mordkovitch-gdz-11-2001 (Алгебра и начала анализа 10-11 класс - Задачник - Мордкович)

А.А. Сапожниковк задачнику «Алгебра и начала анализа. Задачникдля 10-11 кл. общеобразовательных учрежденийА.Г. Мордкович, Л.О. Денищева, Т.А. Корешкова,Т.Н. Мишустина, Е.Е. Тульчинская —М.: «Мнемозина», 2001 г.»Глава 5. Первообразная и интеграл§ 37. Первообразная и неопределенный интеграл984.а) F(x) = x 3 ,f(x) = 3x 2 ,F' ( x ) = 3x 2 ;б) F(x) = x 9 ,F' ( x ) = 9 x 8 ;в) F(x) = x 6F' ( x ) = 6 x 5 ;г) F(x) = x 11F' ( x ) = 11x 10 ;а) F(x) = x 2 + x 3 ;F' ( x ) = 2x + 3x 2 ;б) F(x) = x 4 + x 11 ;F' ( x ) = 4 x 3 + 11x 10 ;в) F(x) = x 7 + x 9 ;F' ( x ) = 7 x 6 + 9 x 8 ;г) F(x) = x 13 + x 19 ;F' ( x ) = 13x 12 + 19 x 18 ;986.а) F(x) = 3 sin x ;б) F(x) = −4 cos x ;в) F(x) = −9 sin x;г) F(x) = 5 cos x ;F' ( x ) = 3 cos x ;F' ( x ) = 4 sin x ;F' ( x ) = −9 cos x ;F' ( x ) = −5 sin x ;987.а) f ( x ) = −985.б) f ( x ) =988.а) f ( x ) =б) f ( x ) =989.1x7x21F( x ) =;F( x ) = x + C ;;2 x6x;а) f ( x ) = 4 x 10 ;б) f ( x ) = −3x 6 ;в) f ( x ) = 5x 7 ;г) f ( x ) = −9 x 19 ;21+C ;x7F( x ) = − + C ;x;2F( x ) = 12 x + C ;4 11x +C ;113F( x ) = − x 7 + C ;75 8F( x ) = x + C ;89 20F( x ) = −x +C;20F( x ) =990.x 3 x 17++C;317x 10 x 34F( x ) =++C;1034а) f ( x ) = x 2 + x 16 ;F( x ) =б) f ( x ) = x 9 + x 33 ;991.в) f ( x ) = x 13 + x 18 ;F( x ) =x 14 x 19++C;1419г) f ( x ) = x + x 14 ;F( x ) =x 2 x 15++C;2151 x2++C;x 2x2111− 2 ; F( x ) = x + + C ;б) f ( x ) =x2 x xа) f ( x ) = −в) f ( x ) = −г) f ( x ) =992.11x212 xF( x ) =+x;+ x 3 ; F( x ) =+1;F( x ) = x + x + C ;а) f ( x ) = 4 x 3 − 6 x 2 ;F( x ) = x 4 − 2x 3 + C ;б) f ( x ) = 13x 6 + 9x 4 ;F(x) = 13г) f ( x ) = 12 x 10F( x ) = x 5 −а) f ( x ) = −3 sin x + 2 cos x ;б) f ( x ) =4−9;cos 2 x2;в) f ( x ) = −4 cos x +sin 2 x5г) f ( x ) = −13 sin x +;cos 2 x994.2sin x⎛а) f ( x ) = sin ⎜ 3x +⎝x7x5+9 +C;75x6+C;212 x 11 3x 8+ 3x 7 ; F( x ) =++C ;118в) f ( x ) = 5x 4 − 3x 5 ;993.1 x4++C;x 4π⎞⎟;6⎠F( x ) = 3 cos x + 2 sin x + C ;F( x ) = −4ctgx − 9 tgx + C ;F( x ) = −4 sin x − 2ctgx + C ;F( x ) = 13 cos x + 5tgx + C .1π⎞⎛F( x ) = − cos⎜ 3x + ⎟ + C ;36⎠⎝3995.996.⎛π⎞б) f ( x ) = cos⎜ − 2 x ⎟ ;4⎝⎠в) f ( x ) = cos(4 x − 3) ;F( x ) =x⎞⎛г) f ( x ) = sin ⎜ 2 − ⎟ ;2⎠⎝x⎞⎛F(x) = 2cos ⎜ 2 − ⎟ + C .2⎠⎝998.;(6 x + 1) 21б) f ( x ) =;(8x − 3) 21в) f ( x ) =;(7 x − 3) 21г) f ( x ) = −;(10 x + 2) 2а) f ( x ) =б) f ( x ) =997.1а) f ( x ) = −17x − 91F( x ) =F(x) = −1+C;7(7x − 3)F(x) =;г) ∫ −в)∫2x5dxx4∫ (x39cos 2 x16sin 2 x15dx = − 9tgx + C ;dx =16ctgx + C ;15+C.x2020dx = −+C.2xx=3 x +C.б)∫ − x 2 dx ==5 x +C.г)∫x4− cos x + C .41 ⎞x 10⎛dx=+ tgx + C .б) ∫ ⎜ x 9 +⎟10cos 2 x ⎠⎝x3в) ∫ ( x 2 + cos x )dx =+ sin x + C .31 ⎞x7⎛dx=− ctgx + C .г) ∫ ⎜ x 6 +⎟7sin 2 x ⎠⎝999.

