1       x – 1 –     x 2 – 1 2 dx =         (x – 1) 3 2 3 2 – x 2 4 + x 2 5 1 = 19 12 – 1 4 (1, 0) (5, 2) Y X O y 2 = x – 1 2y = x – 1 = 4 3 eM© GKK (Ans.) 15. x A‡ÿi mv‡c‡ÿ †hvMRxKiY K‡i x = y 2 Ges y = x – 2 †iLv `y‡Uv w`‡q Ave× †ÿ‡Îi †ÿÎdj wbY©q Ki| [BUET 12-13, 10-11] mgvavb: x = y 2 O Y X A(1, – 1) C(2, 0) B(4, 2) y = x – 2 GLv‡b, x = y 2  x = (x – 2)2  x 2 – 5x + 4 = 0  x = 1, 4 Ges y = – 1, 2  A =  1 0 x dx +  2 1 {– (x – 2)} dx +  4 0 x dx –  4 2 (x – 2) dx =     2 3 x 3 2 1 0 –     x 2 2 – 2x 2 1 +     2 3 x 3 2 4 0 –     x 2 2 – 2x 4 2 = 2 3 + 1 2 + 16 3 – 2 = 9 2 eM© GKK (Ans.) A_ev, y A‡ÿi mv‡c‡ÿ †ÿÎdj =  2 – 1 {(y + 2) – y 2 } dy =     y 2 2 + 2y – y 3 3 2 – 1 = 9 2 eM© GKK (Ans.) 16.   2  3 dx 1 + sinx – cosx [BUET 11-12] mgvavb: I =  dx 1 + sinx – cosx =  dx 1 + 2tan x 2 1 + tan2x 2 – 1 – tan2x 2 1 + tan2x 2 =      1 + tan2 x 2 dx 1 + tan2x 2 + 2 tan x 2 – 1 + tan2 x 2 awi, z = tan x 2 dz = 1 2 sec2x 2 dx x  3  2 z 1 3 1 =  sec2 x 2 dx 2 tan2 x 2 + 2 tan x 2 =  1 2 sec2 x 2 dx tan x 2     tan x 2 + 1