Tiêu đề Integration Varsity Practice Sheet Solution.pdf ✅

†hvMRxKiY  Varsity Practice Sheet Solution 1 10 †hvMRxKiY Integration weMZ mv‡j DU-G Avmv cÖkœvejx 1. y 2 = 4x I y = x Øviv Ave× †ÿ‡Îi †ÿÎdjÑ [DU 23-24, 10-11, 08-09, 05-06; JU 14-15; RU 10-11] 3 8 eM© GKK 8 3 eM© GKK 3 eM© GKK 8 eM© GKK DËi: 8 3 eM© GKK e ̈vL ̈v: y 2 = 4x = 4  1(x)  y 2 = 4ax  a = 1 y = x  y = mx  m = 1  wb‡Y©q †ÿÎdj = 8a2 3m3 = 8  1 2 3  1 3 = 8 3 eM© GKK 2.  x 1 – x 2 dx Gi †hvMR wbY©q Ki- [DU 23-24; RU 23-24; KU 08-09; RUET 05-06] – 1 – x 2 1 – x 2 1 + x2 – 1 + x2 DËi: – 1 – x 2 e ̈vL ̈v:  x 1 – x 2 dx = – 1 2  – 2x 1 – x 2 dx = – 1 2  d(1 – x 2 ) 1 – x 2 = – 1 2  2 1 – x 2 + c = – 1 – x 2 + c 3.  dx e x + e–x = †KvbwU? [DU 22-23, 09-10; KU 19-20, 16-17, 14-15; JU 17-18, 10-11; RU 09-10] sin–1 (ex ) 1 1 + ex cos–1 (e–x ) tan–1 (ex ) DËi: tan–1 (ex ) e ̈vL ̈v:  dx e x + e–x =  e x dx e 2x + 1 =  dz 1 + z2 awi, e x = z  e x dx = dz = tan–1 z + c = tan–1 (ex ) + c 4.  –1 –2 (x + |x|) dx = ? [DU 21-22] 3 0 1 6 DËi: 0 e ̈vL ̈v:  –1 –2 (x + |x|) dx =  –1 –2 (x – x) dx    ⸪ |x| =    – x ; x < 0 x ; x  0 =  –1 –2 0 dx = 0 5.  2 0 |x – 1|dx = ? [DU 20-21] 0 1 2 1 2 DËi: 1 e ̈vL ̈v:  2 0 |x – 1|dx =  1 0 – (x – 1) dx +  2 1 (x – 1) dx =    –  x 2 2 + x 1 0 +     – x + x 2 2 2 1 = – 1 2 + 1 – 2 + 4 2 + 1 – 1 2 = 1 1 2 |x – 1| O X Y    ⸪  |x – 1| =   – (x – 1) ; x < 1 x – 1 ; x ≥ 1 6. [0, 2] e ̈ewa‡Z y = x – 1 Ges y = 0 †iLv Øviv Ave× A‡ji †gvU †ÿÎdj KZ? [DU 19-20]  2 0 (x – 1) dx  2 0 |x – 1| dx 2  2 1 (1 – x) dx 2  2 0 (x – 1) dx DËi:  2 0 |x – 1| dx
2  Higher Math 1st Paper Chapter-10 e ̈vL ̈v: 2 y = x – 1 1 O – 1 O 1 2 y = |x – 1| X Y Y X X Y  wb‡Y©q †ÿÎdj =  2 0 |x – 1| dx [⸪ †ÿÎdj (+ ve)] 7.  dx (e x + e–x ) 2 = ? [DU 19-20] 1 2(e 2x+1 + 1) + c – 1 2(e 2x + 1) + c 1 2e2x + c – 1 2e2x + c DËi: – 1 2(e 2x + 1) + c e ̈vL ̈v:  dx (e x + e–x ) 2 = 1 2  2e2x dx (e 2x + 1) 2 = 1 2  dz z 2 awi, e 2x + 1 = z  2e2x dx = dz = – 1 2 . 1 z + c = – 1 2(e 2x + 1) + c 8.  