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

†hvMRxKiY  Engineering Question Bank Solution (HSC 26) 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.)