2018年高中数学课时跟踪检测十一微积分基本定理新人教A版选修2_2201803074233
高中数学人教A版选修2-2(课时训练):1.6 微积分基本定理 Word版含答案.docx
1.6 微积分基本定理[学习目标]1.直观了解并掌握微积分基本定理的含义. 2.会利用微积分基本定理求函数的定积分. [知识链接]1.导数与定积分有怎样的联系?答 导数与定积分都是微积分学中两个最基本、最重要的概念,运用它们之间的联系,我们可以找出求定积分的方法,求导数与定积分是互为逆运算.2.在下面图(1)、图(2)、图(3)中的三个图形阴影部分的面积分别怎样表示?答 根据定积分与曲边梯形的面积的关系知: 图(1)中S =⎠⎛ab f (x )d x ,图(2)中S =-⎠⎛ab f (x )d x ,图(3)中S =⎠⎛0b f (x )d x -⎠⎛a0f (x )d x .[预习导引] 1.微积分基本定理如果f (x )是区间[a ,b ]上的连续函数,并且F ′(x )=f (x ),那么⎠⎛ab f (x )d x =F (b )-F (a ).2.函数f (x )与其一个原函数的关系 (1)若f (x )=c (c 为常数),则F (x )=cx ; (2)若f (x )=x n (n ≠-1),则F (x )=1n +1·x n +1;(3)若f (x )=1x ,则F (x )=ln_x (x >0);(4)若f (x )=e x ,则F (x )=e x ;(5)若f (x )=a x,则F (x )=a xln a(a >0且a ≠1);(6)若f (x )=sin x ,则F (x )=-cos_x ; (7)若f (x )=cos x ,则F (x )=sin_x .要点一 求简单函数的定积分 例1 计算下列定积分 (1)⎠⎛123d x ; (2)⎠⎛02(2x +3)d x ;(3)⎠⎛3-1(4x -x 2)d x ; (4)⎠⎛12(x -1)5d x .解 (1)因为(3x )′=3,所以⎠⎛123d x =(3x )⎪⎪⎪21=3×2-3×1=3. (2)因为(x 2+3x )′=2x +3, 所以⎠⎛2(2x +3)d x =(x 2+3x )⎪⎪⎪2=22+3×2-(02+3×0)=10. (3)因为⎝⎛⎭⎫2x 2-x33′=4x -x 2, 所以⎠⎛3-1(4x -x 2)d x =⎝⎛⎭⎫2x 2-x 33⎪⎪⎪3-1 =⎝⎛⎭⎫2×32-333-⎣⎡⎦⎤2×(-1)2-(-1)33=203.(4)因为⎣⎡⎦⎤16(x -1)6′=(x -1)5, 所以⎠⎛21(x -1)5d x=16(x -1)6⎪⎪⎪21=16(2-1)6-16(1-1)6 =16. 规律方法 (1)用微积分基本定理求定积分的步骤: ①求f (x )的一个原函数F (x ); ②计算F (b )-F (a ). (2)注意事项:①有时需先化简,再求积分;②f (x )的原函数有无穷多个,如F (x )+c ,计算时,一般只写一个最简单的,不再加任意常数c .跟踪演练1 求下列定积分: (1)∫π20(3x +sin x )d x ;(2)⎠⎛21⎝⎛⎭⎫e x -1x d x . 解 (1)∵⎝⎛⎭⎫32x 2-cos x ′=3x +sin x , ∴∫π20(3x +sin x )d x =⎝⎛⎭⎫32x 2-cos x ⎪⎪⎪⎪π20=⎣⎡⎦⎤32×⎝⎛⎭⎫π22-cos π2-⎝⎛⎭⎫32×0-cos0=3π28+1; (2)∵(e x -ln x )′=e x -1x,∴⎠⎛21(e x-1x )d x =()e x -ln x ⎪⎪⎪21=(e 2-ln2)-(e -0) =e 2-e -ln2.要点二 求较复杂函数的定积分 例2 求下列定积分:(1)⎠⎛41x (1-x )d x ; (2)∫π202cos 2x2d x ;(3)⎠⎛41(2x +1x)d x . 解 (1)∵x (1-x )=x -x , 又∵⎝⎛⎭⎫23x 32-12x 2′=x -x .∴⎠⎛41x (1-x )d x =⎝⎛⎭⎫23x 32-12x 2⎪⎪⎪41 =⎝⎛⎭⎫23×432-12×42-⎝⎛⎭⎫23-12=-176. (2)∵2cos 2x2=1+cos x ,(x +sin x )′=1+cos x ,∴原式=∫π20(1+cos x )d x =(x +sin x )⎪⎪⎪⎪π20=π2+1.(3)∵⎝⎛⎭⎫2xln2+2x ′=2x +1x,∴⎠⎛41(2x +1x)d x =⎝⎛⎭⎫2xln2+2x ⎪⎪⎪41=⎝⎛⎭⎫24ln2+24-⎝⎛⎭⎫2ln2+2=14ln2+2. 规律方法 求较复杂函数的定积分的方法:(1)掌握基本初等函数的导数以及导数的运算法则,正确求解被积函数的原函数,当原函数不易求时,可将被积函数适当变形后求解,具体方法是能化简的化简,不能化简的变为幂函数、正、余弦函数、指数、对数函数与常数的和与差. (2)确定积分区间,分清积分下限与积分上限. 跟踪演练2 计算下列定积分: (1)∫π30(sin x -sin2x )d x ;(2)⎠⎛0ln 2e x (1+e x )d x .