giai tich 12 on tap cuoi nam 2 cot

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%KLM %KNL  %?    5  A O  P-    %KLD;$ A Q %KLD;8 - =R=/2STB& ,U8-PV AQ'B0:'*# &'()5 TP-   # &  8  * 28*V %KNWX=Y5 A ML * *0'Z  R#-B 4  >7  *9  5   T  P  -     V %L?M[  [\T2V]W2=TA^_`^V L [\T2V]=Ta_`^V 2 A0b0 0  *8  5>  F G$ A M  '8  3B &     ,  4 ' , '8$ AN&'()     T0c D;V A%- ' ,L0b0 !  *8    '8 $ % [dT2V]2 W [dT2V]42 [dT2V]2 e$M* L$M 458'88" f58eg$ K()TD;[hJV .i [dT2V]2 W 4[ \T2V]W2=TA^_`^V [dT2V]424[ L \T2V]=Ta_`^V 2 [dT2V]24[[ \T2V]82=TA^_`^V Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n L JI [\T2V]82=TA^_`^V %KNJj3* =Q'()$ A O  P-    %KWD;$ AQ'B8*R#- /2Sk>=R =54/430'( 4UL*'(4UWD;$ A  O  P  -    # &*l[j'(4U$ [dT2V]2 W `l [dT2V]42`l [dT2V]2`l T*9  lm     5nV A%-#& '(4UTD;V$ K(4ULTD;[hJV l[j$ : ;. ;0 "*<%& JI WI JI AQ'(4UL*WTD;V   [  -     [*5$ AX=Y4f *  #+        * B=8 &  6$  T;8  *  'o /#  '  /  f   #+5p2'(8 -V %KNq./01'(4U A%[4*0WTD;V;8 *B& 90:- P28 4!-* $ %KW       $ %KNLj4f *'8 AQ'[0b0R#-     =      L TD;V Aj8m *0*[[$ %KNW      W TD;V AOP- # &      *   8-m AlRU A%[*0 AN&L TD;V A%[*0 AN&$ K(4UWTD;[hqV l[jTD;V lrg X- \T2V= slRU \T2V02  4  *  #+    [ dT2V\T2V*?0dT2V]dIT2V02 ]\T2V02$ .0W [tW202]2 W `l_2rTA^_`^V [tL[0]4`l_rTa_`^V [t80]`l_rTa_`^V W$ L .0J tT82VI02]tTAV02]82`l W 5m5a l[jTD;V Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n W t\T2V02]dT2V`l t\IT2V02]\T2V`l t5\T2V02]5t\T2V02 HI `a A%i-6 $ %KNJJ AO[P-#& $ A%KqTD;V T8* 90:- PV A N & 0 *8 D;$ A J l[jl6- ' ,28$ : ;. ;0 "*<%& qI LqI Aj8m *0qD;*P- $ AM/2S28 *$ %KJ  D        A;8* 8- #  &  *  Q  / '(4UJ$ Aj8'(4UmL **0HD;T[- V %KqF Al8- 8'3HD;$ A=78#1*[- 5&=45>*Q $ AQ'B' =5> L  !"#$ AX/#8- mP- 4*0uD;*L*0 5*88$ A  %i  [  */  01  4 8 Zm ' *8,#$ A%- .0 .92rTa_`^V B tTJ2`W[2V02] JtTV02`WtL[202] AJ82`W42`l AN&'(4U A*0H A%KH A&=45> AlRU5> A*0u []Wt2 W 02`t2 AW[J 02] W[J2 J `J2 L[J `l$ []Jt8202AL[J 2 02 LJ2 ]J2A`l J4J []L[uTW2`JV u `l .0q? \T2V]J2`W[2= 58Ta_`^V ; X!-'v 28$ J$D K(4UJTD;[hHV .0HTD;[huV q$F3  !"