а)1+C.10(10x + 2)27x − 9 + C ;72F( x ) = −42 − 3x + C .3в) ∫ 6cos xdx = 6sin x + C ;3dx1+C ;6(6x + 1)1+C;8(8x − 3)б) ∫ −∫21sin( 4x − 3) + C ;4F(x) = −а) ∫ 4sin xdx = − 4cos x + C ;а)1 ⎛π⎞sin ⎜ − 2x ⎟ + C ;2 ⎝4⎠F( x ) =;42 − 3xF( x ) = −+ sin x )dx =⎛ 1x32⎞⎜⎟xdxx=++C.+∫ ⎜⎝ 2 x ⎟⎠3⎞⎛ 1x2б) ∫ ⎜⎜+C .+ x ⎟⎟dx = x +2⎠⎝2 x1000. а)1 x4⎛ 13⎞+xdx=−++C .⎜⎟∫⎝ x2 ⎠x 41001.

а)б)1 x6⎛ 15⎞−+xdx=⎜⎟∫⎝ x2 ⎠ x + 6 +C .(2 − 9x )6∫ (2 − 9x ) dx = −1002. а)63(7 + 5x )13∫ (7 + 5x ) dx =14б)70б) y =12сos x+C.+C⎛π 1⎞M ⎜ ; ⎟;⎝ 3 4⎠1003. а) y = sin x,Y = − cos x + C ;73113= − + C ; C = ; Y = –cosx + .4442⎛π⎞M ⎜ ; −1⎟ ;4⎝⎠,Y = tgx + C ;−1 = 1 + C ;C = −2 ;Y = –tgx –2.⎛π ⎞⎝⎠111+ С ; С = ; Y = + sin x.222xY = −3ctg + C ; 0 = −3 + C ; C = 3 ;3в) y = cos x, M ⎜ ;1⎟ ; Y = sin x + C ; 1 =6г)y=1sin 2 (x / 3)xY = −3ctg + 3.3⎛ 3π ⎞, M ⎜ ;0 ⎟ ;⎝ 4 ⎠υ = 1+ 2t ;1004.s( t ) = t + t 2 + C ;5 = 2+4+C;C = −1 ;2s( t ) = t + t − 1 .1005.s( t ) =υ = −4 sin 3t ;42cos t + .331006. а) y' = x 4 − 3x 2 ;s( t ) =4cos t + C ;3y=2=4+C;3C=2;3x5− x3 + C .55б) y' = x 12 − 8x 71007. а) y' = sin x + 1 ;б) y' =1009.

