选修2-2 1.6 微积分基本定理练习题
人教版高二数学学案选修2-2练习1.6微积分基本定理

§1.6 微积分基本定理一、基础过关1. 已知物体做变速直线运动的位移函数s =s (t ),那么下列命题正确的是( ) ①它在时间段[a ,b ]内的位移是s =s (t )|b a ;②它在某一时刻t =t 0时,瞬时速度是v =s ′(t 0);③它在时间段[a ,b ]内的位移是s =lim n →∞∑i =1n b -a ns ′(ξi );④它在时间段[a ,b ]内的位移是s =ʃb a s ′(t )d t .A .①B .①②C .①②④D .①②③④2. 若F ′(x )=x 2,则F (x )的解析式不正确的是( ) A .F (x )=13x 3B .F (x )=x 3C .F (x )=13x 3+1D .F (x )=13x 3+c (c 为常数)3. ʃ10(e x +2x )d x 等于( ) A .1 B .e -1C .eD .e +14. 已知f (x )=⎩⎪⎨⎪⎧ x 2,-1≤x ≤0,1,0<x ≤1,则ʃ1-1f (x )d x 的值为( ) A.32 B.43C.23 D .-235. ʃπ20sin 2x 2d x 等于( ) A.π4 B.π2-1C .2 D.π-246.ʃ1-1|x |d x 等于( )A .ʃ1-1x d xB .ʃ1-1(-x )d xC .ʃ0-1(-x )d x +ʃ10x d xD .ʃ0-1x d x +ʃ10(-x )d x二、能力提升7. 设f (x )=⎩⎪⎨⎪⎧lg x ,x >0x +a 03t 2d t ,x ≤0, 若f [f (1)]=1,则a =________.8.设函数f (x )=ax 2+c (a ≠0),若ʃ10f (x )d x =f (x 0),0≤x 0≤1,则x 0的值为________. 9.设f (x )是一次函数,且ʃ10f (x )d x =5,ʃ10xf (x )d x =176,则f (x )的解析式为________. 10.计算下列定积分:(1)ʃ21(e x +1x)d x ;(2)ʃ91x (1+x )d x ; (3)ʃ200(-0.05e -0.05x +1)d x ;(4)ʃ211x (x +1)d x . 11.若函数f (x )=⎩⎪⎨⎪⎧ x 3,x ∈[0,1],x ,x ∈(1,2],2x ,x ∈(2,3].求ʃ30f (x )d x 的值.12.已知f (a )=ʃ10(2ax 2-a 2x )d x ,求f (a )的最大值.三、探究与拓展13.求定积分ʃ3-4|x +a |d x .答案1.D 2.B 3.C 4.B 5.D 6.C7.1 8.339.f (x )=4x +310.解 (1)∵(e x +ln x )′=e x +1x, ∴ʃ21(e x +1x)d x =(e x +ln x )|21=e 2+ln 2-e. (2)∵x (1+x )=x +x ,(12x 2+23x 32)′=x +x , ∴ʃ91x (1+x )d x =(12x 2+23x 32)|91 =1723. (3)∵(e -0.05x +1)′=-0.05e -0.05x +1,∴ʃ200(-0.05e-0.05x +1)d x =e -0.05x +1|200=1-e.(4)∵1x (x +1)=1x -1x +1,(ln x )′ =1x ,(ln(x +1))′=1x +1, ∴ʃ211x (x +1)d x =ln x |21-ln(x +1)|21=2ln 2-ln 3. 11.解 由积分的性质,知:ʃ30f (x )d x =ʃ10f (x )d x +ʃ21f (x )d x +ʃ32f (x )d x=ʃ10x 3d x +ʃ21x d x +ʃ322x d x=x 44|10+23x 32|21+2xln 2|32=14+432-23+8ln 2-4ln 2=-512+432+4ln 2.12.解 ∵(23ax 3-12a 2x 2)′=2ax 2-a 2x , ∴ʃ10(2ax 2-a 2x )d x=(23ax 3-12a 2x 2)|10=23a -12a 2, 即f (a )=23a -12a 2 =-12(a 2-43a +49)+29=-12(a -23)2+29, ∴当a =23时,f (a )有最大值29. 13.解 (1)当-a ≤-4即a ≥4时,原式=ʃ3-4(x +a )d x =(x 22+ax )|3-4=7a -72. (2)当-4<-a <3即-3<a <4时,原式=ʃ-a -4[-(x +a )]d x +ʃ3-a (x +a )d x =(-x 22-ax )|-a -4+(x 22+ax )|3-a =a 22-4a +8+(a 22+3a +92) =a 2-a +252. (3)当-a ≥3即a ≤-3时,原式=ʃ3-4[-(x +a )]d x =(-x 22-ax )|3-4=-7a +72. 综上,得ʃ3-4|x +a |d x=⎩⎪⎨⎪⎧ 7a -72 (a ≥4)a 2-a +252 (-3<a <4)-7a +72 (a ≤-3).。
选修2-2 1.6 微积分基本定理练习题

选修2-2 1.6 微积分基本定理一、选择题1.下列积分正确的是( )A.122713=⎰xdxB.e e dx x e x-=⎰2121 C.()316122ln 0=+⎰dx e e xx 222=⎰-ππxdx D. [答案] A [解析]12232723233227132271312713=-⨯===⎰⎰-x dx x xdx2.=⎪⎭⎫⎝⎛+⎰-dx x x 22421 A.214B.54C.338D.218[答案] A[解析] ⎠⎛2-2⎝ ⎛⎭⎪⎫x 2+1x 4d x =⎠⎛2-2x 2d x +⎠⎛2-21x 4d x=13x 3| 2-2+⎝ ⎛⎭⎪⎫-13x -3| 2-2 =13(x 3-x -3)| 2-2 =13⎝ ⎛⎭⎪⎫8-18-13⎝ ⎛⎭⎪⎫-8+18=214.故应选A. 3.⎰-11|x |d x 等于( )A.⎠⎛1-1x d xB.⎠⎛1-1d xC.⎠⎛0-1(-x )d x +⎠⎛01x d xD.⎠⎛0-1x d x +⎠⎛01(-x )d x[答案] C[解析] ∵|x |=⎩⎪⎨⎪⎧x (x ≥0)-x (x <0)∴⎠⎛1-1|x |d x =⎠⎛0-1|x |d x +⎠⎛01|x |d x=⎠⎛0-1(-x )d x +⎠⎛01x d x ,故应选C.4.设f (x )=⎩⎪⎨⎪⎧x 2(0≤x <1)2-x (1≤x ≤2),则⎠⎛02f (x )d x 等于( )A.34 B.45 C.56D .不存在[答案] C[解析] ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x取F 1(x )=13x 3,F 2(x )=2x -12x 2,则F ′1(x )=x 2,F ′2(x )=2-x∴⎠⎛02f (x )d x =F 1(1)-F 1(0)+F 2(2)-F 2(1)=13-0+2×2-12×22-⎝ ⎛⎭⎪⎫2×1-12×12=56.故应选C.5.⎠⎛ab f ′(3x )d x =( )A .f (b )-f (a )B .f (3b )-f (3a ) C.13[f (3b )-f (3a )] D .3[f (3b )-f (3a )][答案] C[解析] ∵⎣⎢⎡⎦⎥⎤13f (3x )′=f ′(3x ) ∴取F (x )=13f (3x ),则⎠⎛a bf ′(3x )d x =F (b )-F (a )=13[f (3b )-f (3a )].故应选C. 6.⎠⎛03|x 2-4|d x =( )A.213B.223 C.233D.253[答案] C[解析] ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x=⎝ ⎛⎭⎪⎫4x -13x 3| 20+⎝ ⎛⎭⎪⎫13x 3-4x | 32=233.7.θθπd ⎰⎪⎭⎫ ⎝⎛-322sin 21 的值为 ( )A .-32B .-12 C.12D.32[答案] D [解析] ∵1-2sin2θ2=cos θ23sin cos 2sin 213030302===⎪⎭⎫ ⎝⎛-∴⎰⎰πππθθθθd d ,故应选D 8.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .cos xB .sin xC .-cos xD .-sin x[答案] A[解析] F (x )=⎠⎛0x cos t d t =sin t | x0=sin x -sin0=sin x .所以F ′(x )=cos x ,故应选A. 9.若⎠⎛0k (2x -3x 2)d x =0,则k =( )A .0B .1C .0或1D .以上都不对[答案] C[解析] ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)| k 0=k 2-k 3=0,∴k =0或1.10.函数F (x )=⎠⎛0x t (t -4)d t 在[-1,5]上( )A .有最大值0,无最小值B .有最大值0和最小值-323C .有最小值-323,无最大值D .既无最大值也无最小值 [答案] B[解析] F (x )=⎠⎛0x (t 2-4t )d t =⎝ ⎛⎭⎪⎫13t 3-2t 2| x 0=13x 3-2x 2(-1≤x ≤5).F ′(x )=x 2-4x ,由F ′(x )=0得x =0或x =4,列表如下:x (-1,0) 0 (0,4) 4 (4,5) F ′(x ) +-0 +F (x )极大值极小值可见极大值F (0)=0,极小值F (4)=-3.又F (-1)=-73,F (5)=-253∴最大值为0,最小值为-323. 二、填空题11.计算定积分:①⎠⎛1-1x 2d x =________②⎠⎛23⎝ ⎛⎭⎪⎫3x -2x2d x =________ ③⎠⎛02|x 2-1|d x =________ ④⎠⎛0-π2|sin x |d x =________[答案] 23;436;2;1[解析] ①⎠⎛1-1x 2d x =13x 3| 1-1=23.②⎠⎛23⎝⎛⎭⎪⎫3x -2x 2d x =⎝ ⎛⎭⎪⎫32x 2+2x | 32=436.③⎠⎛02|x 2-1|d x =⎠⎛01(1-x 2)d x +⎠⎛12(x 2-1)d x=⎝ ⎛⎭⎪⎫x -13x 3| 10+⎝ ⎛⎭⎪⎫13x 3-x | 21=2. ④()1cos sin sin 02202==-=---⎰⎰πππxdx x dx x12..________2cos 2sin 220=⎪⎭⎫ ⎝⎛+⎰dx x x π[答案] 1+π2[解析]()()12cos sin 12cos 2sin 220220+=-=+=⎪⎭⎫ ⎝⎛+⎰⎰ππππx x dx x dx x x 13.(2010·陕西理,13)从如图所示的长方形区域内任取一个点M (x ,y ),则点M 取自阴影部分的概率为________.[答案] 13[解析] 长方形的面积为S 1=3,S 阴=⎠⎛013x 2dx =x 3| 10=1,则P =S 1S 阴=13. 14.已知f (x )=3x 2+2x +1,若⎠⎛1-1f (x )d x =2f (a )成立,则a =________.[答案] -1或13[解析] 由已知F (x )=x 3+x 2+x ,F (1)=3,F (-1)=-1, ∴⎠⎛1-1f (x )d x =F (1)-F (-1)=4,∴2f (a )=4,∴f (a )=2.即3a 2+2a +1=2.解得a =-1或13.三、解答题15.计算下列定积分:(1)⎠⎛052x d x ;(2)⎠⎛01(x 2-2x )d x ;(3)⎠⎛02(4-2x )(4-x 2)d x ;(4)⎠⎛12x 2+2x -3x d x .[解析] (1)⎠⎛052x d x =x 2| 50=25-0=25.(2)⎠⎛01(x 2-2x )d x =⎠⎛01x 2d x -⎠⎛012x d x=13x 3| 10-x 2| 10=13-1=-23. (3)⎠⎛02(4-2x )(4-x 2)d x =⎠⎛02(16-8x -4x 2+2x 3)d x=⎝⎛⎭⎪⎫16x -4x 2-43x 3+12x 4| 20=32-16-323+8=403.