tan(sin–1 x) 1 – x 2 dx =? [DU 18-19] sec2 (sin–1 x) + c sec(sin–1 x) + c ln|sec(sin–1 x)| + c ln|tan(sin–1 x)| + c DËi: ln|sec(sin–1 x)| + c e ̈vL ̈v:  tan(sin–1 x) 1 – x 2 dx =  tanz dz awi, sin–1 x = z  dx 1 – x 2 = dz = ln|secz| + c = ln|sec(sin–1 x)| + c 9.  e x dx = ? [DU 18-19] 2 3 (e x ) 3 2 + c 1 2 e x + c 2e x 2 + c e x 2 + c DËi: 2e x 2 + c e ̈vL ̈v:  e x dx =  e x 2 dx = e x 2 1 2 + c = 2e x 2 + c 10. y = x 2 , x Aÿ, x = 1 Ges x = 3 Øviv Ave× †ÿ‡Îi †ÿÎdj KZ? [DU 17-18; JU 20-21] 26 3 25 3 8 9 DËi: 26 3 e ̈vL ̈v: y = x 2  wb‡Y©q †ÿÎdj =  3 1 y dx =  3 1 x 2 dx =     x 3 3 3 1 = 3 3 3 – 1 3 3 x = 1 x = 3 Y X X Y O y = x2 = 26 3 eM© GKK 11.  dx x x 2 – 1 = f(x) + c n‡j, f(x) mgvb- [DU 17-18] sin x sin–1 x cos x sec–1 x DËi: sec–1 x e ̈vL ̈v:  dx x x 2 – 1 = sec–1 x + c [m~Î]  f(x) = sec–1 x 12. y = x, y = 0 †iLvØq Ges x 2 + y2 = 16 e„Ë Øviv cÖ_g PZzf©v‡M Ave× †ÿ‡Îi †ÿÎdj KZ? [DU 16-17] 2 eM© GKK 3 eM© GKK 4 eM© GKK 5 eM© GKK DËi: 2 eM© GKK e ̈vL ̈v: x 2 + y2 = 16  x 2 + y2 = 42  e ̈vmva©, r = 4, †K›`a (0, 0) (0, 0) X Y O y = x X Y x = y, y = 0 I e„Ë Øviv Ave× †ÿÎdj e„‡Ëi †gvU †ÿÎd‡ji AvU fv‡Mi GK fvM|  wb‡Y©q †ÿÎdj = 1 8 r 2 = 1 8    (4)2 = 2 eM© GKK
†hvMRxKiY  Varsity Practice Sheet Solution 3 13. hw`  4 1 f(x) dx = 5 nq, Z‡e  1 0 f(3x + 1) dx Gi gvbÑ [DU 16-17] 5 4 4 3 5 3 5 DËi: 5 3 e ̈vL ̈v: awi, 3x + 1 = z  3dx = dz  dx = 1 3 dz x 0 1 z 1 4   1 0 f(3x + 1) dx = 1 3  4 1 f(z) dz = 1 3  5     ⸪  4 1 f(x) dx = 5   1 0 f(3x + 1) dx = 5 3 A_ev,  4 1 f(x) dx = 1 3  –  3  –  3 f(3x + 1) dx = 1 3  1 0 f(3x + 1) dx   1 0 f(3x + 1) dx = 5 3 14.  e x (1 + x) cos2 (xex ) dx = f(x) + c; f(x) = ? [DU 15-16, 14-15, 11-12, 07-08] sin(xex ) tan(xex ) cot(xex ) sec(xex ) DËi: tan(xex ) e ̈vL ̈v:  e x (1 + x) cos2 (xex ) dx =  dz cos2 z awi, xex = z  e x (1 + x) dx = dz =  sec2 z dz = tanz + c = tan(xex ) + c  f(x) = tan(xex ) 15.  10 0 |x – 5| dx = ? [DU 15-16] 25 2 25 50 5 DËi: 25 e ̈vL ̈v:  10 0 |x – 5| dx     ⸪ |x – 5| =    – (x – 5) ; x < 5 x – 5 ; x ≥ 5 =  5 0 – (x – 5) dx +  10 5 (x – 5) dx =    –  x 2 2 + 5x 5 0 +     x 2 2 – 5x 10 5 = – 25 2 + 25 + 100 2 – 50 – 25 2 + 25 = 25 A_ev, O 5 10 X Y 5  10 5 |x – 5| dx = 2  1 2  5  5 = 25 16. y 2 = 16x Ges y = 4x Øviv Ave× †ÿ‡Îi †ÿÎdjÑ [DU 15-16, 11-12, 07-08] 2 3 eM© GKK – 2 3 eM© GKK 3 2 eM© GKK 1 3 eM© GKK DËi: 2 3 eM© GKK e ̈vL ̈v: y 2 = 16x  y 2 = 4  4x  y 2 = 4ax  a = 4 y = 4x  y = mx  m = 4  y 2 = 16x I y = 4x Øviv Ave× †ÿÎdj = 8a2 3m3 = 8  4 2 3  4 3 = 2 3 eM© GKK 17.  