解 (1)sin x -sin2x 的一个原函数是-cos x + 12cos2x ,所以∫π30(sin x -sin2x )d x =⎝⎛⎭⎫-cos x +12cos2x ⎪⎪⎪⎪π30=⎝⎛⎭⎫-12-14-⎝⎛⎭⎫-1+12=-14. (2)∵e x (1+e x )=e x +e 2x , ∴⎝⎛⎭⎫e x +12e 2x ′=e x +e 2x , ∴⎠⎛0ln 2e x (1+e x )d x =⎠⎛0ln 2()e x +e 2x d x=⎝⎛⎭⎫e x +12e 2x ⎪⎪⎪ln2=e ln2+12e 2ln2-e 0-12e 0=2+12×4-1-12=52.要点三 定积分的简单应用例3 已知f (a )=⎠⎛10(2ax 2-a 2x )d x ,求f (a )的最大值.解 ∵⎝⎛⎭⎫23ax 3-12a 2x 2′=2ax 2-a 2x ,∴⎠⎛10(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 -232+29, ∴当a =23时,f (a )有最大值29.规律方法 定积分的应用体现了积分与函数的内在联系,可以通过积分构造新的函数,进而对这一函数进行性质、最值等方面的考查,解题过程中注意体会转化思想的应用. 跟踪演练3 已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛10f (x )d x =-2,求a 、b 、c 的值.解 由f (-1)=2,得a -b +c =2. ① 又f ′(x )=2ax +b ,∴f ′(0)=b =0, ②而⎠⎛10f (x )d x =⎠⎛10(ax 2+bx +c )d x =⎝⎛⎭⎫13ax 3+12bx 2+cx ⎪⎪⎪1=13a +12b +c , ∴13a +12b +c =-2, ③由①②③式得a =6,b =0,c =-4. 要点四 求分段函数的定积分 例4 计算下列定积分:(1)若f (x )=⎩⎪⎨⎪⎧x 2 (x ≤0)cos x -1 (x >0),求∫π2-1f (x )d x ;(2)⎠⎛30|x 2-4|d x .解 (1)∫π2-1f (x )d x =⎠⎛0-1x 2d x +∫π20(cos x -1)d x ,又∵⎝⎛⎭⎫13x 3′=x 2,(sin x -x )′=cos x -1∴原式=13x 3⎪⎪⎪0-1+(sin x -x )⎪⎪⎪⎪π20=⎝⎛⎭⎫0+13+⎝⎛⎭⎫sin π2-π2-(sin0-0) =43-π2.(2)∵|x 2-4|=⎩⎪⎨⎪⎧x 2-4 (x ≥2或x ≤-2),4-x 2 (-2<x <2),又∵⎝⎛⎭⎫13x 3-4x ′=x 2-4,⎝⎛⎭⎫4x -13x 3′=4-x 2, ∴⎠⎛30|x 2-4|d x =⎠⎛20(4-x 2)d x +⎠⎛32(x 2-4)d x=⎝⎛⎭⎫4x -13x 3⎪⎪⎪20+⎝⎛⎭⎫13x 3-4x ⎪⎪⎪32 =⎝⎛⎭⎫8-83-0+(9-12)-⎝⎛⎭⎫83-8=233. 规律方法 (1)求分段函数的定积分时,可利用积分性质将其表示为几段积分和的形式; (2)带绝对值的解析式,先根据绝对值的意义找到分界点,去掉绝对值号,化为分段函数; (3)含有字母参数的绝对值问题要注意分类讨论. 跟踪演练4 求⎠⎛3-3(|2x +3|+|3-2x |)d x .解 ∵|2x +3|+|3-2x |=⎩⎪⎨⎪⎧-4x ,x <-32,6,-32≤x ≤32,4x ,x >32,∴⎠⎛3-3(|2x +3|+|3-2x |)d x=∫-32-3(-4x )d x +∫32-326d x +∫3324x d x=-2x 2⎪⎪⎪⎪-32-3+6x ⎪⎪⎪32-32+2x 2⎪⎪⎪⎪332=45.1.∫π2-π2(1+cos x )d x 等于( )A .πB .2C .π-2D .π+2答案 D解析 ∵(x +sin x )′=1+cos x , ∴⎪⎪∫π2-π2(1+cos x )d x =(x +sin x )π2-π2=π2+sin π2-⎣⎡⎦⎤-π2+sin ⎝⎛⎭⎫-π2=π+2. 2.若⎠⎛1a ⎝⎛⎭⎫2x +1x d x =3+ln2,则a 的值是( ) A .5 B .4 C .3 D .2答案 D解析 ⎠⎛1a ⎝⎛⎭⎫2x +1x d x =⎠⎛1a 2x d x +⎠⎛1a 1xd x =x 2|a 1+ ln x 错误!=a 2-1+ln a =3+ln2,解得a =2. 3.⎠⎛02⎝⎛⎭⎫x 2-23x d x =________. 答案 43解析 ⎠⎛02⎝⎛⎭⎫x 2-23x d x =⎠⎛02x 2d x -⎠⎛0223x d x =x 33⎪⎪⎪⎪20-x 2320=83-43=43. 4.已知f (x )=⎩⎨⎧4x -2π,0≤x ≤π2,cos x ,π2<x ≤π,计算⎠⎛0πf (x )d x .解 ⎠⎛0πf (x )d x =∫π20f (x )d x +错误!f (x )d x=∫π20(4x -2π)d x +错误!cos x d x ,取F 1(x )=2x 2-2πx ,则F 1′(x )=4x -2π; 取F 2(x )=sin x ,则F 2′(x )=cos x .所以∫π20(4x -2π)d x +错误!cos x d x =(2x 2-2πx )错误!+sin x 错误!,即错误!f (x )d x =-错误!π2-1.1.求定积分的一些常用技巧(1)对被积函数,要先化简,再求积分.