(4)⎠⎛12x 2+2x -3x d x =⎠⎛12⎝⎛⎭⎪⎫x +2-3x d x=⎝ ⎛⎭⎪⎫12x 2+2x -3ln x | 21=72-3ln2.16.计算下列定积分:(1)⎰462cos ππxdx (2)dx x x 2321⎰⎪⎪⎭⎫ ⎝⎛+ (3) ()⎰+20sin 3πdx x x (4)⎰b a x dx e [解析] (1)取F (x )=12sin2x ,则F ′(x )=cos2x∴⎪⎭⎫⎝⎛-=⎰6)4(2cos 46ππππF F xdx=12⎝ ⎛⎭⎪⎫1-32=14(2-3). (2)取F (x )=x 22+ln x +2x ,则F ′(x )=x +1x+2.∴⎠⎛23⎝ ⎛⎭⎪⎫x +1x 2d x =⎠⎛23⎝⎛⎭⎪⎫x +1x +2d x=F (3)-F (2)=⎝ ⎛⎭⎪⎫92+ln3+6-⎝ ⎛⎭⎪⎫12×4+ln2+4=92+ln 32. (3)取F (x )=32x 2-cos x ,则F ′(x )=3x +sin x∴()()18302sin 3220+=-⎪⎭⎫⎝⎛=+⎰πππF F dx x x(4)取()xe x F =,则xe x F =)('∴a b b axbax e e e dx e -==⎰17.计算下列定积分:(1)⎠⎛0-4|x +2|d x ;(2)已知f (x )=,求⎠⎛3-1f (x )d x 的值.[解析] (1)∵f (x )=|x +2|=∴⎠⎛0-4|x +2|d x =-⎠⎛-4-2(x +2)d x +⎠⎛0-2(x +2)d x=-⎝ ⎛⎭⎪⎫12x 2+2x | -2-4+⎝ ⎛⎭⎪⎫12x 2+2x | 0-2=2+2=4.(2)∵f (x )=∴⎠⎛3-1f (x )d x =⎠⎛0-1f (x )d x +⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01(1-x )d x +⎠⎛12(x -1)d x=⎝ ⎛⎭⎪⎫x -x 22| 10+⎝ ⎛⎭⎪⎫x 22-x | 21 =12+12=1. 18.(1)已知f (a )=⎠⎛01(2ax 2-a 2x )d x ,求f (a )的最大值;(2)已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛01f (x )d x =-2,求a ,b ,c 的值.[解析] (1)取F (x )=23ax 3-12a 2x 2则F ′(x )=2ax 2-a 2x ∴f (a )=⎠⎛01(2ax 2-a 2x )d x=F (1)-F (0)=23a -12a 2=-12⎝ ⎛⎭⎪⎫a -232+29 ∴当a =23时,f (a )有最大值29.(2)∵f (-1)=2,∴a -b +c =2① 又∵f ′(x )=2ax +b ,∴f ′(0)=b =0② 而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x取F (x )=13ax 3+12bx 2+cx则F ′(x )=ax 2+bx +c∴⎠⎛01f (x )d x =F (1)-F (0)=13a +12b +c =-2③解① ② ③ 得a =6,b =0,c =-4.。
高中数学选修2-2课时作业15:§1.6 微积分基本定理

§1.6 微积分基本定理一、选择题1.ʃ21⎝⎛⎭⎫e x +1x d x 等于( ) A .e 2-ln 2B .e 2-e -ln 2C .e 2+e +ln 2D .e 2-e +ln 2考点 利用微积分基本定理求定积分题点 利用微积分基本定理求定积分[答案] D[解析] ʃ21⎝⎛⎭⎫e x +1x =(e x +ln x )|21 =(e 2+ln 2)-(e +ln 1)=e 2-e +ln 2.2.若π20(sin cos )d x a x x -⎰=2,则实数a 等于( )A .-1B .1C .- 3 D. 3 考点 微积分基本定理的应用题点 利用微积分基本定理求参数[答案] A[解析] π20(sin cos )d x a x x -⎰=(-cos x -a sin x )π20|=0-a -(-1-0)=1-a =2,∴a =-1,故选A.3.若S 1=ʃ21x 2d x ,S 2=ʃ211xd x ,S 3=ʃ21e x d x ,则S 1,S 2,S 3的大小关系为( ) A .S 1<S 2<S 3B .S 2<S 1<S 3C .S 2<S 3<S 1D .S 3<S 2<S 1考点 利用微积分基本定理求定积分题点 利用微积分基本定理求定积分[答案] B[解析] 因为S 1=ʃ21x 2d x =⎪⎪13x 321=13×23-13=73,S 2=ʃ211xd x =ln x |21=ln 2, S 3=ʃ21e x d x =e x |21=e 2-e =e(e -1). 又ln 2<ln e =1,且73<2.5<e(e -1), 所以ln 2<73<e(e -1),即S 2<S 1<S 3. 4.ʃ30|x 2-4|d x 等于( )A.213B.223C.233D.253考点 分段函数的定积分题点 分段函数的定积分[答案] C [解析] ∵|x 2-4|=⎩⎪⎨⎪⎧x 2-4,2≤x ≤3,4-x 2,0≤x ≤2, ∴ʃ30|x 2-4|d x =ʃ32(x 2-4)d x +ʃ20(4-x 2)d x = ⎪⎪⎝⎛⎭⎫13x 3-4x 32+⎪⎪⎝⎛⎭⎫4x -13x 320 =⎣⎡⎦⎤(9-12)-⎝⎛⎭⎫83-8+⎣⎡⎦⎤⎝⎛⎭⎫8-83-0 =-3-83+8+8-83=233. 5.若函数f (x ),g (x )满足ʃ1-1f (x )g (x )d x =0,则称f (x ),g (x )为区间[-1,1]上的一组正交函数.给出三组函数:①f (x )=sin 12x ,g (x )=cos 12x ; ②f (x )=x +1,g (x )=x -1;③f (x )=x ,g (x )=x 2.其中为区间[-1,1]上的正交函数的组数为( )A .0B .1C .2D .3考点 微积分基本定理的应用题点 微积分基本定理的综合应用[答案] C[解析] 对于①,ʃ1-1sin 12x cos 12x d x =ʃ1-112sin x d x =0, 所以①是区间[-1,1]上的一组正交函数;对于②,ʃ1-1(x +1)(x -1)d x =ʃ1-1(x 2-1)d x ≠0,所以②不是区间[-1,1]上的一组正交函数;对于③,ʃ1-1x ·x 2d x =ʃ1-1x 3d x =0,所以③是区间[-1,1]上的一组正交函数.6.若f (x )=x 2+2ʃ10f (x )d x ,则ʃ10f (x )d x 等于( ) A .-13B .-1 C.13 D .1考点 利用微积分基本定理求定积分题点 利用微积分基本定理求定积分[答案] A[解析] ∵f (x )=x 2+2ʃ10f (x )d x ,∴ʃ10f (x )d x =⎪⎪⎝⎛⎭⎫13x 3+2x ʃ10f (x )d x 10 =13+2ʃ10f (x )d x , ∴ʃ10f (x )d x =-13. 二、填空题7.设f (x )=⎩⎪⎨⎪⎧x 2,x ≤0,cos x -1,x >0,则ʃ1-1f (x )d x =________. 考点 分段函数的定积分题点 分段函数的定积分[答案] sin 1-23[解析] ʃ1-1f (x )d x =ʃ0-1x 2d x +ʃ10(cos x -1)d x=⎪⎪13x 30-1+(sin x -x )|10=⎣⎡⎦⎤13×03-13×(-1)3+[(sin 1-1)-(sin 0-0)] =sin 1-23. 8.已知f (x )=3x 2+2x +1,若ʃ1-1f (x )d x =2f (a )成立,则a =________.考点 微积分基本定理的应用题点 利用微积分基本定理求参数[答案] -1或13[解析] ʃ1-1f (x )d x =(x 3+x 2+x )|1-1=4, 2f (a )=6a 2+4a +2,由题意得6a 2+4a +2=4,解得a =-1或13. 9.从如图所示的长方形区域内任取一个点M (x ,y ),则点M 取自阴影部分的概率为________.考点 微积分基本定理的应用题点 微积分基本定理的综合应用[答案] 13[解析] 长方形的面积为S 1=3,S 阴=ʃ103x 2d x =x 3|10=1,则P =S 阴S 1=13. 10.设f (x )=⎩⎪⎨⎪⎧lg x ,x >0,x +ʃa 03t 2d t ,x ≤0,若f (f (1))=1,则a =____________. 考点 微积分基本定理的应用题点 利用微积分基本定理求参数[答案] 1[解析] 因为x =1>0,所以f (1)=lg 1=0.又当x ≤0时,f (x )=x +ʃa 03t 2d t =x +t 3|a 0=x +a 3,所以f (0)=a 3.因为f (f (1))=1,所以a 3=1,解得a =1.11.设f (x )是一次函数,且ʃ10f (x )d x =5,ʃ10xf (x )d x =176,则f (x )的[解析]式为________. 考点 微积分基本定理的应用题点 利用微积分基本定理求参数[答案] f (x )=4x +3[解析] ∵f (x )是一次函数,∴设f (x )=ax +b (a ≠0),∴ʃ10f (x )d x =ʃ10(ax +b )d x =ʃ10ax d x +ʃ10b d x=12a +b =5, ʃ10xf (x )d x =ʃ10x (ax +b )d x=ʃ10(ax 2)d x +ʃ10bx d x =13a +12b =176. ∴⎩⎨⎧ 12a +b =5,13a +12b =176,解得⎩⎪⎨⎪⎧a =4,b =3. ∴f (x )=4x +3. 12.已知α∈⎣⎡⎦⎤0,π2,则当ʃα0(cos x -sin x )d x 取最大值时,α=________. 考点 微积分基本定理的应用题点 微积分基本定理的综合应用[答案] π4[解析] ʃα0(cos x -sin x )d x =(sin x +cos x )|α0=sin α+cos α-1=2sin ⎝⎛⎭⎫α+π4-1. ∵α∈⎣⎡⎦⎤0,π2,则α+π4∈⎣⎡⎦⎤π4,34π, 当α+π4=π2,即α=π4时, 2sin ⎝⎛⎭⎫α+π4-1取得最大值. 三、解答题13.已知f (x )=ʃx -a (12t +4a )d t ,F (a )=ʃ10[f (x )+3a 2]d x ,求函数F (a )的最小值.考点 微积分基本定理的应用题点 微积分基本定理的综合应用解 因为f (x )=ʃx -a (12t +4a )d t =(6t 2+4at )|x -a=6x 2+4ax -(6a 2-4a 2)=6x 2+4ax -2a 2,F (a )=ʃ10[f (x )+3a 2]d x =ʃ10(6x 2+4ax +a 2)d x=(2x 3+2ax 2+a 2x )|10=a 2+2a +2=(a +1)2+1≥1.所以当a =-1时,F (a )取到最小值为1.四、探究与拓展14.已知函数f (x )=⎩⎨⎧ (x +1)2,-1≤x ≤0,1-x 2,0<x ≤1,则ʃ1-1f (x )d x 等于( ) A.3π-812B.4+3π12C.4+π4D.-4+3π12 考点 分段函数的定积分题点 分段函数的定积分[答案] B[解析] ʃ1-1f (x )d x =ʃ0-1(x +1)2d x +ʃ101-x 2d x ,ʃ0-1(x +1)2d x = ⎪⎪13(x +1)30-1=13, ʃ101-x 2d x 以原点为圆心,以1为半径的圆的面积的四分之一,故ʃ101-x 2d x =π4, 故ʃ1-1f (x )d x =13+π4=4+3π12. 15.已知f ′(x )是f (x )在(0,+∞)上的导数,满足xf ′(x )+2f (x )=1x2,且ʃ21[x 2f (x )-ln x ]d x =1. (1)求f (x )的[解析]式;(2)当x >0时,证明不等式2ln x ≤e x 2-2. 考点 微积分基本定理的应用题点 微积分基本定理的综合应用(1)解 由xf ′(x )+2f (x )=1x2,得 x 2f ′(x )+2xf (x )=1x, 即[x 2f (x )]′=1x, 所以x 2f (x )=ln x +c (c 为常数),即x 2f (x )-ln x =c .又ʃ21[x 2f (x )-ln x ]d x =1,即ʃ21c d x =1,所以cx |21=1,所以2c -c =1,所以c =1.所以x 2f (x )=ln x +1,所以f (x )=ln x +1x 2. (2)证明 由(1)知f (x )=ln x +1x 2(x >0), 所以f ′(x )=1x ×x 2-2x (ln x +1)x 4=-2ln x -1x 3, 当f ′(x )=0时,x =12e-,f ′(x )>0时,0<x <12e -, f ′(x )<0时,x >12e -,所以f (x )在(0,12e -)上单调递增,在(12e-,+∞)上单调递减. 所以f (x )max = 12(e )f -=e 2, 所以f (x )=ln x +1x 2≤e 2, 即2ln x ≤e x 2-2.。