а) y ' =13x2−91010.υ=y=−− 4x ;+ sin x;x2б) y' = −13 x 2++C.x24y = − − 2x 2 + C .x+x;x245x2− cos x ;6;2t + 1x 13− x8 + C .13y = − cos x + x + C .y = sin x − 9 x + C .б) y' = cos x − 9 ;1008. а) y' =y=;y=9− cos x + C.xy=5− sin x + C .xs( t ) = 6 2 t + 1 + C ;s(0) = 6 + C = 3 ;C = −3 ;s( t ) = 6 2 t + 1 − 3 .1011. a ( t ) = 2( t + 1) 2 ; υ( t ) =221( t + 1) 3 + C ; υ(0) = + C1 = 1 ; C1 = ;333211115( t + 1) 3 + ; s(t) = (t + 1) 4 + t + C2 ; s(0) = + C2 = 1 ; C2 = ;636633115s( t ) = ( t + 1) 4 + t + .636υ( t ) =221012. а) f ( x ) = sin x + cos x = 1 ;x2F( x ) = x + C .x2F( x ) = − cos x + C .б) f (x) = 2sin cos = sin x ;в) f ( x ) = 1 + tg 2 x =1F( x ) = tgx + C .;cos 2 x1;г) f ( x ) = 1 + ctg 2 x =sin 2 xxx1013. а) g(x) = 8sin cos = 4sin x;22F( x ) = −ctgx + C .⎛π ⎞M ⎜ ;3 ⎟ ;⎝2 ⎠G ( x ) = −4 cos x + C ; C = 3 ; G ( x ) = −4 cos x + 3 .x2б) g(x) = 2cos 2 − 1 = cos x,6⎛π ⎞M ⎜ ;16 ⎟ ;⎝3⎠G ( x ) = sin x + C ; 16 =x2в) g(x) = cos 2 − sin 2333+ C ; C = 16 −; G ( x ) = sin x + 16 −.222x= cos x,2M ( 0;7 ) ;G ( x ) = sin x + C ; 7 = 0 + C ; G ( x ) = sin x + 7 .г) g(x) = 1 − 2sin 2⎛π ⎞M ⎜ ;15 ⎟ ;⎝2 ⎠x= cos x,2G ( x ) = sin x + C ; 15 = 1 + C ; C = 14 ; G ( x ) = sin x + 14 .1014.

а)б)∫ (tg∫ (cos∫ (ctg222)x + 1 dx = ∫)1cos 2 xdx = tgx + C .x − sin 2 x dx = ∫ cos 2 x =)x + 1 dx = ∫1sin 2 x + C .21dx = −ctgx + C .sin 2 x11г) ∫ sin x cos xdx = ∫ sin 2xdx = − cos 2 x + C .24в)1111015. а) ∫ sin 2x sin 6xdx= ∫ (cos 4x − cos8x)dx= − sin 4x − sin 8x+C *.2816111б) ∫ sin 4x cos3xdx = ∫ (sin 7x + sin x)dx = − cos x − cos 7x + C *.2214111в) ∫ cos3x cos5xdx= ∫ (cos8x+ cos 2x)dx= sin2x+ sin 8x + C.241611⎛11⎞г) ∫ sin 2x cos8xdx= ∫ (cos 6x − cos10x)dx= ⎜ sin 6x − sin10x ⎟ +C .22⎝ 610⎠11⎛1 1⎞1016. а) ∫ sin 2 xdx = ∫ ⎜ − cos 2x ⎟ dx = x − sin 2x + C .24⎝2 2⎠21⎛1 1⎞⎛1 1⎞4б) ∫ sin xdx= ∫ ⎜ − cos 2x ⎟ dx= ∫ ⎜ − cos 2x+ cos 2 2x ⎟dx =22424⎝⎠⎝⎠1 11111⎛1 1⎞= ∫ ⎜ − cos 2x+ + cos 4x ⎟dx= x − sin 2x+ x+ sin 4x+C =8 8448 32⎝4 2⎠=3x 11− sin 2 x +sin 4 x + C .8 43211⎛1 1⎞2224⎝⎠1 13x 11⎛1 1⎞г) ∫ cos 4 xdx = ∫ ⎜ + cos 2x + + cos 4x ⎟dx = + sin 2x+ sin 4x+C .8 88 432⎝4 2⎠в) ∫ cos 2 xdx = ∫ ⎜ + cos 2x ⎟ dx = x + sin 2x + C .71017.