(2)若被积函数是分段函数,依据定积分“对区间的可加性”,分段积分再求和. (3)对于含有绝对值符号的被积函数,要去掉绝对值符号才能积分.2.由于定积分的值可取正值,也可取负值,还可以取0,而面积是正值,因此不要把面积理解为被积函数对应图形在某几个区间上的定积分之和,而是在x 轴下方的图形面积要取定积分的相反数.一、基础达标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 )⎪⎪ba ; 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 =li m n →∞i =1nb -ans ′(ξi ); ④它在时间段[a ,b ]内的位移是s =⎠⎛ab s ′(t )d t .A .①B .①②C .①②④D .①②③④答案 D2.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)答案 B解析 若F (x )=x 3,则F ′(x )=3x 2,这与F ′(x )=x 2不一致,故选B. 3.⎠⎛01(e x +2x )d x 等于( )A .1B .e -1C .eD .e +1答案 C解析 ⎠⎛01(e x +2x )d x =(e x +x 2)|10=(e 1+12)-(e 0+02)=e. 4.已知f (x )=⎩⎪⎨⎪⎧x 2,-1≤x ≤0,1,0<x ≤1,则⎠⎛1-1f (x )d x 的值为( )A.32 B .43C .23D .-23答案 B解析 ⎠⎛1-1f (x )d x =⎠⎛0-1x 2d x +⎠⎛011d x =⎪⎪x 330-1+1=13+1=43,故选B. 5.设函数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________.答案33解析 由已知得13a +c =ax 20+c ,∴x 20=13,又∵0≤x 0≤1,∴x 0=33. 6.(2013·湖南)若⎠⎛0T x 2d x =9,则常数T 的值为________.答案 3解析 ⎠⎛0T x 2d x =⎪⎪13x 3T 0=13T 3=9,即T 3=27,解得T =3. 7.已知⎠⎛1-1(x 3+ax +3a -b )d x =2a +6且f (t )=⎠⎛0t (x 3+ax +3a -b )d x 为偶函数,求a ,b 的值.解 ∵f (x )=x 3+ax 为奇函数, ∴⎠⎛1-1(x 3+ax )d x =0,∴⎠⎛1-1(x 3+ax +3a -b )d x=⎠⎛1-1(x 3+ax )d x +⎠⎛1-1(3a -b )d x=0+(3a -b )[1-(-1)]=6a -2b . ∴6a -2b =2a +6,即2a -b =3, ①又f (t )=⎪⎪⎣⎡⎦⎤x 44+a 2x 2+(3a -b )x t0 =t 44+at 22+(3a -b )t 为偶函数, ∴3a -b =0,②由①②得a =-3,b =-9. 二、能力提升8.∫π20sin 2x2d x 等于( )A.π4 B .π2-1C .2D .π-24答案 D解析 ∫π20sin 2x 2d x =∫π201-cos x 2d x =⎪⎪12(x -sin x )π20=π-24,故选D. 9.(2013·江西)若S 1=⎠⎛12x 2d x ,S 2=⎠⎛121x d x ,S 3=⎠⎛12e 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答案 B 解析 S 1=⎠⎛12x 2d x =13x 3⎪⎪21=73,S 2=⎪⎪⎪⎠⎛121x d x =ln x 21=ln2<1,S 3=⎠⎛12e x d x =e x |21=e 2-e =e(e -1)>73,所以S 2<S 1<S 3,选B.10.设f (x )=⎩⎪⎨⎪⎧lg x ,x >0,x +⎠⎛0a 3t 2d t ,x ≤0.若f [f (1)]=1,则a =________.答案 1解析 因为x =1>0,所以f (1)=lg1=0.又x ≤0时,f (x )=x +⎠⎛0a 3t 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⎠⎛01f (x )d x =5,⎠⎛01xf (x )d x =176,求f (x )的解析式.解 ∵f (x )是一次函数,设f (x )=ax +b (a ≠0),则 ⎠⎛01f (x )d x =⎠⎛01(ax +b )d x =⎠⎛01ax d x +⎠⎛01b d x =12a +b =5, ⎠⎛01xf (x )d x =⎠⎛01x (ax +b )d x =⎠⎛01(ax 2)d x +⎠⎛a1b x d x =13a +12b =176. 由⎩⎨⎧12a +b =513a +12b =176,得⎩⎪⎨⎪⎧a =4b =3.即f (x )=4x +3.12.若函数f (x )=⎩⎪⎨⎪⎧x 3,x ∈[0,1],x ,x ∈(1,2],2x ,x ∈(2,3].求⎠⎛03f (x )d x 的值.