人教版高中数学选修2-21.6微积分基本定理练习

【成才之路】 2015-2016 学年高中数学 1.6 微积分基本定理练习 新人教 A 版选修 2-2一、选择题π π 与曲线 y = cosx 所围成的关闭图1. (2015 广·西柳州市模拟 )由直线 x =- , x = , y = 033形的面积为 ()1 B . 1A. 23D.3C . 2[答案 ]Dππ3,选 D.[分析 ]由题意得, S =23cosxdx = 2sinx|3= 02.(2013 景·德镇市高二质检 ) 若曲线 y = x 与直线 x = a 、y = 0 所围成关闭图形的面积为 a 2,则正实数 a 为 ()45A. 9B . 945 C .3D . 3[答案 ] A[分析 ]由题意知,axdx = a 2,31,∴33∵ (2x 2 ) =′ xa xdx =2x 2 |0a=2a 2 ,323 3342 2.∴ a2 = a ,∴ a =393.由曲线 y = x 2 和直线 x = 0,x = 1, y = t 2, t ∈ (0,1)所围成的图形 (暗影部分 )的面积的最小值为 ()11A. 4B . 312 C.2D . 3[答案 ]Ay = x 2[分析 ]2得,x =t ,故 S =t 221 2 221 3 t 1321由 y = t(t - x )dx + (x - t )d x =( t x - x )|0+ ( x - t x)|tx>00 t33=4t 3- t 2+ 1,3 321令 S ′= 4t - 2t = 0,∵ 0<t<1,∴ t = ,1 1易知当 t = 时, S min = .242x,4.设 f(x)=x2f(x)dx 等于 ()2-xx 则34A. 4B . 55C .6D .不存在[答案 ] C[分析 ]21 22(2- x)dx ,f(x)dx =x dx +0 011 312,取 F 1(x)= x , F 2(x)= 2x - x32则 F ′ 2,F ′1(x)= x 2(x)= 2-x ,∴ 2f(x)dx = F 1(1)- F 1(0)+ F 2(2)- F 2(1)1 1 21 25 = - 0+ 2×2- ×2- 2×1- ×1= .故应选 C.32265.(2014 ~ 2015 ·河南周口市高二期末 )已知函数 f(x)= x n+ mx 的导函数 f ′(x)= 2x + 2,则 3f(- x)dx = ()1A . 0B . 322C .- 3D . 3[答案] D[分析 ]∵ f(x)= x n +mx 的导函数 f ′(x)=2x + 2,∴ nx n -1+m =2x + 2,解得 n = 2, m = 2,∴ f(x)=x 2+ 2x ,∴ f(- x)= x 2- 2x ,∴ 3f(- x)dx =,则3(x 2- 2x)dx = (13x 3- x 2)|31= 9-9-13+ 1=23,应选 D.11π2θ6. 3 1- 2sin2 d θ的值为 ()31A .- 2B .- 213C .2D . 2[答案] D2θ[分析 ] ∵ 1-2sin 2= cos θ,∴π1- 2sin2θπ π 3 3 d θ=cos θd θ= sin θ|3=,故应选 D.23 02二、填空题7.计算定积分:12①x dx = ________________- 1② 3 3x - 22 dx = ________________x2③ 2|x 2-1|dx = ________________0 0④ |sinx|dx = ________________π 2[答案 ]2 ② 43 ③2④ 1①3621 3 12 [分析 ]① 1 x dx =3x |-1 = 3.-1② 32 dx =3 2 2 3 433x - 2 x + x |2 = 6 .2x 22 -1|dx =22 2 13 1+1 32= 2.③ 2|x1(1- x )dx +(x - 1)dx = x - x |0 x - x |1 0133④|sinx|dx =(- sinx)dx = cosx| -π= 1.-π -π 2228.从如下图的长方形地区内任取一个点M(x ,y),则点 M 取自暗影部分的概率为 ________________ .1 [答案 ]3[分析 ]长方形的面积为S1=3,S阴=23 1S1113x dx= x |0= 1,则 P== .0S阴39.已知 f(x)= 3x2+ 2x+ 1,若1 f(x)dx= 2f(a)建立,则 a=________________.-11[答案 ]-1 或3[分析 ]由已知32F(x)= x+ x + x, F(1) = 3, F(- 1)=- 1,∴1 f(x)dx=F(1)- F(- 1)= 4,-1∴2f(a)= 4,∴ f(a)= 2.即 3a2+ 2a+ 1= 2.解得 a=- 1 或1 . 3三、解答题10.计算以下定积分:(1)22(2)2x2+ 2x- 3 (4- 2x)(4 - x )dx;x dx. 01[分析 ](1)2(4- 2x)(4- x2)dx=2(16- 8x- 4x2+ 2x3)dx 00=24 314232+8=40.16x- 4x- x +2x|0=32-16-333(2)x2+ 2x- 3x+ 2-3dx=1227- 3ln2. 2xdx=2x2x + 2x-3ln x |1=112一、选择题11.函数 F(x)=x costdt 的导数是 ()A . F′(x)=cosx B. F ′(x)= sinx C.F ′(x)=- cosx D. F ′(x)=- sinx [答案 ]A[分析 ]xF(x)=x costdt= sint|0= sinx- sin0= sinx.因此 F ′(x)= cosx,故应选 A.12. (2015 江·西教课质量监测)若直线 l 1: x+ay- 1= 0 与 l2: 4x- 2y+3= 0 垂直,则积分3的值为 ()a ( x + sin x- 5)dx-aA . 6+ 2sin 2C.20B.- 6-2cos 2D.- 20[答案 ]D[分析 ]由 l 1⊥ l 2 得 4- 2a = 0 即 a = 2,∴原式=2(x 3+ sin x -5)dx =2(x 3+ sin x)dx + 2(- 5)dx = 0- 20=- 20.- 2- 2- 2[评论 ]若 f(x)为奇函数,定义域 (-a , a), a>0,则af(x)dx = 0.-a13.若 S 1=2x 2dx , S 2=21dx , S 3= 2e x dx ,则 S 1, S 2, S 3 的大小关系为 ()1x11A . S 1<S 2<S 3B . S 2<S 1<S 3C .S 2<S 3<S 1D . S 3<S 2<S 1[答案 ] B[分析 ]S 1=2 2 x3 2 7x dx = 3 |1= .13S 2=21dx = lnx|12 = ln2- ln1=ln2.x1S 3= 2 x x 2 2e dx = e |1= e - e = e(e -1) . 1∵ e>2.7,∴ S 3>3>S 1>S 2.应选 B.14. (2015 河·南高考适应性测试 )定义在 R 上的可导函数y = f(x),假如存在 x 0∈ [a , b],b fx x使得 f( x 0)=a建立,则称 x 0 为函数 f(x) 在区间 [a ,b]上的 “均匀值点 ”,那么函数 f( x)b - a=x 3-3x 在区间 [ -2,2] 上 “均匀值点 ”的个数为 ()A . 1B . 2C .3D . 4[答案 ]C23- 3xx 14 3 2 2x-2x - x -2[分析 ]= 4 2= 0,即 x 03- 3x 0= 0,解得:由已知得: f( x 0)=44x 0= 0 或 x 0= ± 3,∴ f(x)的均匀值点有 3 个,应选 C.二、填空题π15. (2014 绍·兴模拟 )-2(x + cosx)dx =________________.π2[答案 ]2π1 2π[分析 ]+sinx)| π- 2 (x + cosx)dx = ( x-2 =2.22π216. (2014 山·东省菏泽市期中 )函数 y =x 2与 y = kx(k>0) 的图象所围成的暗影部分的面积为9,则 k =________________. 2[答案 ] 3 [分析 ]y = kx , x = 0,x = k ,由y = x 2,解得或y = 0,y = k 2.由题意得,k(kx - x 2)dx = (1kx 2- 1x 3)|0k = 1k 3- 1k 3= 1k 3= 9,∴ k = 3.232362三、解答题17.已知 f(x)=ax 2 +bx + c(a ≠0),且 f(- 1)=2, f ′(0)= 0, 1f(x)dx =- 2,求 a 、 b 、 c的值.[分析 ]∵ f(- 1)= 2,∴ a - b + c = 2.①又∵ f ′(x)= 2ax + b ,∴ f ′(0)=b = 0② 而 1f(x)dx = 1(ax 2+ bx +c)dx ,取 F(x)=1ax 3+1bx 2+ cx ,32 则 F ′(x)= ax 2+ bx + c ,∴111f(x)dx = F(1)- F(0) = 3a + 2b + c =- 2③解①②③得 a =6, b = 0,c =- 4.18.如图,直线 y = kx 分抛物线 y =x - x 2 与 x 轴所围成图形为面积相等的两部分,求k的值.[分析 ]抛物线 y = x - x 2 与 x 轴两交点的横坐标 x 1= 0, x 2= 1,因此,抛物线与x 轴所围图形的面积23S = 1(x - x 2 )dx = (x- x)|01= 1- 1= 1.232 36抛物线 y = x - x 2与直线 y = kx 两交点的横坐标为x ′= 0, x ′=1- k ,因此 S=1-k 2122(x - x1- k 2 - x 3 1-k1 3,-kx)dx = ( 2x 3 )|0 = (1 -k)6又知 S = 1,因此 (1-k)3=1.62313= 1-4于是 k=1-2 2.。
人教新课标版数学高二选修2-2检测 1.6微积分基本定理