# F TD;[hwV .0u L [txW2 W `yz02=Ta_`^V  J {2 W [tTJ82AJ 2AL V02=TA^_`^V [tWTW2`JV H 02 Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n J tx\T2V|T2Vz02]t\T2V02|tT2V02 0[]t2[8202 ]A4[82[`l 0[t202 5=>  ! - Kiến thứcN 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C% 432W2 V x e x       −   43 x e x ? ?        − ;J;/#W ;*qBk B4L+VV0V V$ ;,U  n m n m aa = _ nm n m a a a − =  c b c a c ba += + ( ) ∫ ± dxxgxf MNMN ] ∫ ∫ ± dxxgdxxf MNMN 0V$8]• ∫ + dxbaxf MN ]• V M?MNN  xx −+ 8'3 B/#78 ,U;. $8] ( ) [ ] MC%NC% ?  baba −++ V ∫ + dxbaxf MN CbaxF a ++= MN  V'k*# O4? V dxxxxdx x xx ∫ ∫         ++= ++ − J  P  J ? J  ] H w W J u J J u J H w W x x x C + + + V ∫ − dx e x x ? ] ∫ ∫ − −       dxedx e x x ? ] W 4W L T4W LV x x C e + − + − 0V M?C%QNC% ?  J@C=C% xxxx += ( ) ∫ ∫ ∫ +=⇒ xdxxdxxdxx ?C%QC% ?  J@C=C% Cxx +       +−= ?@CQ@C     V J W L W x e C − − + Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n } x B x A ? − + + = %v3*#= 'ƒCW* = ( ) M?MNN ? xx BAxBA −+ +++− Kƒ' ,      = = ⇔    =+ =+− J ? J   >? 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M=NMNM=N −+ = ff  ?Q= W$DTV] W `‡W_   ∈ xL_Hz J$*?DITV]W`L MDTV43 f(t) = Wt + L S ?Q>?QMNM=N =−=−= SS K()?8 ‘l8]\T2V415p'k 0='8x_z$%?#’9 C'(]\T2V=18* ' !’2]_2]' ,-4 "7%""-%&@%&T%qwD;=LaWV“ ?;+%"%&"`-R"'"[% ‘l8\T2V441='8x_ z$;ƒdT2V43 \T2V='8x_z$% dTV‡dTV' ,-4#+Q' T#+2'(='8x_zV \T2V5U T V b a f x dx ∫ Vậy: T V T V T V T V b b a a f x dx F x F b F a = = − ∫ lRU]8"”> 9 T V a_ T V T V a b a a a b f x dx f x dx f x dx = = − ∫ ∫ ∫ .iWV a?J JJ ?  ?  J? =−== ∫ xdxx V ∫ =−=−== e e entldt t   >8%8%  M/2S ` T V b a f x dx ∫ Š#13*8\ /5p#13*82 $ `M\T2V41*5p+ ='8x_z? Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n La [...]... cos 2 x = ? ⇒ sin 2 xcox 2 x = ? 2 1 2 a ) ∫ 1 − x dx = ∫ (1 − x )dx + ∫ ( x − 1)dx 0 0 1 1 2   x2  x2   + =x−    2 − x = 1  2 0   1 π 2 π 2 0 1 sin 2 x = (1 − cos 2 x ) 2 20 b) sin 2 xdx = 1 (1 − cos 2 x )dx ∫ ∫ Giải câu b) π a + b a b am = + , n = a m −n c c c a π 2 1 1 = ( x − sin 2 x ) = 4 2 2 0 ln 2 ∫ c) 0 1 cos x = (1 + cos 2 x ) 2 sin 2 x cos 2 x 1 = (sin 2 x + sin 2 x cos 2 x... (a + b )2 = a2 + 2ab + b2 (2x + 1 )2 = 4x2 + 4x + 1 1 ∫1+ x 2 dx = 0 π 4 π 4 1 ∫ 1 + tan 0 2 dt π = ∫ dt = 2 4 x cos t 0 1 1 0 0 ( ) 2 2 HĐ4 : a) ∫ (2 x + 1) dx = ∫ 4 x + 4 x + 1 dx 1 u = 2x + 1 ; du = u’dx = 2dx  4x3  