解 由积分的性质,知:经典小初高讲义小初高优秀教案 ⎠⎛03f (x )d x =⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01x 3d x +⎠⎛12x d x +⎠⎛232x d x =x 44⎪⎪⎪⎪10+23x 3221 ⎪⎪+2x ln232 =14+432-23+8ln2-4ln2=-512+432+4ln2. 三、探究与创新13.求定积分⎠⎛3-4|x +a |d x . 解 (1)当-a ≤-4即a ≥4时,原式=⎠⎛3-4(x +a )d x = ⎪⎪⎝⎛⎭⎫x 22+ax 3-4=7a -72. (2)当-4<-a <3即-3<a <4时, 原式=⎠⎛-4-a [-(x +a )]d x +⎠⎛3-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=⎠⎛3-4[-(x +a )]d x = ⎪⎪⎝⎛⎭⎫-x 22-ax 3-4= -7a +72. 综上,得⎠⎛3-4|x +a |d x =⎩⎪⎨⎪⎧ 7a -72(a ≥4),a 2-a +252(-3<a <4),-7a +72(a ≤-3).。
高中数学人教A版选修2-2(课时训练):1.6 微积分基本定理 Word版含答案
1.6 微积分基本定理[学习目标]1.直观了解并掌握微积分基本定理的含义. 2.会利用微积分基本定理求函数的定积分. [知识链接]1.导数与定积分有怎样的联系?答 导数与定积分都是微积分学中两个最基本、最重要的概念,运用它们之间的联系,我们可以找出求定积分的方法,求导数与定积分是互为逆运算.2.在下面图(1)、图(2)、图(3)中的三个图形阴影部分的面积分别怎样表示?答 根据定积分与曲边梯形的面积的关系知: 图(1)中S =⎠⎛ab f (x )d x ,图(2)中S =-⎠⎛ab f (x )d x ,图(3)中S =⎠⎛0b f (x )d x -⎠⎛a0f (x )d x .[预习导引] 1.微积分基本定理如果f (x )是区间[a ,b ]上的连续函数,并且F ′(x )=f (x ),那么⎠⎛ab f (x )d x =F (b )-F (a ).2.函数f (x )与其一个原函数的关系 (1)若f (x )=c (c 为常数),则F (x )=cx ; (2)若f (x )=x n (n ≠-1),则F (x )=1n +1·x n +1;(3)若f (x )=1x ,则F (x )=ln_x (x >0);(4)若f (x )=e x ,则F (x )=e x ;(5)若f (x )=a x,则F (x )=a xln a(a >0且a ≠1);(6)若f (x )=sin x ,则F (x )=-cos_x ; (7)若f (x )=cos x ,则F (x )=sin_x .要点一 求简单函数的定积分 例1 计算下列定积分 (1)⎠⎛123d x ; (2)⎠⎛02(2x +3)d x ;(3)⎠⎛3-1(4x -x 2)d x ; (4)⎠⎛12(x -1)5d x .解 (1)因为(3x )′=3,所以⎠⎛123d x =(3x )⎪⎪⎪21=3×2-3×1=3. (2)因为(x 2+3x )′=2x +3, 所以⎠⎛02(2x +3)d x =(x 2+3x )⎪⎪⎪2=22+3×2-(02+3×0)=10.(3)因为⎝⎛⎭⎫2x 2-x33′=4x -x 2, 所以⎠⎛3-1(4x -x 2)d x =⎝⎛⎭⎫2x 2-x 33⎪⎪⎪3-1=⎝⎛⎭⎫2×32-333-⎣⎡⎦⎤2×(-1)2-(-1)33=203. (4)因为⎣⎡⎦⎤16(x -1)6′=(x -1)5, 所以⎠⎛21(x -1)5d x=16(x -1)6⎪⎪⎪21=16(2-1)6-16(1-1)6 =16. 规律方法 (1)用微积分基本定理求定积分的步骤: ①求f (x )的一个原函数F (x ); ②计算F (b )-F (a ). (2)注意事项:①有时需先化简,再求积分;②f (x )的原函数有无穷多个,如F (x )+c ,计算时,一般只写一个最简单的,不再加任意常数c .跟踪演练1 求下列定积分: (1)∫π20(3x +sin x )d x ;(2)⎠⎛21⎝⎛⎭⎫e x -1x d x . 解 (1)∵⎝⎛⎭⎫32x 2-cos x ′=3x +sin x , ∴∫π20(3x +sin x )d x =⎝⎛⎭⎫32x 2-cos x ⎪⎪⎪⎪π20=⎣⎡⎦⎤32×⎝⎛⎭⎫π22-cos π2-⎝⎛⎭⎫32×0-cos 0=3π28+1; (2)∵(e x -ln x )′=e x -1x,∴⎠⎛21(e x-1x )d x =()e x -ln x ⎪⎪⎪21=(e 2-ln 2)-(e -0) =e 2-e -ln 2.要点二 求较复杂函数的定积分 例2 求下列定积分:(1)⎠⎛41x (1-x )d x ; (2)∫π202cos 2x2d x ;(3)⎠⎛41(2x +1x)d x . 解 (1)∵x (1-x )=x -x , 又∵⎝⎛⎭⎫23x 32-12x 2′=x -x . ∴⎠⎛41x (1-x )d x =⎝⎛⎭⎫23x 32-12x 2⎪⎪⎪41 =⎝⎛⎭⎫23×432-12×42-⎝⎛⎭⎫23-12=-176. (2)∵2cos 2x2=1+cos x ,(x +sin x )′=1+cos x ,∴原式=∫π20(1+cos x )d x =(x +sin x )⎪⎪⎪⎪π20=π2+1.