一、选择题1.设a =⎠⎛01x 13d x ,b =⎠⎛01x 2d x ,c =⎠⎛01x 3d x ,则a ,b ,c 的大小关系是( ) A .a >b >c B .c >a >b C .a >c >bD .c >b >a【解析】 ∵a =⎠⎛01x 13d x =⎪⎪⎪10=34;b =⎠⎛01x 2d x =x 33⎪⎪⎪10=13,c =⎠⎛01x 3d x =x 44⎪⎪⎪1=14,∴a >b >c .【答案】 A2.已知f (x )=⎩⎪⎨⎪⎧x 2,x ∈[0,1),1x ,x ∈[1,e 2)(其中e 为自然对数的底数),则f (x )d x的值为( )A.43 B.53 C.73 D.83【解析】 f (x )d x =⎠⎛01x 2d x +∫e 211x d x =13x 3⎪⎪⎪ 10+ln x ⎪⎪⎪e 21=13+2=73.【答案】 C3.(2013·安阳高二检测)⎠⎛01|1-x |d x =( )A .0B .1C .2D .-2 【解析】 ⎠⎛01(1-x )d x +⎠⎛12(x -1)d x=(x -12x 2)⎪⎪⎪10+(12x 2-x )⎪⎪⎪21=(1-12)+(12×4-2)-(12-1) =1. 【答案】 B4.设f (x )=ax 2+c (a ≠0),若⎠⎛01f (x )d x =f (x 0),0≤x 0≤1,则x 0的值为( )A .-33B.33C .-13 D.13【解析】 ∵⎠⎛01f (x )d x =(13ax 3+cx )⎪⎪⎪10=13a +c ,∴ax 20+c =13a +c , ∴x 20=13.∵0≤x 0≤1.∴x 0=33.【答案】 B5.若⎠⎛0k (2x -3x 2)d x =0,则k 等于( )A .0B .1C .0或1D .不确定【解析】 ∵⎠⎛0k (2x -3x 2)d x =(x 2-x 3)⎪⎪⎪k0=k 2-k 3,∴k 2-k 3=0,即k =0或k =1, 又当k =0时,不合题意,∴k =1. 【答案】 B 二、填空题6.已知函数f (a )=⎠⎛0a sin x d x ,则f [f (π2)]=________.【解析】 ∵f (a )=⎠⎛0a sin x d x =(-cos x )⎪⎪⎪a0=1-cos a .∴f (π2)=1-cos π2=1. ∴f [f (π2)]=f (1)=1-cos 1. 【答案】 1-cos 17.(2012·江西高考)计算定积分(x 2+sin x )d x =________.【解析】 ∵(13x 3-cos x )′=x 2+sin x , ∴(x 2+sin x )d x =(13x 3-cos x )⎪⎪⎪1-1=23. 【答案】 238.(2013·福州高二检测)⎠⎛12(1x +1x 2)d x =________.【解析】 ∵⎠⎛12(1x +1x 2)d x =(ln x -1x )⎪⎪⎪21=(ln 2-12)-(ln 1-1)=ln 2+12. 【答案】 ln 2+12 三、解答题9.计算下列定积分: (1)⎠⎛121x (x +1)d x ; (2)(cos x +2x )d x .【解】 (1)∵⎠⎛121x (x +1)d x =⎠⎛12(1x -1x +1)d x=[ln x -ln(x +1)]⎪⎪⎪21=ln 43.(2)(cos x +2x )d x =(sin x +2x ln 2)⎪⎪⎪⎪π2-π210.(1)设f (x )=⎩⎨⎧x 2,x ≤0,cos x -1,x >0,求f (x )d x .(2)求x 2d x (a >0).【解】 (1) f (x )d x =x 2d x +⎠⎛01(cos x -1)d x =13x 3⎪⎪⎪0-1+(sin x -x )⎪⎪⎪10=sin 1-23.(2)由x 2=⎩⎪⎨⎪⎧x ,x ≥0,-x ,x <0,得11.已知f (x )=⎩⎨⎧2x +1,x ∈[-2,2],1+x 2,x ∈(2,4],求使⎠⎛k 3f (x )d x =403恒成立的k 值.【解】 (1)当k ∈(2,3]时,⎠⎛k 3f (x )d x =⎠⎛k3(1+x 2)d x =(x +13x 3)⎪⎪⎪3k=3+13×33-(k +13k 3)=403 整理得k 3+3k +4=0, 即k 3+k 2-k 2+3k +4=0, ∴(k +1)(k 2-k +4)=0, ∴k =-1.而k ∈(2,3],∴k =-1舍去. (2)当k ∈[-2,2]时,⎠⎛k 3f (x )d x =⎠⎛k 2(2x +1)d x +⎠⎛23(1+x 2)d x =(x 2+x )⎪⎪⎪2k +(x +13x 3)⎪⎪⎪32=(22+2)-(k 2+k )+(3+13×33)-(2+13×23) =403-(k 2+k )=403, ∴k 2+k =0,解得k =0或k =-1, 综上所述,k =0或k =-1.。
人教新课标A版高中选修2-2数学1.6微积分基本定理同步练习C卷

(1) 求等待开垦土地的面积;
(2) 如何确定点C的位置,才能使得整块土地总价值最大.
23. (5分) 已知F(x)= (t2+2t-8)dt,(x>0).
(1)求F(x)的单调区间;
(2)求函数F(x)在[1,3]上的最值.
24. (5分) 计算由抛物线y=3-(x-1)2 和y= 所围成的图形的面积.
人教新课标A版选修2-2数学1.6微积分基本定理同步练习C卷
姓名:________班级:________ 成绩:________
一、 选择题 (共15题;共30分)
1. (2分) (2016高二下·桂林开学考) 的值是( )
A .
B .
C .
D .
2. (2分) 若 dx=2,则λ等于( )
A . 0
25. (10分) (2017高二下·钦州港期末) 设y=f(x)是二次函数,方程f(x)=0有两个相等的实根,且f′(x)=2x+2.
(1) 求y=f(x)的表达式;
(2) 求y=f(x)的图象与两坐标轴所围成封闭图形的面积.
参考答案
一、 选择题 (共15题;共30分)
1-1、
2-1、
3-1、
4-1、
A .
B .
C .
D .
二、 填空题 (共5题;共5分)
16. (1分) (2018·呼和浩特模拟) ________.
17. (1分) (2016高二下·昌平期中) 设 ,则 =________.
18. (1分) (2019高二下·荆门期末) 关于曲线C: ,给出下列五个命题:
①曲线C关于直线y=x对称;②曲线C关于点 对称;③曲线C上的点到原点距离的最小值为 ;④当 时,曲线C上所有点处的切线斜率为负数;⑤曲线C与两坐标轴所围成图形的面积是 .上述命题中,为真命题的是________.(将所有真命题的编号填在横线上)
#11-12学年高中数学 1.6 微积分基本定理同步练习 新人教A版选修2-2

选修2-2 1.6 微积分基本定理一、选择题1.下列积分正确的是( )[答案] AA.214B.54 C.338D.218[答案] A[解析] ⎠⎛2-2⎝ ⎛⎭⎪⎫x 2+1x 4d x =⎠⎛2-2x 2d x +⎠⎛2-21x 4d x=13x 3| 2-2+⎝ ⎛⎭⎪⎫-13x -3| 2-2 =13(x 3-x -3)| 2-2 =13⎝ ⎛⎭⎪⎫8-18-13⎝ ⎛⎭⎪⎫-8+18=214.故应选A.3.⎠⎛1-1|x |d x 等于( )A.⎠⎛1-1x d xB.⎠⎛1-1d xC.⎠⎛0-1(-x )d x +⎠⎛01x d xD.⎠⎛0-1x d x +⎠⎛01(-x )d x[答案] C[解析] ∵|x |=⎩⎪⎨⎪⎧x (x ≥0)-x (x <0)∴⎠⎛1-1|x |d x =⎠⎛0-1|x |d x +⎠⎛01|x |d x=⎠⎛0-1(-x )d x +⎠⎛01x d x ,故应选C.4.设f (x )=⎩⎪⎨⎪⎧x 2(0≤x <1)2-x (1≤x ≤2),则⎠⎛02f (x )d x 等于( )A.34B.45C.56D .不存在[答案] C[解析] ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x取F 1(x )=13x 3,F 2(x )=2x -12x 2,则F ′1(x )=x 2,F ′2(x )=2-x∴⎠⎛02f (x )d x =F 1(1)-F 1(0)+F 2(2)-F 2(1)=13-0+2×2-12×22-⎝ ⎛⎭⎪⎫2×1-12×12=56.故应选C.5.⎠⎛ab f ′(3x )d x =( )A .f (b )-f (a )B .f (3b )-f (3a ) C.13[f (3b )-f (3a )]D .3[f (3b )-f (3a )][答案] C[解析] ∵⎣⎢⎡⎦⎥⎤13f (3x )′=f ′(3x ) ∴取F (x )=13f (3x ),则⎠⎛abf ′(3x )d x =F (b )-F (a )=13[f (3b )-f (3a )].故应选C. 6.⎠⎛03|x 2-4|d x =( )A.213B.223 C.233D.253[答案] C[解析] ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x=⎝ ⎛⎭⎪⎫4x -13x 3| 20+⎝ ⎛⎭⎪⎫13x 3-4x | 32=233.A .-32B .-12C.12D.32[答案] D[解析] ∵1-2sin2θ2=cos θ8.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .cos xB .sin xC .-cos xD .-sin x[答案] A[解析] F (x )=⎠⎛0x cos t d t =sin t | x0=sin x -sin0=sin x .所以F ′(x )=cos x ,故应选A. 9.若⎠⎛0k (2x -3x 2)d x =0,则k =( )A .0B .1C .0或1D .以上都不对[答案] C[解析] ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)| k 0=k 2-k 3=0,∴k =0或1.10.函数F (x )=⎠⎛0x t (t -4)d t 在[-1,5]上( )A .有最大值0,无最小值B .有最大值0和最小值-323C .有最小值-323,无最大值D .既无最大值也无最小值 [答案] B[解析] F (x )=⎠⎛0x (t 2-4t )d t =⎝ ⎛⎭⎪⎫13t 3-2t 2| x 0=13x 3-2x 2(-1≤x ≤5).F ′(x )=x 2-4x ,由F ′(x )=0得x =0或x =4,列表如下:可见极大值F (0)=0,极小值F (4)=-3.又F (-1)=-73,F (5)=-253∴最大值为0,最小值为-323. 二、填空题 11.计算定积分: ①⎠⎛1-1x 2d x =________②⎠⎛23⎝ ⎛⎭⎪⎫3x -2x 2d x =________③⎠⎛02|x 2-1|d x =________ ④⎠⎛0-π2|sin x |d x =________[答案] 23;436;2;1[解析] ①⎠⎛1-1x 2d x =13x 3| 1-1=23.②⎠⎛23⎝⎛⎭⎪⎫3x -2x 2d x =⎝ ⎛⎭⎪⎫32x 2+2x | 32=436.③⎠⎛02|x 2-1|d x =⎠⎛01(1-x 2)d x +⎠⎛12(x 2-1)d x=⎝ ⎛⎭⎪⎫x -13x 3| 10+⎝ ⎛⎭⎪⎫13x 3-x | 21=2.[答案] 1+π213.(2010·陕西理,13)从如图所示的长方形区域内任取一个点M (x ,y ),则点M 取自阴影部分的概率为________.[答案] 13[解析] 长方形的面积为S 1=3,S 阴=⎠⎛013x 2dx =x 3| 10=1,则P =S 1S 阴=13. 14.已知f (x )=3x 2+2x +1,若⎠⎛1-1f (x )d x =2f (a )成立,则a =________.[答案] -1或13[解析] 由已知F (x )=x 3+x 2+x ,F (1)=3,F (-1)=-1, ∴⎠⎛1-1f (x )d x =F (1)-F (-1)=4,∴2f (a )=4,∴f (a )=2.即3a 2+2a +1=2.解得a =-1或13.三、解答题15.计算下列定积分: (1)⎠⎛052x d x ;(2)⎠⎛01(x 2-2x )d x ;(3)⎠⎛02(4-2x )(4-x 2)d x ;(4)⎠⎛12x 2+2x -3x d x .[解析] (1)⎠⎛052x d x =x 2| 50=25-0=25.(2)⎠⎛01(x 2-2x )d x =⎠⎛01x 2d x -⎠⎛012x d x=13x 3| 10-x 2| 10=13-1=-23. (3)⎠⎛02(4-2x )(4-x 2)d x =⎠⎛02(16-8x -4x 2+2x 3)d x=⎝⎛⎭⎪⎫16x -4x 2-43x 3+12x 4| 20=32-16-323+8=403.(4)⎠⎛12x 2+2x -3x d x =⎠⎛12⎝⎛⎭⎪⎫x +2-3x d x=⎝ ⎛⎭⎪⎫12x 2+2x -3ln x | 21=72-3ln2.16.计算下列定积分:[解析] (1)取F (x )=12sin2x ,则F ′(x )=cos2x=12⎝ ⎛⎭⎪⎫1-32=14(2-3). (2)取F (x )=x 22+ln x +2x ,则F ′(x )=x +1x+2.∴⎠⎛23⎝ ⎛⎭⎪⎫x +1x 2d x =⎠⎛23⎝⎛⎭⎪⎫x +1x +2d x=F (3)-F (2)=⎝ ⎛⎭⎪⎫92+ln3+6-⎝ ⎛⎭⎪⎫12×4+ln2+4=92+ln 32. (3)取F (x )=32x 2-cos x ,则F ′(x )=3x +sin x17.