13 = + 2x2 + x  =  3  3  0 b) u = 2x + 1 ⇒ du = 2dx Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n 12 u (1) c/ Tính: ∫ (2x + 1)2dx = g (u ) du và so u (0) 1 2 u du 2 c) u(0)=1, u(1) = 3 sánh với kết... 2) dx =  −  3  2 của 2 nhóm Nhận xét và sửa chữa   −1 2 −1 Bài tập tương tự: Tính Ghi nhận bài tập về nhà c) y = ( x − 6) 2 , y = 6 x − x 2 diện tích hình phẳng • ( x − 6) 2 − (6 x − x 2 ) = 0 ⇔ 2 x 2 − 18 x + 36 = 0 giới hạn bởi các đường Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n 22 a) y = x 2 , y = x + 2 và x = 2, x = 4 6 2 • S = ∫ ( 2 x − 18 x + 36)dx c) y = ( x − 6) 2 , y = 6 x − x 2 và x = 1, x = 5... hành cách chia đa thức 3 a) ∫ ( 1 + 3 x ) 2 dx Đặt u = 1+ 3x ⇒ du = 3dx 0 +x=0 ⇒ u=1 +x=1 ⇒ u=4 1 3 4 5 1 2 ∫ ( 1 + 3 x ) dx = 3 ∫ u 2 du = 15 u 2 0 1 b) x3 −1 x2 + x +1 = x +1 x2 −1 x2 + x + 1 x2 + x 1 Ghi bảng 1 3 2 1 2 ∫ 0 4 =4 1 2 15 1 2 x3 −1 1   dx = ∫  x + dx 2 x +1 x −1 0 1 x+1 x  x2 2 1 3 = + ln( x + 1)  = + ln  2  2  0 8 2 ln(1 + x ) dx x2 1 c) ∫ Gợi ý: dùng pp tích phân từng... tan2t = cos 2 t nên đặt x = tan t Hãy áp 2 I = ∫ (2 x + 1) dx 0 a/ Hãy tính I bằng cách khai triển (2x + 1 )2 b/ Đặt u = 2x + 1 Biến đổi (2x + 1)2dx thành g(u)du 2 dx π π 1 < t < ⇒ dx = dt 2 2 cos 2 t + khi x = 0 ⇒ t = 0 π x =1 ⇒ t = 4 + Đặt x = tan t , - 1 1 1 ∫1+ x 0 dụng quy tắc trên giải vd5 Hoạt động 4 :Cho ∫ f ( x )dx = α f (ϕ (t ))ϕ ' (t )dt ∫ Tiến hành HĐ4 (a + b )2 = a2 + 2ab + b2 (2x +... x cos 2 x ) 2 1 1 = sin 2 x + sin 4 x 2 4 2 e 2 x +1 + 1 dx = ex ln 2 ln 2 0 0 x +1 ∫ e dx + ∫e −x dx ln 2 1  1 1  =  e x +1 − x  = e ln 2+ 1 − ln 2 − e + 1 = e + 2 e 0 e  d) π 2 ∫ sin 2 x cos xdx = 0 π π 1 1 ∫ sin 2 xdx + 4 ∫ sin 4 xdx 20 0 π 1 1  = − cos 2 x + cos 4 x  = 0 16 4 0 HĐ4: Giải bài tập 3 Hoạt động của Gv Hãy nhắc lại quy tắc tính tích phân bằng đổi biến dạng 2 Đặc u = x +... Trình bày lời giải Nhận xét đánh giá 1 − x 2 = 1 − sin 2 t = cos t x = 0 ⇒ sint = 0 ⇒ t = 0 π x = 1 ⇒ sint = 1 ⇒ t = 2 Khi đó 1 ∫ 0 π 2 1 − x 2 dx = ∫ cos 2 tdt = 0 π 2 1 (1 + cos 2t )dt 2 0 Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n 16 π 2 1 1 = ( t + sin 2t ) 2 2 0 = π 4 Tiết :57 Luyệ n tậ p 2 TÍ CH PHÂN 1,Mục tiêu: 1 Kiến thức: Luyện giải các bài tập về tính tích phân 2 Kĩ năng: Vận dụng thành thạo các phương... 