а) ∫б) ∫dx22sin x cos xcos 2xdx22sin x cos x=∫=∫sin 2 x+ cos 2 x1 ⎞⎛ 1dx= ∫ ⎜+ 2 ⎟ dx=tgx-ctgx+C .2sin x cos x⎝ cos x sin x ⎠2cos 2 x − sin 2 xsin 2 x cos 2 x1018. а) f ( x ) = 2 x + 3 ;21 ⎞⎛ 1dx = ∫ ⎜ 2 −⎟ dx =-ctgx − tgx+C .⎝ sin x cos 2 x ⎠б) f (x) = 12 ( 3x − 1)3 ;2F(x) = x + 3x + C ;f(x) = 0 ⇔ x = –3/2;F(–3/2) = 9/4 – 9/2 + C = 0;C = 9/4;F(x) = x 2 + 3x + 9 / 4 .F( x ) = (3x − 1)4 + C ;f(x) = 0 ⇔ x = 1/3;F(1/3) = C = 0 ;1019. а) f (x) = 2x,б) f (x) = 3x 3 ,y = x + 2,2F(x) = x + C ;F( x ) = (3x − 1)4 .y = x 02 + C + 2x 0 (x − x 0 ) =y = 3/ 4x 04 + C + 3x 03 (x − x 0 ) =2 xx 0 − x 02=+C ;2x 0 = 1 ;x 0 = 1/ 2 ;1y = x− +C = x+2;49C= ;49F( x ) = x 2 + .4= 3x 0 3 x − 21 4x0 + C;43x 0 3 = 3 ;x0 =1;y = 3x − (9 / 4) + C = 3x + 2 ;17;4317F( x ) = x 4 +.44C=1020.

y = 3 cos 3x + 6 sin 6 x ;Y = sin 3x − cos 6x + C ;3π6 = sin− cos 3π + C ;26 = −1 + 1 + C ;C=6;Y = sin 3x − cos 6x + 6 ;π⎛π⎞Y⎜ ⎟ = sin − cos π + 6 = 1 + 1 + 6 = 8 ;62⎝ ⎠§ 38. Определенный интеграл11021. а) ∫ x 3dx =2−38x4412−3y = 3x + 2;F(x) = 3/ 4x 4 + C ;=1161 465.−= − =4 81 ⋅ 4 4 81 3243б)dx∫ x23112= − +1= .x133=−1x5в) ∫ x dx =5−129г)dx∫24x4=−1=2 x9432 1 33+ =.5 5 5= 6−4 = 2 .π1022. а)ππ2∫ sin xdx = − cosxπ2π2в)∫π−cos xdx = sin xπ2−π2= 1 .

б)π4dx∫π cos 2 x−= tg xπ4−π4= 1+1 = 2 .4π2= 1+1 = 2 .г)2dx∫π sin 2 x= −ctg xπ2π4=1.4π21023. а)∫π−π3πcos 2xdx =11sin 2 x 2 π = .22−44ππ⎞ 32ππ 10 3⎛dx = −5ctg ⎜ x + ⎟ = −5ctg+ 5ctg =б) ∫.π3⎠0333⎝0 sin 2 ⎛ x + ⎞⎜⎟3⎠⎝5πв)∫π2 sin2π3г)xxdx = −6 cos337∫π cos 2 3xππ2= −3 + 3 3 .πdx =77 7tg3x 03 = 0 + = .33 3451024. а)∫13б)∫1dx2x − 1dx10 − 3x=−5= 2x − 1 = 3 − 1 = 2 .1210 − 3x3313=−24+2= .33392∫1025. а)4 x 5 − 3x 4 + x 3 − 1x2121 ⎞⎛dx = ∫ ⎜ 4 x 3 − 3x 2 + x − 2 ⎟dx =x ⎠1⎝2⎛x2 1 ⎞11= ⎜ x 4 − x3 ++ ⎟ = 16 − 8 + 2 + − 1 + 1 − − 1 = 9 .⎜⎟2x22⎝⎠1−1 5x 7−1− 4x 6 + 2x−1 ⎛2 ⎞2⎞⎛dx = ∫ ⎜ 5x 4 − 4x 3 + 2 ⎟ dx = ⎜ x 5 − x 4 − ⎟ =3x ⎠ −2⎝xx ⎠−2−2 ⎝= −1 − 1 + 2 + 32 + 16 − 1 = 47б) ∫3 6x 43− 4x 3 +7x 2 − 13⎛1 ⎞1⎞⎛dx= ∫ ⎜ 6x 2 − 4x+7 − 2 ⎟ dx = ⎜ 2x 3 − 2x 2 + 7x + ⎟ =2x⎠2⎝xx ⎠22⎝115= 54 − 18 + 21 + − 16 + 8 − 14 − = 34 .326в) ∫−1 3x 6г) ∫− 4x 5 − 7x 4 + 3x 2−1 ⎛3 ⎞dx = ∫ ⎜ 3x 2 − 4x − 7 + 2 ⎟ dx =x ⎠−2 ⎝x4−2−13⎞3⎛= ⎜ x 3 − 2x 2 − 7x − ⎟ = −1 − 2 + 7 + 3 + 8 + 8 − 14 − = 7,5.x ⎠ −22⎝(0)331026.