(3)∵⎝⎛⎭⎫2xln 2+2x ′=2x +1x,∴⎠⎛41(2x+1x)d x =⎝⎛⎭⎫2x ln 2+2x ⎪⎪⎪41 =⎝⎛⎭⎫24ln 2+24-⎝⎛⎭⎫2ln 2+2=14ln 2+2. 规律方法 求较复杂函数的定积分的方法:(1)掌握基本初等函数的导数以及导数的运算法则,正确求解被积函数的原函数,当原函数不易求时,可将被积函数适当变形后求解,具体方法是能化简的化简,不能化简的变为幂函数、正、余弦函数、指数、对数函数与常数的和与差. (2)确定积分区间,分清积分下限与积分上限. 跟踪演练2 计算下列定积分: (1)∫π30(sin x -sin 2x )d x ;(2)⎠⎛0ln 2e x (1+e x )d x .解 (1)sin x -sin 2x 的一个原函数是-cos x + 12cos 2x ,所以∫π30(sin x -sin 2x )d x =⎝⎛⎭⎫-cos x +12cos 2x ⎪⎪⎪⎪π30=⎝⎛⎭⎫-12-14-⎝⎛⎭⎫-1+12=-14. (2)∵e x (1+e x )=e x +e 2x , ∴⎝⎛⎭⎫e x +12e 2x ′=e x +e 2x , ∴⎠⎛0ln 2e x (1+e x )d x =⎠⎛0ln 2()e x+e2xd x=⎝⎛⎭⎫e x +12e 2x ⎪⎪⎪ln 2=e ln 2+12e 2ln 2-e 0-12e 0=2+12×4-1-12=52.要点三 定积分的简单应用例3 已知f (a )=⎠⎛10(2ax 2-a 2x )d x ,求f (a )的最大值.解 ∵⎝⎛⎭⎫23ax 3-12a 2x 2′=2ax 2-a 2x , ∴⎠⎛10(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 -232+29, ∴当a =23时,f (a )有最大值29.规律方法 定积分的应用体现了积分与函数的内在联系,可以通过积分构造新的函数,进而对这一函数进行性质、最值等方面的考查,解题过程中注意体会转化思想的应用. 跟踪演练3 已知f (x )=ax 2+bx +c (a ≠0),且f (-1)=2,f ′(0)=0,⎠⎛10f (x )d x =-2,求a 、b 、c 的值.解 由f (-1)=2,得a -b +c =2. ① 又f ′(x )=2ax +b ,∴f ′(0)=b =0, ②而⎠⎛10f (x )d x =⎠⎛10(ax 2+bx +c )d x =⎝⎛⎭⎫13ax 3+12bx 2+cx ⎪⎪⎪1=13a +12b +c , ∴13a +12b +c =-2, ③由①②③式得a =6,b =0,c =-4. 要点四 求分段函数的定积分 例4 计算下列定积分:(1)若f (x )=⎩⎪⎨⎪⎧x 2 (x ≤0)cos x -1 (x >0),求∫π2-1f (x )d x ;(2)⎠⎛30|x 2-4|d x .解 (1)∫π2-1f (x )d x =⎠⎛0-1x 2d x +∫π20(cos x -1)d x ,又∵⎝⎛⎭⎫13x 3′=x 2,(sin x -x )′=cos x -1 ∴原式=13x 3⎪⎪⎪0-1+(sin x -x )⎪⎪⎪⎪π20=⎝⎛⎭⎫0+13+⎝⎛⎭⎫sin π2-π2-(sin 0-0) =43-π2.(2)∵|x 2-4|=⎩⎪⎨⎪⎧x 2-4 (x ≥2或x ≤-2),4-x 2(-2<x <2), 又∵⎝⎛⎭⎫13x 3-4x ′=x 2-4,⎝⎛⎭⎫4x -13x 3′=4-x 2, ∴⎠⎛30|x 2-4|d x =⎠⎛20(4-x 2)d x +⎠⎛32(x 2-4)d x=⎝⎛⎭⎫4x -13x 3⎪⎪⎪20+⎝⎛⎭⎫13x 3-4x ⎪⎪⎪32 =⎝⎛⎭⎫8-83-0+(9-12)-⎝⎛⎭⎫83-8=233. 规律方法 (1)求分段函数的定积分时,可利用积分性质将其表示为几段积分和的形式; (2)带绝对值的解析式,先根据绝对值的意义找到分界点,去掉绝对值号,化为分段函数; (3)含有字母参数的绝对值问题要注意分类讨论. 跟踪演练4 求⎠⎛3-3(|2x +3|+|3-2x |)d x .解 ∵|2x +3|+|3-2x |=⎩⎪⎨⎪⎧-4x ,x <-32,6,-32≤x ≤32,4x ,x >32,∴⎠⎛3-3(|2x +3|+|3-2x |)d x=∫-32-3(-4x )d x +∫32-326d x +∫3324x d x=-2x 2⎪⎪⎪⎪-32-3+6x ⎪⎪⎪32-32+2x 2⎪⎪⎪⎪332=45.1.∫π2-π2(1+cos x )d x 等于( )A .πB .2C .π-2D .π+2答案 D解析 ∵(x +sin x )′=1+cos x , ∴⎪⎪∫π2-π2(1+cos x )d x =(x +sin x )π2-π2=π2+sin π2-⎣⎡⎦⎤-π2+sin ⎝⎛⎭⎫-π2=π+2. 2.若⎠⎛1a ⎝⎛⎭⎫2x +1x d x =3+ln 2,则a 的值是( ) A .5 B .4 C .3 D .2答案 D解析 ⎠⎛1a ⎝⎛⎭⎫2x +1x d x =⎠⎛1a 2x d x +⎠⎛1a 1xd x =x 2|a 1+ ln x ⎪⎪a1=a 2-1+ln a =3+ln 2,解得a =2.3.