计算下列定积分: (1)⎠⎛0-4|x +2|d x ;(2)已知f (x )=,求⎠⎛3-1f (x )d x 的值.[解析] (1)∵f (x )=|x +2|=∴⎠⎛0-4|x +2|d x =-⎠⎛-4-2(x +2)d x +⎠⎛0-2(x +2)d x=-⎝ ⎛⎭⎪⎫12x 2+2x | -2-4+⎝ ⎛⎭⎪⎫12x 2+2x | 0-2=2+2=4.(2)∵f (x )=∴⎠⎛3-1f (x )d x =⎠⎛0-1f (x )d x +⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01(1-x )d x +⎠⎛12(x -1)d x=⎝ ⎛⎭⎪⎫x -x 22| 10+⎝ ⎛⎭⎪⎫x 22-x | 21 =12+12=1. 18.(1)已知f (a )=⎠⎛01(2ax 2-a 2x )d x ,求f (a )的最大值;(2)已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛01f (x )d x =-2,求a ,b ,c 的值.[解析] (1)取F (x )=23ax 3-12a 2x 2则F ′(x )=2ax 2-a 2x ∴f (a )=⎠⎛01(2ax 2-a 2x )d x=F (1)-F (0)=23a -12a 2=-12⎝ ⎛⎭⎪⎫a -232+29∴当a =23时,f (a )有最大值29.(2)∵f (-1)=2,∴a -b +c =2① 又∵f ′(x )=2ax +b ,∴f ′(0)=b =0② 而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x取F (x )=13ax 3+12bx 2+cx则F ′(x )=ax 2+bx +c∴⎠⎛01f (x )d x =F (1)-F (0)=13a +12b +c =-2③解①②③得a =6,b =0,c =-4.。
高中数学 专题1.6 微积分基本定理练习(含解析)新人教A版选修2-2(2021年整理)

微积分基本定理(时间:25分,满分50分)班级 姓名 得分1。
ʃ1,0(e x +2x )d x 等于( )A .1B .e -1C .eD .e +1【答案】 C【解析】 ʃ错误!(e x +2x )d x =(e x +x 2)|错误!=(e 1+12)-(e 0+02)=e.2.sin 2错误!d x 等于( )A.错误!B.错误!-1C .2D 。
错误! 【答案】 D3。
若ʃ错误!(2x +k )d x =2,则k =( )A 。
1B 。
2 C.3 D.4【答案】 A【解析】 ∵ʃ1,0(2x +k )d x =(x 2+kx )|错误!=1+k =2,∴k =1。
4.已知,若成立,则a = . A.或 B 。
C. D 。
0【答案】 A【解析】取,则,, 所以,所以,所以。
即,解得或.5.等于( )20⎰()2321f x x x =++()()112f xd x f a -=⎰1-13131-()32F x x x x =++()13F =()11F -=-()()()11114f x F d xF ---==⎰()24f a =()2f a =23212a a ++=1a =-1311x dx -⎰A 。
B. C 。
D 。
【答案】C【解析】|x |=∴=,选C 。
6.由直线x =0、x =1、y =0和曲线y =x 2+2x 围成的图形的面积为( ).【答案】A=n (n +1)(2n +1)+ =+=,∴所求面积S =. 7.设f (x )是一次函数,且ʃ1,0f (x )d x =5,ʃ错误!xf (x )d x =错误!,则f (x )的解析式为________.【答案】 f (x )=4x +3 【解析】 ∵f (x )是一次函数,设f (x )=ax +b (a ≠0),则ʃ错误!f (x )d x =ʃ错误!(ax +b )d x =ʃ错误!ax d x +ʃ错误!b d x =错误!a +b =5,ʃ错误!xf (x )d x =ʃ错误!x (ax +b )d x =ʃ错误!(ax 2)d x +ʃ错误!bx d x =错误!a +错误!b =错误!。
高中数学 16 微积分基本定理同步练习 新人教A版选修2-2 试题

心尺引州丑巴孔市中潭学校选修2-2 1.6 微积分根本定理一、选择题1.以下积分正确的选项是( )[答案] AA.214B.54C.338D.218[答案] A[解析] ⎠⎛2-2⎝⎛⎭⎫x 2+1x 4d x =⎠⎛2-2x 2d x +⎠⎛2-21x 4d x=13x 3| 2-2+⎝⎛⎭⎫-13x -3| 2-2=13(x 3-x -3)| 2-2=13⎝⎛⎭⎫8-18-13⎝⎛⎭⎫-8+18=214.故应选A.3.⎠⎛1-1|x |d x 等于( )A.⎠⎛1-1x d xB.⎠⎛1-1d xC.⎠⎛0-1(-x )d x +⎠⎛01x d x D.⎠⎛0-1x d x +⎠⎛01(-x )d x [答案] C[解析] ∵|x |=⎩⎨⎧ x (x ≥0)-x (x <0)∴⎠⎛1-1|x |d x =⎠⎛0-1|x |d x +⎠⎛01|x |d x=⎠⎛0-1(-x )d x +⎠⎛01x d x ,故应选C.4.设f (x )=⎩⎨⎧ x 2(0≤x <1)2-x (1≤x ≤2),那么⎠⎛02f (x )d x 等于() A.34 B.45C.56 D .不存在[答案] C[解析] ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x 取F 1(x )=13x 3,F 2(x )=2x -12x 2, 那么F ′1(x )=x 2,F ′2(x )=2-x ∴⎠⎛02f (x )d x =F 1(1)-F 1(0)+F 2(2)-F 2(1) =13-0+2×2-12×22-⎝⎛⎭⎫2×1-12×12=56.故应选C. 5.⎠⎛ab f ′(3x )d x =( ) A .f (b )-f (a )B .f (3b )-f (3a ) C.13[f (3b )-f (3a )] D .3[f (3b )-f (3a )] [答案] C[解析] ∵⎣⎡⎦⎤13f (3x )′=f ′(3x ) ∴取F (x )=13f (3x ),那么 ⎠⎛a bf ′(3x )d x =F (b )-F (a )=13[f (3b )-f (3a )].故应选C. 6.⎠⎛03|x 2-4|d x =( ) A.213B.223C.233D.253 [答案] C[解析] ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x =⎝⎛⎭⎫4x -13x 3| 20+⎝⎛⎭⎫13x 3-4x | 32=233. A .-32 B .-12C.12D.32[答案] D[解析] ∵1-2sin 2θ2=cos θ8.函数F (x )=⎠⎛0x cos t d t 的导数是( ) A .cos xB .sin xC .-cos xD .-sin x[答案] A [解析] F (x )=⎠⎛0x cos t d t =sin t | x0=sin x -sin0=sin x . 所以F ′(x )=cos x ,故应选A.9.假设⎠⎛0k (2x -3x 2)d x =0,那么k =( ) A .0B .1C .0或1D .以上都不对 [答案] C[解析] ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)| k 0=k 2-k 3=0, ∴k =0或1.10.函数F (x )=⎠⎛0x t (t -4)d t 在[-1,5]上( ) A .有最大值0,无最小值B .有最大值0和最小值-323C .有最小值-323,无最大值 D .既无最大值也无最小值[答案] B[解析] F (x )=⎠⎛0x (t 2-4t )d t =⎝⎛⎭⎫13t 3-2t 2| x 0=13x 3-2x 2(-1≤x ≤5). F ′(x )=x 2-4x ,由F ′(x )=0得x =0或x =4,列表如下:F ′(x )+ 0 - 0 +F (x ) 极大值极小值 可见极大值F (0)=0,极小值F (4)=-3. 又F (-1)=-73,F (5)=-253∴最大值为0,最小值为-323. 二、填空题11.计算定积分:①⎠⎛1-1x 2d x =________ ②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =________ ③⎠⎛02|x 2-1|d x =________ ④⎠⎛0-π2|sin x |d x =________[答案] 23;436;2;1 [解析] ①⎠⎛1-1x 2d x =13x 3| 1-1=23. ②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =⎝⎛⎭⎫32x 2+2x | 32=436. ③⎠⎛02|x 2-1|d x =⎠⎛01(1-x 2)d x +⎠⎛12(x 2-1)d x =⎝⎛⎭⎫x -13x 3| 10+⎝⎛⎭⎫13x 3-x | 21=2. [答案] 1+π213.(2021·理,13)从如下列图的长方形区域内任取一个点M (x ,y ),那么点M 取自阴影局部的概率为________.[答案] 13[解析] 长方形的面积为S 1=3,S 阴=⎠⎛013x 2dx =x 3| 10=1,那么P =S 1S 阴=13.14.f (x )=3x 2+2x +1,假设⎠⎛1-1f (x )d x =2f (a )成立,那么a =________. [答案] -1或13[解析] 由F (x )=x 3+x 2+x ,F (1)=3,F (-1)=-1, ∴⎠⎛1-1f (x )d x =F (1)-F (-1)=4, ∴2f (a )=4,∴f (a )=2.即3a 2+2a +1=2.解得a =-1或13. 三、解答题15.计算以下定积分:(1)⎠⎛052x d x ;(2)⎠⎛01(x 2-2x )d x ; (3)⎠⎛02(4-2x )(4-x 2)d x ;(4)⎠⎛12x 2+2x -3x d x . [解析] (1)⎠⎛052x d x =x 2| 50=25-0=25. (2)⎠⎛01(x 2-2x )d x =⎠⎛01x 2d x -⎠⎛012x d x =13x 3| 10-x 2| 10=13-1=-23. (3)⎠⎛02(4-2x )(4-x 2)d x =⎠⎛02(16-8x -4x 2+2x 3)d x =⎝⎛⎭⎫16x -4x 2-43x 3+12x 4| 20 =32-16-323+8=403. (4)⎠⎛12x 2+2x -3x d x =⎠⎛12⎝⎛⎭⎫x +2-3x d x =⎝⎛⎭⎫12x 2+2x -3ln x | 21=72-3ln2. 16.计算以下定积分:[解析] (1)取F (x )=12sin2x ,那么F ′(x )=cos2x =12⎝⎛⎭⎫1-32=14(2-3).(2)取F (x )=x 22+ln x +2x ,那么 F ′(x )=x +1x+2. ∴⎠⎛23⎝⎛⎭⎫x +1x 2d x =⎠⎛23⎝⎛⎭⎫x +1x +2d x =F (3)-F (2)=⎝⎛⎭⎫92+ln3+6-⎝⎛⎭⎫12×4+ln2+4 =92+ln 32. (3)取F (x )=32x 2-cos x ,那么F ′(x )=3x +sin x 17.计算以下定积分:(1)⎠⎛0-4|x +2|d x ; (2)f (x )=,求⎠⎛3-1f (x )d x 的值. [解析] (1)∵f (x )=|x +2|=∴⎠⎛0-4|x +2|d x =-⎠⎛-4-2(x +2)d x +⎠⎛0-2(x +2)d x =-⎝⎛⎭⎫12x 2+2x | -2-4+⎝⎛⎭⎫12x 2+2x | 0-2 =2+2=4.(2)∵f (x )= ∴⎠⎛3-1f (x )d x =⎠⎛0-1f (x )d x +⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01(1-x )d x +⎠⎛12(x -1)d x =⎝⎛⎭⎫x -x 22| 10+⎝⎛⎭⎫x 22-x | 21 =12+12=1.18.(1)f (a )=⎠⎛01(2ax 2-a 2x )d x ,求f (a )的最大值; (2)f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛01f (x )d x =-2,求a ,b ,c 的值. [解析] (1)取F (x )=23ax 3-12a 2x 2 那么F ′(x )=2ax 2-a 2x ∴f (a )=⎠⎛01(2ax 2-a 2x )d x =F (1)-F (0)=23a -12a 2 =-12⎝⎛⎭⎫a -232+29 ∴当a =23时,f (a )有最大值29. (2)∵f (-1)=2,∴a -b +c =2①又∵f ′(x )=2ax +b ,∴f ′(0)=b =0②而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x 取F (x )=13ax 3+12bx 2+cx 那么F ′(x )=ax 2+bx +c ∴⎠⎛01f (x )d x =F (1)-F (0)=13a +12b +c =-2③ 解①②③得a =6,b =0,c =-4.。