x 0 f(p) f(p) dp = ? [DU 15-16] 1 2 f 2 (x) 1 2 x 2 1 2 [{f(x)}2 – {f(0)}2 ] f(x) – f(0) DËi: 1 2 [{f(x)}2 – {f(0)}2 ] e ̈vL ̈v: I =  f(p) f(p) dp =  z dz awi, f(p) = z  f(p) dp = dz = 1 2 z 2 + c = 1 2 {f(p)}2 + c   x 0 f(p) f(p) dp =     1 2 {f(p)} 2 x 0 = 1 2 [{f(x)}2 – {f(0)}2 ]
4  Higher Math 1st Paper Chapter-10 18. y = x Ges y = x 2 Øviv Ave× †ÿ‡Îi †ÿÎdj (eM© GK‡K)Ñ [DU 14-15] 5 6 1 6 – 1 6 1 3 DËi: 1 6 e ̈vL ̈v: y = x 2  x 2 = 4  1 4 y  x 2 = 4ay  a = 1 4 y = x  y = mx  m = 1  y = x 2 I y = x Øviv Ave× †ÿ‡Îi †ÿÎdj = 8 3 a 2m 3 = 8 3      1 4 2  (1)3 = 1 6 eM© GKK 19.  1 0 ln(x + 1) x + 1 dx = ? [DU 14-15; RU 23-24, 17-18] 1 2 (ln2)2 1 2 ln2  0 DËi: 1 2 (ln2)2 e ̈vL ̈v:  1 0 ln(x + 1) x + 1 dx =  ln2 0 z dz = 1 2 [z ] 2 ln2 0 awi, ln(x + 1) = z  dx x + 1 = dz x 0 1 z 0 ln2 = 1 2 (ln2)2 20. y = – a 2 – x 2 I y = 0 Øviv Ave× †ÿ‡Îi †ÿÎdjÑ [DU 13-14] 1 4 a 2 1 2 a 2 a 2 1 2 a 2 DËi: 1 2 a 2 e ̈vL ̈v: y = – a 2 – x 2  y 2 = a2 – x 2  x 2 + y2 = a2 †K›`a (0, 0) I e ̈vmva©, r = a  y = – a 2 – x 2 I y = 0 Øviv Ave× †ÿ‡Îi †ÿÎdj n‡e x 2 + y2 = a2 e„‡Ëi †ÿÎd‡ji A‡a©K| y = – a 2 – x 2 X Y X Y O  wb‡Y©q †ÿÎdj = 1 2 r 2 = 1 2 a 2 eM© GKK 21. abvZ¥K x Gi Rb ̈ F(x) =  x 1 lnt dt n‡j F(x) = ? [DU 13-14] 1 x lnx xlnx xlnx – x DËi: lnx e ̈vL ̈v: F(x) =  x 1 lnt dt = [tlnt – t] x 1 = xlnx – x + 1  F(x) = lnx + 1 – 1 + 0 = lnx 22.  1 0 dx 2x – x 2 Gi gvb n‡e- [DU 08-09; Agri. Guccho 20-21; RU 19-20, 17-18, 11-12; JU 14-15, 11-12, 10-11; KU 04-05] –  2  4 3 4 5 2 DËi: mwVK DËi †bB e ̈vL ̈v:  1 0 dx 2x – x 2 =  1 0 dx 1 – (x 2 – 2x + 1) =  1 0 dx 1 – (x – 1) 2 = [sin ] –1 (x – 1) 1 0 = 0 –    –   2 =  2  mwVK DËi †bB| 23.  1 0 x 1 – x 2 dx Gi gvbÑ [DU 07-08; KU 13-14; JU 09-10] 1 2  2 1  2 DËi: 1 e ̈vL ̈v:  1 0 x 1 – x 2 dx = – 1 2  1 0 – 2x 1 – x 2 dx = – 1 2 [2 1 – x ] 2 1 0     ⸪  f(x) f(x) dx = 2 f(x) + c = – (0 – 1) = 1