⎠⎛02⎝⎛⎭⎫x 2-23x d x =________. 答案 43解析 ⎠⎛02⎝⎛⎭⎫x 2-23x d x =⎠⎛02x 2d x -⎠⎛0223x d x =x 33⎪⎪⎪⎪20-x 2320=83-43=43. 4.已知f (x )=⎩⎨⎧4x -2π,0≤x ≤π2,cos x ,π2<x ≤π,计算⎠⎛0πf (x )d x .解 ⎠⎛0πf (x )d x =∫π20f (x )d x +错误!f (x )d x=∫π20(4x -2π)d x +错误!cos x d x ,取F 1(x )=2x 2-2πx ,则F 1′(x )=4x -2π; 取F 2(x )=sin x ,则F 2′(x )=cos x .所以∫π20(4x -2π)d x +错误!cos x d x =(2x 2-2πx )错误!+sin x ⎪⎪⎪ππ2=-12π2-1,即⎠⎛0πf (x )d x =-12π2-1.1.求定积分的一些常用技巧(1)对被积函数,要先化简,再求积分.(2)若被积函数是分段函数,依据定积分“对区间的可加性”,分段积分再求和.(3)对于含有绝对值符号的被积函数,要去掉绝对值符号才能积分.2.由于定积分的值可取正值,也可取负值,还可以取0,而面积是正值,因此不要把面积理解为被积函数对应图形在某几个区间上的定积分之和,而是在x 轴下方的图形面积要取定积分的相反数.一、基础达标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 )⎪⎪ba ; 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 =li m n→∞∑i =1n b -ans ′(ξi ); ④它在时间段[a ,b ]内的位移是s =⎠⎛ab s ′(t )d t .A .①B .①②C .①②④D .①②③④答案 D2.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)答案 B解析 若F (x )=x 3,则F ′(x )=3x 2,这与F ′(x )=x 2不一致,故选B. 3.⎠⎛01(e x +2x )d x 等于( )A .1B .e -1C .eD .e +1答案 C解析 ⎠⎛01(e x +2x )d x =(e x +x 2)|10=(e 1+12)-(e 0+02)=e.4.已知f (x )=⎩⎪⎨⎪⎧x 2,-1≤x ≤0,1,0<x ≤1,则⎠⎛1-1f (x )d x 的值为( )A.32 B .43C .23D .-23答案 B解析 ⎠⎛1-1f (x )d x =⎠⎛0-1x 2d x +⎠⎛011d x =⎪⎪x 330-1+1=13+1=43,故选B. 5.设函数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________.答案33解析 由已知得13a +c =ax 20+c ,∴x 20=13,又∵0≤x 0≤1,∴x 0=33. 6.(2013·湖南)若⎠⎛0T x 2d x =9,则常数T 的值为________.答案 3解析 ⎠⎛0T x 2d x =⎪⎪13x 3T 0=13T 3=9,即T 3=27,解得T =3. 7.已知⎠⎛1-1(x 3+ax +3a -b )d x =2a +6且f (t )=⎠⎛0t (x 3+ax +3a -b )d x 为偶函数,求a ,b 的值.解 ∵f (x )=x 3+ax 为奇函数, ∴⎠⎛1-1(x 3+ax )d x =0,∴⎠⎛1-1(x 3+ax +3a -b )d x=⎠⎛1-1(x 3+ax )d x +⎠⎛1-1(3a -b )d x=0+(3a -b )[1-(-1)]=6a -2b . ∴6a -2b =2a +6,即2a -b =3,①又f (t )=⎪⎪⎣⎡⎦⎤x 44+a2x 2+(3a -b )x t 0 =t 44+at 22+(3a -b )t 为偶函数, ∴3a -b =0,②由①②得a =-3,b =-9. 二、能力提升8.∫π20sin 2x2d x 等于( )A.π4B .π2-1C .2D .π-24答案 D解析 ∫π20sin 2x 2d x =∫π201-cos x 2d x =⎪⎪12(x -sin x )π20=π-24,故选D. 9.(2013·江西)若S 1=⎠⎛12x 2d x ,S 2=⎠⎛121x d x ,S 3=⎠⎛12e 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答案 B解析 S 1=⎠⎛12x 2d x =13x 3⎪⎪21=73,S 2=⎪⎪⎪⎠⎛121x d x =ln x 21=ln 2<1,S 3=⎠⎛12e x d x =e x |21=e 2-e =e(e -1)>73,所以S 2<S 1<S 3,选B.10.设f (x )=⎩⎪⎨⎪⎧lg x ,x >0,x +⎠⎛0a 3t 2d t ,x ≤0.若f [f (1)]=1,则a =________.答案 1解析 因为x =1>0,所以f (1)=lg 1=0.