人教版数学选修2-2第1章1.6微积分基本定理学业分层测评

学业分层测评(十一)(建议用时:45分钟)[学业达标]一、选择题 1.⎠⎛241x d x 等于( ) A .-2ln 2 B .2ln 2 C .-ln 2D .ln 2【解析】 ⎠⎛241x d x =ln x ⎪⎪42=ln 4-ln 2=ln 2. 【答案】 D 2.设a =⎠⎛01x 13d x ,b =⎠⎛01x 2d x ,c =⎠⎛01x 3d x ,则a ,b ,c 的大小关系是( ) A .a >b >c B .c >a >b C .a >c >bD .c >b >a【解析】 ∵a =⎠⎛01x 13d x =x 4343⎪⎪⎪10=34,b =⎠⎛01x 2d x =x 33⎪⎪⎪10=13,c =⎠⎛01x 3d x =x 44⎪⎪⎪10=14, ∴a >b >c . 【答案】 A3.已知积分⎠⎛01(kx +1)d x =k ,则实数k =( ) A .2 B .-2 C .1D .-1【解析】 ⎠⎛01(kx +1)d x =⎝ ⎛⎭⎪⎫12kx 2+x ⎪⎪⎪10=12k +1=k ,∴k =2.【答案】 A4.已知f (x )=2-|x |,则⎠⎛-12f (x )d x =( )A .3B .4C .72D .92【解析】 因为f (x )=2-|x |=⎩⎪⎨⎪⎧2+x , x ≤0,2-x , x ≥0,所以⎠⎛-12f (x )d x =⎠⎛-10(2+x )d x+⎠⎛02(2-x )d x =⎝ ⎛⎭⎪⎫2x +x 22⎪⎪⎪0-1+⎝ ⎛⎭⎪⎫2x -x 22⎪⎪20=32+2=72.【答案】 C5.设f (x )=⎩⎨⎧x 2,0≤x <1,2-x ,1<x ≤2,则⎠⎛02f (x )d x =( )A.23 B.34 C.45D.56【解析】 ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x=13x 3⎪⎪ 10+⎝ ⎛⎭⎪⎫2x -12x 2⎪⎪21=13+12=56. 【答案】 D 二、填空题6.若⎠⎛0k (2x -3x 2)d x =0,则k 等于__________.【导学号:62952053】【解析】 ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)|k 0=k 2-k 3=0,∴k =0(舍)或k =1.【答案】 17.设抛物线C :y =x 2与直线l :y =1围成的封闭图形为P ,则图形P 的面积S 等于____________ .【解析】 由⎩⎪⎨⎪⎧y =x 2,y =1,得x =±1.如图,由对称性可知,S =2(1×1-⎠⎛01x 2d x )=2⎝⎛⎭⎪⎫1×1-13x3| 10=43.【答案】438.已知f(x)=⎩⎪⎨⎪⎧lg x,x>0,x+⎠⎛a3t2d t,x≤0,若f(f(1))=1,则a=__________.【解析】因为f(1)=lg 1=0,且⎠⎛a3t2d t=t3|a0=a3-03=a3,所以f(0)=0+a3=1,所以a=1.【答案】 1三、解答题9.计算下列定积分.(1)⎠⎛121x(x+1)d x;(2)⎠⎜⎛-π2π2(cos x+2x)d x.【解】(1)∵⎠⎛121x(x+1)d x=⎠⎛12⎝⎛⎭⎪⎫1x-1x+1d x=[ln x-ln(x+1)]|21=ln 43.(2)⎠⎜⎛-π2π2(cos x+2x)d x=⎝⎛⎭⎪⎫sin x+2xln 2⎪⎪⎪⎪π2-π2=2+1ln 2(2π2-2-π2).10.设f (x )=ax 2+bx +c (a ≠0),f (1)=4,f ′(1)=1,⎠⎛01f (x )d x =196,求f (x ).【导学号:62952054】【解】 因为f (1)=4,所以a +b +c =4, ①f ′(x )=2ax +b ,因为f ′(1)=1,所以2a +b =1, ②⎠⎛01f (x )d x =⎝⎛⎭⎪⎫13ax 3+12bx 2+cx | 10 =13a +12b +c =196,③由①②③可得a =-1,b =3,c =2. 所以f (x )=-x 2+3x +2.[能力提升]1.若⎠⎛1a ⎝ ⎛⎭⎪⎫2x -1x d x =3-ln 2,且a >1,则a 的值为( )A .6B .4C .3D .2【解析】 ⎠⎛1a ⎝ ⎛⎭⎪⎫2x -1x d x =(x 2-ln x )|a 1 =a 2-ln a -1,故有a 2-ln a -1=3-ln 2, 解得a =2. 【答案】 D2.如图1-6-4所示,在边长为1的正方形OABC 中任取一点P ,则点P 恰好取自阴影部分的概率为( )图1-6-4A.14B.15C.16D.17【解析】 因为S 正方形=1, S 阴影=⎠⎛1(x -x )d x =⎝ ⎛⎭⎪⎫23x 32-12x 2| 10=23-12=16,所以点P 恰好取自阴影部分的概率为161=16. 【答案】 C3.计算:⎠⎛-22(2|x |+1)d x =__________.【导学号:62952055】【解析】 ⎠⎛-22 (2|x |+1)d x =⎠⎛-20(-2x +1)d x +⎠⎛02(2x +1)d x =(-x 2+x )|0-2+(x 2+x )|20 =-(-4-2)+(4+2)=12. 【答案】 124.已知f (x )=⎠⎛-a x (12t +4a )d t ,F (a )=⎠⎛01[f (x )+3a 2]d x ,求函数F (a )的最小值.【解】 因为f (x )=⎠⎛-a x (12t +4a )d t =(6t 2+4at )|x -a=6x 2+4ax -(6a 2-4a 2)=6x 2+4ax -2a 2, F (a )=⎠⎛01[f (x )+3a 2]d x =⎠⎛01(6x 2+4ax +a 2)d x=(2x 3+2ax 2+a 2x )|10=2+2a +a 2=a 2+2a +2=(a +1)2+1≥1. 所以当a =-1时,F (a )的最小值为1.。
人教版A版高中数学选修2-2第一章+1.6《微积分基本定理》【练习】(教师版)

1.6 微积分基本定理一、选择题1.10(2)e x x dx +⎰等于( )A.1B.e 1-C.eD.e+1 【答案】C【解析】Q 被积函数22()e ,e x x x x c c +++的原函数为为常数1121200(2)()e (1e e 0.e )()e x x x dx x ∴+=+=+-+=⎰|2.11x dx -⎰等于( )A. 11xdx -⎰B. 11dx -⎰C.()011x dx xdx --+⎰⎰D.()0110xdx x dx -+-⎰⎰【答案】C【解析】|x |=()()0,0,x x x x ⎧≥⎪⎨-<⎪⎩∴=()0110x dx xdx --+⎰⎰,选C.3.若1123ln 2,a x dx x ⎛⎫+=+ ⎪⎝⎭⎰A .6 B .4 C .3 D .2 【答案】D【解析】22111111122ln 1ln 3ln 2a a a aa x dx xdx dx x xa a x x ⎛⎫+=+=+=-+=+ ⎪⎝⎭⎰⎰⎰,解得a =2.4.设113a x dx =⎰,12b x dx =⎰,130c x dx =⎰,则a ,b ,c 的大小关系是( )A.c >a >bB.a >b >cC.a =b >cD.a >c >b【答案】B【解析】14113303344a x dx x===⎰,123101133b x dx x ===⎰,134101144c x dx x ===⎰,因为113434<<,所以a >b >c .5.设f (x )是一次函数,且()105f x dx =⎰,()1176xf x dx =⎰,则f (x )的解析式为( ) A .f (x )=4x +3 B .f (x )=3x +4 C .f (x )=-4x +2 D .f (x )=-3x +4 【答案】A【解析】∵f (x )是一次函数,∴设f (x )=ax +b (a ≠0),则()()11f x dx ax b dx =+=⎰⎰11152axdx bdx a b +=+=⎰⎰,()()1100xf x dx x ax b dx =+=⎰⎰()()1120ax dx bx dx +⎰⎰=1117.326a b +=由15,21117,326a b a b ⎧+=⎪⎪⎨⎪+=⎪⎩解得a =4,b =3,故f (x )=4x +3.6.已知分段函数()21,0,e ,0,x x x f x x -⎧+≤⎪=⎨>⎪⎩则()312f x dx -⎰等于( )A.13e+B.2e -D.12e-【答案】C【解析】121x x <≤⇒-<-21-<, 根据定积分性质可知(()33122f x f x dx --⎰()23221212e x x dx dx -⎡⎤=+-+⎣⎦⎰⎰()32221245e x d x x x dx --=++⎰⎰()3222312125e 3x x x x -⎛⎫=-++- ⎪⎝⎭()23228181025e e 33--⎡⎤⎛⎫⎛⎫=-+--+-- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦713e =-.二、填空题 7.计算定积分()121sin xx dx -+⎰= .【答案】23【解析】()12311112sin cos 33x x dx x x --⎛⎫+=-= ⎪⎝⎭⎰.8.已知()2321f x x x =++,若()()112f x dx f a -=⎰成立,则a = .【答案】1-或13【解析】取()32F x x x x =++,则()13F =,()11F -=-, 所以()()()11114f x F dx F ---==⎰,所以()24f a =,所以()2f a =.即23212a a ++=,解得1a =-或13.三、解答题 9.计算下列定积分:(1)52xdx ⎰;(2)()1202x x dx -⎰;(3)()()220424x x dx --⎰.【解析】(1)5502250252xdx x ==-=⎰.(2)()1112231210001122213x x dx x dx xdx x x -=-=-=-=-⎰⎰⎰.(3)()()(2220424x x dx --=⎰⎰22344116432x x x x ⎛⎫== ⎪⎝⎭--+10.已知f (x )()1f x dx ⎰=-2.(1)求f (x )的解析式;(2)求f (x )在[-1,1]上的最大值与最小值. 【解析】(1)设f (x )=ax 2+bx +c (a ≠0), 则f ′(x )=2ax +b .由f (-1)=2,f ′(0)=0,得2,0,a b c b -+=⎧⎨=⎩即2,0,c a b =-⎧⎨=⎩ ∴f (x )=ax 2+(2-a ).又()10f x dx ⎰=()1202ax a dx ⎡⎤+-⎣⎦⎰ =[13ax 3+(2-a )x ]10=2-23a =-2,∴a =6, 从而f (x )=6x 2-4.(2)∵f(x)=6x2-4,x∈[-1,1].∴当x=0时,f(x)min=-4;当x=±1时,f(x)max=2.。
数学人教A版选修2-2自我小测:1.6 微积分基本定理 Word版含解析