2  2 ∫ sin x dx = 2  ∫ sin x dx + ∫ sin x dx  =   0 π 0  π 2   = 2  ∫ sin xdx − ∫ sin xdx  = - - - = 4 2   π 0  2 2 -0 2 1 − cos 2 x dx = ∫ 2 sin 2 x dx π  sin x , nêìu0 ≤ x ≤ π sin x =  - sinx, nêìuπ ≤ x ≤ 2 sin x IV Củng cố: + Gv nhắc lại các khái niệm và quy tắc trong bài để Hs khắc sâu kiến thức Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n 11 + Dặn BTVN: ®äc SGK, trang 109, 110 2 TÍCH... bảng a) y = 1 – x , y = 0 1 – x2 = 0 ⇔ x = - 1; x = 1 2 1 ( V = π ∫ 1− x −1 ) 2 2 1 dx = π ∫ (1 − 2 x 2 + x 4 )dx −1 1  2x3 x5  16π  = = π x − +   3 5  −1 15  b) y = cosx, y = 0, x = 0, x = π π ππ 2 V = π ∫ cos xdx = ∫ (1 + cos 2 x )dx 0 2 π π 0 π π 2 = x + sin 2 x = 2 0 4 2 0 c) y = tanx, y = 0, x = 0, x = π 4 π 4 π 4  1  V = π ∫ tan 2 xdx = π ∫  − 1 dx 2  0 0  cos x π  π 4 = π (... 3 2 luyện giải bài tập: Hđ1 Giải bài tập 1: Tính diện tích hình phẳng giới hạn bởi các đường Hoạt động của Gv Hoạt động của Hs Ghi bảng 2 Chia hs thành 2 nhóm a) y = x , y = x + 2 mỡi nhóm giải một câu • x2 – (x + 2) = 0 ⇒ x2 – x – 2 = 0 Cho tiến hành hoạt động Tiến hành hoạt động ⇒ x = - 1, x = 2 nhóm nhóm 2 2  x3 x2  9 2 Hãy nhận xét bài làm Trình bày lời giải − 2x  = • S = ∫ ( x − x − 2) dx . 90:- T4"#4 L4PV AM/2S *28 5>$ %Kwl AOP-4 `K() `N Z## m'8 * # Z ## Q#P$ t 420 2 ]24 2A2` AL0b 0$ A78 P8* [K" ]2 W *0*] 820 2 B0]W2 02* ] 2 08'B t2 W 820 2 ]2 W 2A tW 220 2 K" ]2* 0* ]2 02 0] 02* ]A 82 t 220 2]A2 82` t 820 2 ]A2 82` 2`l ./5> ]2 W 2 AWTA2 82` 2 `lV AM478P 8*$ .iLa [t2 W 820 2 ; X!-'v 28 $ q$%. 4! ‚-*  2 4!$ AQ*0hP- A tT282V I  02 ] 2 8` lL t8 2 02 ]D 2 `lW i8'B t2 2 02 ]A 2 8 2 ` 2 `lTl]AlL` lWV AN&'(4U Al6'(4U A*01 [K"€] 2 0*]7 2  02 ./0] 02 *]7 2 t27 2  02 ] 2 $7 2 At7 2  07A 2 7 2 A7 2 `l [K"] 2 0*]8 02 0] 02 *] 2 i8'B t 2 8 2 02 ] 2  2 A t 02 ] 2  2 `8 2 `l [K"]4 2 0*] 02 0]L[W 02 *] 2 i8'B W$N. !"# F TD;[hwV .0u L [txW2 W `yz 02 =Ta_`^V  J {2 W [tTJ8 2 AJ 2AL V 02 =TA^_`^V [tWTW2`JV H 02 Gi¸o ¸n Gi¶i tÝch 12 c¬ b¶n J txT2V|T2Vz 02] tT2V 02 |tT2V 02 0[]t 2[ 8 2 02 ]A4[8 2[ `l

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