又x ≤0时,f (x )=x +⎠⎛0a 3t 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⎠⎛01f (x )d x =5,⎠⎛01xf (x )d x =176,求f (x )的解析式.解 ∵f (x )是一次函数,设f (x )=ax +b (a ≠0),则 ⎠⎛01f (x )d x =⎠⎛01(ax +b )d x =⎠⎛01ax d x +⎠⎛01b d x =12a +b =5, ⎠⎛01xf (x )d x =⎠⎛01x (ax +b )d x =⎠⎛01(ax 2)d x +⎠⎛a 1b x d x =13a +12b =176. 由⎩⎨⎧12a +b =513a +12b =176,得⎩⎪⎨⎪⎧a =4b =3.即f (x )=4x +3.12.若函数f (x )=⎩⎪⎨⎪⎧x 3,x ∈[0,1],x ,x ∈(1,2],2x ,x ∈(2,3].求⎠⎛03f (x )d x 的值.解 由积分的性质,知:⎠⎛03f (x )d x =⎠⎛01f (x )d x +⎠⎛12f (x )d x +⎠⎛23f (x )d x =⎠⎛01x 3d x +⎠⎛12x d x +⎠⎛232x d x =x 44⎪⎪⎪⎪10+23x 3221⎪⎪+2x ln 232 =14+432-23+8ln 2-4ln 2=-512+432+4ln 2. 三、探究与创新13.求定积分⎠⎛3-4|x +a |d x . 解 (1)当-a ≤-4即a ≥4时,原式=⎠⎛3-4(x +a )d x = ⎪⎪⎝⎛⎭⎫x 22+ax 3-4=7a -72. (2)当-4<-a <3即-3<a <4时,原式=⎠⎛-4-a [-(x +a )]d x +⎠⎛3-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=⎠⎛3-4[-(x +a )]d x = ⎪⎪⎝⎛⎭⎫-x 22-ax 3-4= -7a +72. 综上,得⎠⎛3-4|x +a |d x =⎩⎪⎨⎪⎧ 7a -72(a ≥4),a 2-a +252(-3<a <4),-7a +72(a ≤-3).高中数学学习技巧:在学习的过程中逐步做到:提出问题,实验探究,展开讨论,形成新知,应用反思。
1 定积分的概念、微积分基本定理(重点练)高二数学(理)十分钟同步课堂专练(人教A版选修2-2)
1.5~1.6 定积分的概念、微积分基本定理重点练一、单选题1.)10x dx =⎰( )A .22π+B .12π+ C .122π-D .142π- 2.已知函数()e3211(1)2f x x dx x f x x'=⋅--⎰,则()()11f f '+=( ) A .-1B .1C .-2D .23.已知311tan 4e dx x πα⎛⎫+=- ⎪⎝⎭⎰,则2sin cos cos sin αααα+=-( ) A .4-B .4C .5D .5-4.已知()()ln xxf x e e -=+,201sin 2a xdx π=⎰, 1.112b ⎛⎫= ⎪⎝⎭,2log 3c =,则下列选项中正确的是( ) A .()()()f a f b f c >> B .()()()f a f c f b >> C .()()()f c f a f b >>D .()()()f c f b f a >>二、填空题5.011edx x-+=⎰⎰______________.6.如图,在边长为1的正方形中随机撒一粒黄豆,则它落在阴影部分的概率为_______.三、解答题7.计算下列各式的值.(1)()0sin cos d x x x π-⎰;(2)1x⎰.参考答案1.【答案】D【解析】由定积分的运算法则,可得)111()x dx dx x dx =+-⎰⎰⎰,又由1dx ⎰相当于是以(1,0)为圆心,以1为半径的圆的面积的14,如图所示,可得104dx π=⎰, 又因为021011()1()|22x d x x --=-=⎰,所以)1110001()42x dx dx x dx π=+-=-⎰⎰⎰. 故选D.2.【答案】A【解析】因为e111ln |1edx x x ==⎰,所以()()3212f x x x f x '=--,所以()()232'12f x x xf '=--,令1x =,得()()13212f f ''=--,解得1(1)3f '=,所以321()23f x x x x =--,14(1)1233f =--=-, ()()1411133f f ⎛⎫'+=+-=- ⎪⎝⎭,故选A . 3.【答案】D【解析】由()()()331311ln ln ln13e e dx x C e C C x ⎰=+=+-+=,则tan 1tan 341tan πααα+⎛⎫+==- ⎪-⎝⎭,则tan 2α=,由2sin cos 2tan 15cos sin 1tan αααααα++==---故选D. 4.【答案】C【解析】()()ln xxf x e e-=+,x ∈R ,则()()()ln xx f x ee f x --=+=,所以()f x 为R 上的偶函数,并且()x xx xe ef x e e---'=+,则[)0,x ∈+∞时,()0f x '≥,当且仅当0x =时,“=”成立, 所以()f x 在[)0,x ∈+∞上单调递增,在(],0x ∈-∞上单调递减,()220111sin cos 222a xdx x ππ==-=⎰1.111110222b ⎛⎫⎛⎫<=<= ⎪ ⎪⎝⎭⎝⎭,221log log 332c ==-, 又()22111log 3log 3222f c f f f ⎛⎫⎛⎫⎛⎫=-=> ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,所以()()()f c f a f b >>.