а) υ(t)=3t 2 − 4t+1 ; S(3)= ∫ 3t 3 − 4t+1 dt=t 2 − 2t 2 +t =27 − 18+3=12.б) υ( t ) =133 8 2 612dt =5t + 1 = − = .0 5 5555t + 1; S(3) = ∫5t + 103(0)3203в) υ( t ) = 4t − 6 t ; S(3) = ∫ 4t 3 − 6t 2 dt =t 4 − 2t 3 = 81 − 54 = 27г) υ( t ) =137t + 4; S(3) = ∫001210 4 6dt =7t + 4 = − = .077 7 77t + 436⎛ x3 x 2⎞61027. а) ρ(x) = x 2 − x − 1, l = 6; ∫ (x 2 − x − 1)dx = ⎜−− x⎟= 48.⎜⎟020⎝ 3⎠б) ρ(x) =31(x + 3)231, l = 3; ∫0 (x+ 3)2dx = −11 1 1=− + = .x+306 3 6⎛ x32⎞2828в) ρ(x) = − x 2 + 6x, l = 2; ∫ (− x 2 + 6x)dx = ⎜ − + 3x 2 ⎟ = − + 12 = .⎜ 3⎟330⎝⎠0г) ρ(x) =1011(2x + 1)2, l = 1; ∫120 (2x + 1)1dx = −11 1 1=− + = .2(2x + 1) 06 2 33⋅3=10,5 (в ответе задачника опечатка).231028.

а) ∫ f (x)dx=3 ⋅ 1+3 ⋅ 1+−23б)3 2⋅2= 6,5 .2∫ f (x)dx = 3 ⋅ 2 ⋅−241029. а) y = x 2 , y = 0, x = 4; S = ∫ x 2dx =0x33464.3=03б) y=x , y=0, x= − 3, x=1;01−30S=- ∫ x 3dx+ ∫ x 3dx= -x4401+−3x481 1 82 41= + = = .44 4 4 20x3в) y = x , y = 0, x = −3; S = ∫ x dx =3−30202= 9.−3x5г) y = x , y = 0, x = -1, x = 2; S = ∫ x dx =5−12424=−132 1 33+ = .5 5 52⎛ x4⎞+ 2x ⎟ = 8.1030. а) y = x + 2, y = 0, x = 0, x = 2; S = ∫ (x + 2)dx = ⎜⎜⎟0⎝ 4⎠02334⎛ x3⎞6432б) y = − x + 4x, y = 0; S = ∫ (− x + 4x)dx = ⎜⎜ − + 2x 2 ⎟⎟ = − + 32 = .330⎝ 3⎠04222⎛x3 ⎞32в) y = 4 − x , y = 0; S = ∫ (4 − x )dx = ⎜ 4x − ⎟ = .⎜⎟3−2⎝⎠ −2 3222⎛ x4 ⎞+x ⎟⎟⎝ 4⎠0г) y = − x 3 + 1, y = 0, x = 0, x = −2; S= ∫ (− x 3 +1)dx= ⎜ −⎜−21031. а) y =12, y = 0, x = 1, x = 2; S = ∫x91, y = 0, x = 1, x = 9; S = ∫б) y =x141, y = 0, x = 1, x = 4; S = ∫в) y =x1г) y =1x2120=4 + 2=6.−22dx = −111= − +1 = .x1221x91dx = 2 x = 6 − 2 = 4.1x41dx = 2 x = 4 − 2 = 2.1x−1, y = 0, x = −1, x = −3; S = ∫21−3 xdx = −21x−1=1−−31 2= .3 311πππ21032.