自我小测1.下列等式不正确的是( )A .11-⎰12 013x d x =0 B .11-⎰2 014d x =4 028 C .11-⎰ 2 014x 3d x =0 D .11-⎰cos x d x =0 2.若e 为自然对数的底数,则32⎰e 2-x d x =( ) A .1e -1 B .1-1eC .1-eD .e -13.53⎰x 2+1xd x 等于( ) A .8-ln 53 B .8+ln 53C .16-ln 53D .16+ln 534.0k⎰(2x -3x 2)d x =0,则k 等于( ) A .1 B .0C .0或1D .不确定5.下列定积分不大于0的是( )A .11-⎰|x |d x B .11-⎰(1-|x |)d x C .11-⎰|x -1|d x D .11-⎰(|x |-1)d x6.计算:21⎰⎝ ⎛⎭⎪⎫1x +1x 2d x =__________. 7.设函数f (x )=ax 2+c (a ≠0),若10⎰f (x )d x =f (x 0),0≤x 0≤1,则x 0的值为__________. 8.若f (x )=⎩⎪⎨⎪⎧ -e x ,x >1,|x |,x ≤1,则2⎰f (x )d x =__________. 9.计算定积分: (1)17π3π3⎰(2sin x -3cos x )d x ; (2)a a -⎰x 2d x (a >0); (3)21⎰1x (x +1)d x .⎰f(x)d x=5,10⎰xf(x)d x=176,求f(x)的解析式.10.设f(x)是一次函数,且1参考答案1.解析:11-⎰ 2 013x d x =2 013x 2211|-=2 0132-2 0132=0, 11-⎰2 014d x =2 014x 11|-=4 028, 11-⎰2 014x 3d x =2 0144x 411|-=0, 11-⎰cos x d x =sin x 11|-=sin 1-sin(-1)=2sin 1.故D 不正确. 答案:D2.解析:32⎰e 2-x d x =-e 2-x 32|=-e -1+e 0=1-1e . 答案:B3.解析:53⎰x 2+1x d x =53⎰x d x +53⎰1x d x =12x 253|+ln x 53| =12(52-32)+ln 5-ln 3=8+ln 53,故选B . 答案:B4.解析:∵0k ⎰(2x -3x 2)d x =(x 2-x 3)0|k =k 2-k 3, ∴k 2-k 3=0,即k =0或k =1.又∵k =0时不合题意,∴k =1.答案:A5.解析:11-⎰|x |d x =01-⎰(-x )d x +10⎰x d x=-12x 201|-+12x 201|=1, 11-⎰(1-|x |)d x =01-⎰(1+x )d x +10⎰(1-x )d x =1, 11-⎰|x -1|d x =11-⎰(1-x )d x =2,11-⎰(|x |-1)d x =11--⎰(1-|x |)d x =-1.答案:D6.解析:21⎰⎝ ⎛⎭⎪⎫1x +1x 2d x =⎝ ⎛⎭⎪⎫ln x -1x 21| =⎝⎛⎭⎪⎫ln 2-12-(ln 1-1)=ln 2+12. 答案:ln 2+127.解析:10⎰f (x )d x =10⎰(ax 2+c )d x =⎝ ⎛⎭⎪⎫13ax 3+cx 10| =a 3+c =ax 20+c , ∵0≤x 0≤1,∴x 0=33. 答案:33 8.解析:20⎰f (x )d x =10⎰|x |d x +21⎰(-e x )d x =10⎰x d x +21⎰(-e x )d x =12x 210|+(-e x )21| =12-e 2+e. 答案:12-e 2+e 9.解:(1)17π3π3⎰(2sin x -3cos x )d x =217π3π3⎰sin x d x -317π3π3⎰cos x d x =2(-cos x )17π3π3|-3sin x 17π3π3|=2⎝ ⎛⎭⎪⎫-cos 17π3+cos π3-3⎝ ⎛⎭⎪⎫sin 17π3-sin π3 =2⎝ ⎛⎭⎪⎫-12+12-3⎝ ⎛⎭⎪⎫-32-32=3 3. (2)由x 2=⎩⎪⎨⎪⎧ x ,x ≥0,-x ,x <0,得a a -⎰x 2d x =0a⎰x d x +0a -⎰(-x )d x =12x 20|a -12x 20|a -=a 2. (3)f (x )=1x (x +1)=1x -1x +1, 取F (x )=ln x -ln(x +1)=lnx x +1,则F ′(x )=1x -1x +1.所以21⎰1x (x +1)d x =21⎰⎝ ⎛⎭⎪⎫1x -1x +1d x =ln xx +121|=ln 43.10.解:设f (x )=ax +b (a ≠0), 则10⎰f (x )d x =10⎰(ax +b )d x =10⎰ax d x +10⎰b d x =12a +b =5, 10⎰xf (x )d x =10⎰x (ax +b )d x=10⎰ax 2d x +1⎰bx d x =13a +12b =176. 由⎩⎪⎨⎪⎧ 12a +b =5,13a +12b =176,解得a =4,b =3. 故f (x )=4x +3.。
人教A版选修2-2 1.6 微积分基本定理.docx

选修2-2 1.6 微积分基本定理一、选择题1.下列积分正确的是( )[答案] AA.214 B.54 C.338D.218[答案] A[解析] ⎠⎛2-2⎝⎛⎭⎫x 2+1x 4d x =⎠⎛2-2x 2d x +⎠⎛2-21x 4d x =13x 3| 2-2+⎝⎛⎭⎫-13x -3| 2-2 =13(x 3-x -3)| 2-2 =13⎝⎛⎭⎫8-18-13⎝⎛⎭⎫-8+18=214.故应选A.3.⎠⎛1-1|x |d x 等于( )A.⎠⎛1-1x d xB.⎠⎛1-1d xC.⎠⎛0-1(-x )d x +⎠⎛01x d xD.⎠⎛0-1x d x +⎠⎛01(-x )d x[答案] C[解析] ∵|x |=⎩⎪⎨⎪⎧x (x ≥0)-x (x <0)∴⎠⎛1-1|x |d x =⎠⎛0-1|x |d x +⎠⎛01|x |d x=⎠⎛0-1(-x )d x +⎠⎛01x d x ,故应选C.4.设f (x )=⎩⎪⎨⎪⎧x 2 (0≤x <1)2-x (1≤x ≤2),则⎠⎛02f (x )d x 等于( )A.34 B.45 C.56D .不存在[答案] C[解析] ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x取F 1(x )=13x 3,F 2(x )=2x -12x 2,则F ′1(x )=x 2,F ′2(x )=2-x ∴⎠⎛02f (x )d x =F 1(1)-F 1(0)+F 2(2)-F 2(1)=13-0+2×2-12×22-⎝⎛⎭⎫2×1-12×12=56.故应选C. 5.⎠⎛ab f ′(3x )d x =( )A .f (b )-f (a )B .f (3b )-f (3a ) C.13[f (3b )-f (3a )]D .3[f (3b )-f (3a )][答案] C[解析] ∵⎣⎡⎦⎤13f (3x )′=f ′(3x ) ∴取F (x )=13f (3x ),则⎠⎛ab f ′(3x )d x =F (b )-F (a )=13[f (3b )-f (3a )].故应选C. 6.⎠⎛03|x 2-4|d x =( )A.213 B.223 C.233D.253[答案] C[解析] ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x=⎝⎛⎭⎫4x -13x 3| 20+⎝⎛⎭⎫13x 3-4x | 32=233.A .-32 B .-12C.12D.32[答案] D[解析] ∵1-2sin 2θ2=cos θ8.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .cos xB .sin xC .-cos xD .-sin x[答案] A[解析] F (x )=⎠⎛0x cos t d t =sin t | x0=sin x -sin0=sin x .所以F ′(x )=cos x ,故应选A. 9.若⎠⎛0k (2x -3x 2)d x =0,则k =( )A .0B .1C .0或1D .以上都不对[答案] C[解析] ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)| k0=k 2-k 3=0,∴k =0或1.10.函数F (x )=⎠⎛0x t (t -4)d t 在[-1,5]上( )A .有最大值0,无最小值B .有最大值0和最小值-323C .有最小值-323,无最大值D .既无最大值也无最小值 [答案] B[解析] F (x )=⎠⎛0x (t 2-4t )d t =⎝⎛⎭⎫13t 3-2t 2| x0=13x 3-2x 2(-1≤x ≤5). F ′(x )=x 2-4x ,由F ′(x )=0得x =0或x =4,列表如下:x (-1,0) 0 (0,4) 4 (4,5) F ′(x ) +0 - 0 +F (x )极大值极小值可见极大值F (0)=0,极小值F (4)=-323.又F (-1)=-73,F (5)=-253∴最大值为0,最小值为-323. 二、填空题 11.计算定积分: ①⎠⎛1-1x 2d x =________②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =________ ③⎠⎛02|x 2-1|d x =________ ④⎠⎛0-π2|sin x |d x =________[答案] 23;436;2;1[解析] ①⎠⎛1-1x 2d x =13x 3| 1-1=23.②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =⎝⎛⎭⎫32x 2+2x | 32=436.③⎠⎛02|x 2-1|d x =⎠⎛01(1-x 2)d x +⎠⎛12(x 2-1)d x=⎝⎛⎭⎫x -13x 3| 10+⎝⎛⎭⎫13x 3-x | 21=2.[答案] 1+π213.(2010·陕西理,13)从如图所示的长方形区域内任取一个点M (x ,y ),则点M 取自阴影部分的概率为________.[答案] 13[解析] 长方形的面积为S 1=3,S 阴=⎠⎛013x 2dx =x 3| 10=1,则P =S 1S 阴=13.14.已知f (x )=3x 2+2x +1,若⎠⎛1-1f (x )d x =2f (a )成立,则a =________.[答案] -1或13[解析] 由已知F (x )=x 3+x 2+x ,F (1)=3,F (-1)=-1, ∴⎠⎛1-1f (x )d x =F (1)-F (-1)=4,∴2f (a )=4,∴f (a )=2.即3a 2+2a +1=2.解得a =-1或13.三、解答题15.计算下列定积分: (1)⎠⎛052x d x ;(2)⎠⎛01(x 2-2x )d x ;(3)⎠⎛02(4-2x )(4-x 2)d x ;(4)⎠⎛12x 2+2x -3x d x .[解析] (1)⎠⎛052x d x =x 2| 50=25-0=25.(2)⎠⎛01(x 2-2x )d x =⎠⎛01x 2d x -⎠⎛012x d x=13x 3| 10-x 2| 10=13-1=-23.(3)⎠⎛02(4-2x )(4-x 2)d x =⎠⎛02(16-8x -4x 2+2x 3)d x=⎝⎛⎭⎫16x -4x 2-43x 3+12x 4| 20 =32-16-323+8=403.(4)⎠⎛12x 2+2x -3x d x =⎠⎛12⎝⎛⎭⎫x +2-3x d x =⎝⎛⎭⎫12x 2+2x -3ln x | 21=72-3ln2. 16.计算下列定积分:[解析] (1)取F (x )=12sin2x ,则F ′(x )=cos2x=12⎝⎛⎭⎫1-32=14(2-3). (2)取F (x )=x 22+ln x +2x ,则F ′(x )=x +1x+2.∴⎠⎛23⎝⎛⎭⎫x +1x 2d x =⎠⎛23⎝⎛⎭⎫x +1x +2d x =F (3)-F (2)=⎝⎛⎭⎫92+ln3+6-⎝⎛⎭⎫12×4+ln2+4 =92+ln 32. (3)取F (x )=32x 2-cos x ,则F ′(x )=3x +sin x17.计算下列定积分: (1)⎠⎛0-4|x +2|d x ;(2)已知f (x )=,求⎠⎛3-1f (x )d x 的值.