Tiêu đề Integration Engineering Practice Sheet Solution.pdf ✅

†hvMRxKiY  Engineering Practice Sheet Solution 1 10 †hvMRxKiY Integration WRITTEN weMZ mv‡j BUET-G Avmv cÖkœvejx 1. x = e2y + 1 , x = e– y + 1 , y = – 2 Ges y = 1 Øviv Ave× †ÿ‡Îi †ÿÎdj wbY©q Ki| [BUET 23-24] mgvavb: X (0, 0) (e, 0) x = e2y + 1 y = 1 y = – 2 x = e–y + 1 Y  wb‡Y©q †ÿÎdj =  0 – 2 (e–y + 1 – e 2y + 1)dy +  1 0 (e2y + 1 – e –y + 1)dy =     – e 1–y – 1 2 e 2y + 1 0 – 2 +     1 2 e 2y + 1 + e1 – y 0 1 = 22.998 eM© GKK (Ans.) 2. y = 1 x – 1, x-Aÿ, x = 1 2 Ges x = 2 †iLv ̧wj Øviv mxgve× †ÿ‡Îi †ÿÎdj wbY©q Ki| [BUET 22-23] mgvavb: x = 1 x= 1 2 x = 2 y= 1 x Y – 1 X  wb‡Y©q †ÿÎdj =  1 1 2 y dx +  2 1 – y dx =  1 1 2     1 x – 1 dx –  2 1     1 x – 1 dx = 1 2 eM© GKK (Ans.) 3. ( tanx + cotx) dx wbY©q Ki| [BUET 21-22] mgvavb:  ( tanx + cotx) dx =       sinx  cosx + cosx sinx dx =  sinx + cosx sinxcosx dx = 2  (sinx + cosx) 2sinxcosx dx = 2  (sinx + cosx) 1 – (sinx – cosx) 2 dx = 2  dz 1 – z 2 = 2 sin–1 z + c awi, sinx – cosx = z  1 – 2sinxcosx = z2  (sinx + cosx) dx = dz = 2 sin–1 (sinx – cosx) + c (Ans.) 4. y = 1 – x 2 †iLv I X A‡ÿi g‡a ̈ [0, 2] e ̈ewa‡Z Ave× †ÿ‡Îi †gvU †ÿÎdj wbY©q Ki| †jLwPÎwU AvuK| AZtci gvb wbY©q Ki:  2 0 (1 – x 2 ) dx [BUET 21-22] mgvavb: (1, 0) (0, 1) Y X (2, 0)  wb‡Y©q †ÿÎdj =  1 0 (1 – x 2 ) dx +  2 1 {– (1 – x 2 )} dx = 2 3 + 4 3 = 2 eM© GKK (Ans.) Avevi,  2 0 (1 – x 2 ) dx =     x – x 3 3 2 0 = 2 – 8 3 = – 2 3 (Ans.)
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.)
†hvMRxKiY  Engineering Practice Sheet Solution 3 8. x = b †iLvwU y = (1 – x)2 , y = 0 Ges x = 0 Øviv Ave× †ÿ·K R1 (0  x  b) Ges R2 (b  x  1) AskØq Ggbfv‡e wef3 K‡i †hb, R1 – R2 = 1 4 nq| b Gi gvb wbY©q Ki| [BUET 18-19; MIST 18-19; RUET 15-16] mgvavb: R1 =  b 0 (1 – x)2 dx =     – (1 – x) 3 3 b 0 = – (1 – b) 3 3 + 1 3 Ges, R2 =  1 b (1 – x)2 dx (1,0) x = b (0,1) (0,0) Y X =    –  (1 – x) 3 3 1 b = (1 – b) 3 3  R1 – R2 = – 2(1 – b) 3 3 + 1 3 = 1 4  – 2 3 (1 – b)3 = 1 4 – 1 3  – 2 3 (1 – b)3 = – 1 12  (1 – b)3 = 1 8  1 – b = 1 2  b = 1 2 (Ans.) 9. gvb wbY©q Ki:  e mtan–1x (1 + x2 ) 2 dx [BUET 16-17] mgvavb: awi, tan–1 x =   x = tan  dx = sec2  d  I =  e m (1 + tan2 ) 2 sec2  d =  e m sec2  d =  e m cos2  d = 1 2  e m (1 + cos2) d = 1 2  e m d + 1 2 e m cos2 d = 1 2me m + 1 2 e m m 2 + 22 {mcos2 + 2sin2} + c [ ] ∵  e axcosbx dx = e ax a 2 + b2 (acosbx + bsinbx) = 1 2me mtan–1x + e mtan–1x 2(m 2 + 4) {mcos(2tan–1 x) + 2sin(2tan–1 x)} + c (Ans.) 10. y 2 = ax Ges x 2 + y2 = 4ax †iLv؇qi AšÍeZ©x GjvKvi †ÿÎdj wbY©q Ki| [BUET 16-17] mgvavb: †Q`we›`y wbY©q: x 2 + ax = 4ax  x = 0, 3a  y = 0,  3a  P  (3a, 3a) X Y O P(3a 3a)  †ÿÎdj = 2 ×  3a 0 ( 4ax – x 2 – ax) dx = 2 ×  3a 0 ( 4a2 – (x – 2a) 2 – ax) dx = 2 ×       ( ) x – 2a 2 4ax – x 2 + 2a2 sin–1 ( ) x – 2a 2a – 2 3a (ax) 3 2 3a 0 = 2 ×     a 2 × 3a + 2a2 ×  6 – 2 3a 2 + a 2 =     2 3  – 3 3 + 2 a 2 =     8 3  – 3 3 a 2 eM© GKK (Ans.) 11. †hvMRxKiY Ki:  dx (cosx + sinx) [BUET 15-16; MIST 15-16] mgvavb:  dx (cosx + sinx) =  dx 1 – tan2 x 2 1 + tan2 x 2 + 2 tan x 2 1 + tan2 x 2 =  sec2 x 2 dx     – tan2 x 2 + 2tan x 2 + 1 =  2dz (– z 2 + 2z + 1) = – 2  dz (z 2 – 2z + 1) – 2 awi tan x 2 = z  2 dz = sec2x 2 dx = – 2  dz (z – 1) 2 – ( 2) 2 = – 2  1 2 2 ln z – 1 – 2 z – 1 + 2 + c = – 1 2 ln tan x 2 – 2 – 1 tan x 2 + 2 – 1 + c (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