а) y = sin x, y = 0, x = ; S = ∫ sin xdx = − cos x 2 = 1.020π6π3б) y = cos 2x, y = 0, x = - , x = ;π3π1333+=S = ∫ cos 2xdx = sin 2 x 3 π =.2442−π−66ππ4ππв) y = cos x, y = 0, x = - , x = ; S = ∫ cos xdx = sin x 4 π = 2.44−π−ππ2x244x2г) y = sin , y = 0, x = , x = π; S = ∫ sin dx = −2cosπ2x2ππ2= 2.1ππ1033. а) y = 1 + cos x, y = 0, x = - , x = ;222π2π1π 1 π 1⎛ 1⎞⎛⎞2S = ∫ ⎜1 + cos x ⎟ dx = ⎜ x + sin x ⎟= + + + = π +1π 2 2 2 222⎠⎝⎠π⎝−−22б) y = 1 − sin 2x, y = 0, x = 0, x = π;ππ11 1⎛⎞S = ∫ (1 − sin 2x ) dx = ⎜ x + cos 2x ⎟ = π + − = π.22 2⎝⎠00π2в) y = 2 − 2sin x, y = 0, x = 0, x = ;π2πS = ∫ ( 2 − 2sin x ) dx = ( 2x + 2cos x ) 2 = π − 2.00x2г) y = 2 + cos , y = 0, x = 0, x =2π3 ⎛2π;32πx⎞x ⎞ 3 4π⎛S = ∫ ⎜ 2 + cos ⎟ dx = ⎜ 2x + 2sin ⎟ =+ 3.22⎠03⎠⎝0 ⎝2x41034.

а) ∫ x dx =402312= 4 ; S = 2 ⋅ 8 − 4 = 12 .0ππ2πππб) S = ⋅ 1 − ∫ sin xdx = + cos x 02 = − 1.22202в) S = 16 − ∫ x 2dx =16 −−2x332−28 8 32= 16 − − = .3 3 3ππг) S = ∫ sin xdx = − cos x 0 = 1 + 1 = 2.01035. а) y = x, y = -0,5x + 5, x = -1, x = 3;333x2⎛ 1⎞S= ∫ (−0,5x+5)dx − ∫ xdx= ⎜ − x 2 +5x ⎟ −⎝ 4⎠ −1 2−1−13−1919 1= − +15+ +5 − + =14.442 2б) y = 2x, y = x-2, x = 4;44−2−2S = ∫ 2xdx − ∫ (x − 2)dx = x 24⎛ x2⎞−⎜− 2x ⎟ = 16 − 4 − 8 + 8 + 2 + 4 = 18.⎟−2 ⎜ 2⎝⎠ −24x4в) y = − x, y = 3 − , x = −2, x = 1;1⎛x⎞x2 ⎞⎛S = ∫ ⎜ 3 − ⎟ dx − ∫ − xdx = ⎜ 3x −⎟⎜4⎠8 ⎟⎠−2 ⎝−2⎝1г) y = 1 − xy = 3 − 2x1x=01⎛ x2 ⎞11 17− ⎜ − ⎟ =3 − +6+ + − 2=7 .⎜ 2 ⎟82 28⎠ −2−2 ⎝222⎞2 ⎛ x2S = ∫ (3 − 2x)dx − ∫ (1 − x)dx = (3x − x 2 ) + ⎜− x⎟ = 6− 4+ 2− 2 = 2⎜⎟000⎝ 2⎠021036.