[解析] (1)∵f (x )=|x +2|=∴⎠⎛0-4|x +2|d x =-⎠⎛-4-2(x +2)d x +⎠⎛0-2(x +2)d x=-⎝⎛⎭⎫12x 2+2x | -2-4+⎝⎛⎭⎫12x 2+2x | 0-2 =2+2=4.(2)∵f (x )=∴⎠⎛3-1f (x )d x =⎠⎛0-1f (x )d x +⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01(1-x )d x +⎠⎛12(x -1)d x=⎝⎛⎭⎫x -x 22| 10+⎝⎛⎭⎫x 22-x | 21 =12+12=1. 18.(1)已知f (a )=⎠⎛01(2ax 2-a 2x )d x ,求f (a )的最大值;(2)已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛01f (x )d x =-2,求a ,b ,c的值.[解析] (1)取F (x )=23ax 3-12a 2x 2则F ′(x )=2ax 2-a 2x ∴f (a )=⎠⎛01(2ax 2-a 2x )d x=F (1)-F (0)=23a -12a 2=-12⎝⎛⎭⎫a -232+29∴当a =23时,f (a )有最大值29.(2)∵f (-1)=2,∴a -b +c =2① 又∵f ′(x )=2ax +b ,∴f ′(0)=b =0② 而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x取F (x )=13ax 3+12bx 2+cx则F ′(x )=ax 2+bx +c∴⎠⎛01f (x )d x =F (1)-F (0)=13a +12b +c =-2③解①②③得a =6,b =0,c =-4.。
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选修2-2 1.6 微积分基本定理一、选择题1.下列积分正确的是( )A.122713=⎰xdxB.e e dx x e x-=⎰2121 C.()316122ln 0=+⎰dx e e xx 222=⎰-ππxdx D.[答案] A [解析]12232723233227132271312713=-⨯===⎰⎰-x dx x xdx2.=⎪⎭⎫⎝⎛+⎰-dx x x 22421 A.214 B.54 C.338D.218[答案] A[解析] ⎠⎛2-2⎝⎛⎭⎫x 2+1x 4d x =⎠⎛2-2x 2d x +⎠⎛2-21x 4d x =13x 3| 2-2+⎝⎛⎭⎫-13x -3| 2-2=13(x 3-x -3)| 2-2 =13⎝⎛⎭⎫8-18-13⎝⎛⎭⎫-8+18=214.故应选A. 3.⎰-11|x |d x 等于( )A.⎠⎛1-1x d xB.⎠⎛1-1d xC.⎠⎛0-1(-x )d x +⎠⎛01x d xD.⎠⎛0-1x d x +⎠⎛01(-x )d x[答案] C[解析] ∵|x |=⎩⎪⎨⎪⎧x (x ≥0)-x (x <0)∴⎠⎛1-1|x |d x =⎠⎛0-1|x |d x +⎠⎛01|x |d x=⎠⎛0-1(-x )d x +⎠⎛01x d x ,故应选C.4.设f (x )=⎩⎪⎨⎪⎧x 2 (0≤x <1)2-x (1≤x ≤2),则⎠⎛02f (x )d x 等于( )A.34 B.45 C.56D .不存在[答案] C[解析] ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x取F 1(x )=13x 3,F 2(x )=2x -12x 2, 则F ′1(x )=x 2,F ′2(x )=2-x∴⎠⎛02f (x )d x =F 1(1)-F 1(0)+F 2(2)-F 2(1)=13-0+2×2-12×22-⎝⎛⎭⎫2×1-12×12=56.故应选C.5.⎠⎛ab f ′(3x )d x =( )A .f (b )-f (a )B .f (3b )-f (3a )C.13[f (3b )-f (3a )] D .3[f (3b )-f (3a )][答案] C[解析] ∵⎣⎡⎦⎤13f (3x )′=f ′(3x ) ∴取F (x )=13f (3x ),则⎠⎛ab f ′(3x )d x =F (b )-F (a )=13[f (3b )-f (3a )].故应选C.6.⎠⎛03|x 2-4|d x =( )A.213 B.223 C.233 D.253[答案] C[解析] ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x=⎝⎛⎭⎫4x -13x 3| 20+⎝⎛⎭⎫13x 3-4x | 32=233. 7.θθπd ⎰⎪⎭⎫ ⎝⎛-322sin 21 的值为 ( )A .-32 B .-12 C.12D.32[答案] D[解析] ∵1-2sin 2θ2=cos θ23sin cos 2sin 21330302===⎪⎭⎫ ⎝⎛-∴⎰⎰πππθθθθd d ,故应选D 8.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .cos xB .sin xC .-cos xD .-sin x[答案] A[解析] F (x )=⎠⎛0x cos t d t =sin t | x0=sin x -sin0=sin x . 所以F ′(x )=cos x ,故应选A. 9.若⎠⎛0k (2x -3x 2)d x =0,则k =( )A .0B .1C .0或1D .以上都不对[答案] C[解析] ⎠⎛0k (2x -3x 2)d x =(x 2-x 3)| k=k 2-k 3=0, ∴k =0或1.10.函数F (x )=⎠⎛0x t (t -4)d t 在[-1,5]上( )A .有最大值0,无最小值B .有最大值0和最小值-323 C .有最小值-323,无最大值 D .既无最大值也无最小值 [答案] B[解析] F (x )=⎠⎛0x (t 2-4t )d t =⎝⎛⎭⎫13t 3-2t 2| x0=13x 3-2x 2(-1≤x ≤5).F ′(x )=x 2-4x ,由F ′(x )=0得x =0或x =4,列表如下:x (-1,0) 0(0,4) 4(4,5) F ′(x ) +0 -0 +F (x )极大值极小值可见极大值F (0)=0,极小值F (4)=-323. 又F (-1)=-73,F (5)=-253 ∴最大值为0,最小值为-323.二、填空题11.计算定积分: ①⎠⎛1-1x 2d x =________②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =________③⎠⎛02|x 2-1|d x =________④⎠⎛0-π2|sin x |d x =________[答案] 23;436;2;1[解析] ①⎠⎛1-1x 2d x =13x 3| 1-1=23. ②⎠⎛23⎝⎛⎭⎫3x -2x 2d x =⎝⎛⎭⎫32x 2+2x | 32=436.③⎠⎛02|x 2-1|d x =⎠⎛01(1-x 2)d x +⎠⎛12(x 2-1)d x=⎝⎛⎭⎫x -13x 3| 10+⎝⎛⎭⎫13x 3-x | 21=2.④()1cos sin sin 0222==-=---⎰⎰πππxdx x dx x12..________2cos 2sin 220=⎪⎭⎫ ⎝⎛+⎰dx x x π[答案] 1+π2[解析]()()12cos sin 12cos 2sin 220220+=-=+=⎪⎭⎫ ⎝⎛+⎰⎰ππππx x dx x dx x x 13.(2010·陕西理,13)从如图所示的长方形区域内任取一个点M (x ,y ),则点M 取自阴影部分的概率为________.[答案] 13[解析] 长方形的面积为S 1=3,S 阴=⎠⎛013x 2dx =x 3| 10=1,则P =S 1S 阴=13. 14.已知f (x )=3x 2+2x +1,若⎠⎛1-1f (x )d x =2f (a )成立,则a =________.[答案] -1或13[解析] 由已知F (x )=x 3+x 2+x ,F (1)=3,F (-1)=-1, ∴⎠⎛1-1f (x )d x =F (1)-F (-1)=4,∴2f (a )=4,∴f (a )=2.即3a 2+2a +1=2.解得a =-1或13.三、解答题15.计算下列定积分: (1)⎠⎛052x d x ;(2)⎠⎛01(x 2-2x )d x ;(3)⎠⎛2(4-2x )(4-x 2)d x ;(4)⎠⎛12x 2+2x -3x d x .[解析] (1)⎠⎛052x d x =x 2| 50=25-0=25. (2)⎠⎛01(x 2-2x )d x =⎠⎛01x 2d x -⎠⎛012x d x=13x 3| 10-x 2| 10=13-1=-23. (3)⎠⎛02(4-2x )(4-x 2)d x =⎠⎛02(16-8x -4x 2+2x 3)d x=⎝⎛⎭⎫16x -4x 2-43x 3+12x 4| 20 =32-16-323+8=403.(4)⎠⎛12x 2+2x -3x d x =⎠⎛12⎝⎛⎭⎫x +2-3x d x =⎝⎛⎭⎫12x 2+2x -3ln x | 21=72-3ln2.16.计算下列定积分:(1)⎰462cos ππxdx (2)dx x x 2321⎰⎪⎪⎭⎫⎝⎛+ (3) ()⎰+2sin 3πdx x x (4)⎰bax dx e[解析] (1)取F (x )=12sin2x ,则F ′(x )=cos2x∴⎪⎭⎫⎝⎛-=⎰6)4(2cos 46ππππF F xdx=12⎝ ⎛⎭⎪⎫1-32=14(2-3).(2)取F (x )=x 22+ln x +2x ,则 F ′(x )=x +1x +2.∴⎠⎛23⎝ ⎛⎭⎪⎫x +1x 2d x =⎠⎛23⎝⎛⎭⎫x +1x +2d x =F (3)-F (2)=⎝⎛⎭⎫92+ln3+6-⎝⎛⎭⎫12×4+ln2+4 =92+ln 32.(3)取F (x )=32x 2-cos x ,则F ′(x )=3x +sin x∴()()18302sin 3220+=-⎪⎭⎫⎝⎛=+⎰πππF F dx x x(4)取()xe x F =,则xe x F =)('∴a b baxbax e e e dx e -==⎰17.计算下列定积分:(1)⎠⎛0-4|x +2|d x ;(2)已知f (x )=,求⎠⎛3-1f (x )d x 的值.[解析] (1)∵f (x )=|x +2|=∴⎠⎛0-4|x +2|d x =-⎠⎛-4-2(x +2)d x +⎠⎛0-2(x +2)d x=-⎝⎛⎭⎫12x 2+2x | -2-4+⎝⎛⎭⎫12x 2+2x | 0-2=2+2=4.(2)∵f (x )=∴⎠⎛3-1f (x )d x =⎠⎛0-1f (x )d x +⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01(1-x )d x +⎠⎛12(x -1)d x=⎝⎛⎭⎫x -x 22| 10+⎝⎛⎭⎫x 22-x | 21 =12+12=1.18.(1)已知f (a )=⎠⎛01(2ax 2-a 2x )d x ,求f (a )的最大值;(2)已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛01f (x )d x =-2,求a ,b ,c 的值.[解析] (1)取F (x )=23ax 3-12a 2x 2 则F ′(x )=2ax 2-a 2x ∴f (a )=⎠⎛01(2ax 2-a 2x )d x=F (1)-F (0)=23a -12a 2=-12⎝⎛⎭⎫a -232+29 ∴当a =23时,f (a )有最大值29.(2)∵f (-1)=2,∴a -b +c =2① 又∵f ′(x )=2ax +b ,∴f ′(0)=b =0② 而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x取F (x )=13ax 3+12bx 2+cx 则F ′(x )=ax 2+bx +c∴⎠⎛01f (x )d x =F (1)-F (0)=13a +12b +c =-2③解① ② ③ 得a =6,b =0,c =-4.。