故选C 5.【答案】21π+【解析】11edx x⎰=ln 1e x ln ln1101e =-=-=,因为2-⎰表示的是圆224x y +=在x 轴及其上方的面积,所以2-⎰21222ππ=⨯⨯=,所以11edx x⎰2-+⎰=12π+.故填21π+. 6.【答案】13【解析】由题意,结合定积分可得阴影部分的面积为31120021(1()|33S dx x x ==-=⎰, 由几何概型的计算公式可得,黄豆在阴影部分的概率为113113p ==⨯. 故填137.【答案】(1) 2;(2) π【解析】(1)由题得()0sin cos d (cos sin )|(cos sin )(cos 0sin 0)x x x x x ππππ-=--=-----⎰=10102-++=;(2)令22(1)4(13,0)y x y x y =∴-+=≤≤≥,因为1x ⎰等于1,3,x x x ==轴和曲线ADB 所围成的曲边梯形的面积,如图扇形ACB , 扇形ACB 的面积为212=4ππ⨯⨯,所以1x π=⎰.。
高中数学课时跟踪检测(十五)微积分基本定理(含解析)北师大版选修22
高中数学课时跟踪检测(十五)微积分基本定理(含解析)北师大版选修22课时跟踪检测(十五) 微积分基本定理1.下列积分值等于1的是( ) A. ⎠⎛01x d x B. ⎠⎛01(x +1)d x C. ⎠⎛011d xD. ⎠⎛0112d x解析:选C ⎠⎛011d x =x ⎪⎪⎪1=1.2.⎠⎛01(e x+2x )d x =( )A .1B .e -1C .eD .e +1解析:选C ⎠⎛01(e x+2x )d x =(e x+x 2) ⎪⎪⎪1=(e 1+1)-e 0=e.3. ⎠⎛03|x 2-4|d x =( )A.213 B.223 C.233D.253解析:选C ⎠⎛03|x 2-4|d x =⎠⎛02(4-x 2)d x +⎠⎛23(x 2-4)d x =⎝ ⎛⎭⎪⎫4x -13x 3⎪⎪⎪2+⎝ ⎛⎭⎪⎫13x 3-4x ⎪⎪⎪32=233,故选C. 4.函数F (x )=⎠⎛0 xt (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解析:选 B F (x )=⎠⎛0x(t 2-4t )d t =⎝ ⎛⎭⎪⎫13t 3-2t 2⎪⎪⎪x=13x 3-2x 2(-1≤x ≤5).F ′(x )=x 2-4x ,由F ′(x )=0,得x =0或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)=-3.又F (-1)=-3,F (5)=-3,所以最大值为0,最小值为-323.5.若⎠⎛-a ax 2d x =18(a >0),则a =________.解析:⎠⎛-a ax 2d x =x 33⎪⎪⎪a-a=a 33--a33=18⇒a =3.答案:36.设f (x )=⎩⎪⎨⎪⎧lg x , x >0,x +⎠⎛0a 3t 2d t ,x ≤0,若f (f (1))=1,则a =________.解析:显然f (1)=lg 1=0,f (0)=0+⎠⎛0 a3t 2d t =t 3⎪⎪⎪a=1,得a =1.答案:17.求下列定积分: (1) ⎠⎛122x 2+x +1xd x ; (2) ⎠⎛0π2sin ⎝⎛⎭⎪⎫x +π4d x .解:(1) ⎠⎛122x 2+x +1xd x =⎠⎛12(2x +1x +1)d x=⎠⎛122x d x +⎠⎛121x d x +∫211d x=x 2⎪⎪⎪21+ln x ⎪⎪⎪ 21+x ⎪⎪⎪21=(4-1)+ln 2-ln 1+2-1 =4+ln 2.(2)∵2sin(x +π4)=2⎝ ⎛⎭⎪⎫sin x ·22+cos x ·22=sin x +cos x ,(-cos x +sin x )′=sin x +cos x , ∴⎠⎛0π2sin(x +π4)d x =⎠⎛0 π (sin x +cos x )d x =(-cos x +sin x ) ⎪⎪⎪π=(-cos π+sin π)-(-cos 0+sin 0)=2.8.A ,B 两站相距7.2 km ,一辆电车从A 站开往B 站,电车开出t s 后到达途中C 点,这一段的速度为1.2t m/s ,到C 点的速度为24 m/s ,从C 点到B 站前的D 点这段路程做匀速行驶,从D 点开始刹车,经t s 后,速度为(24-1.2t ) m/s ,在B 站恰好停车,试求:(1)A ,C 间的距离; (2)B ,D 间的距离.解:(1)设从A 到C 的时间为t 1 s ,则1.2t 1=24,解得t 1=20,则AC =⎠⎛0 201.2t d t =0.6t 2⎪⎪⎪20=240(m).即A ,C 间的距离为240 m.(2)设从D 到B 的时间为t 2 s ,则24-1.2t 2=0, 解得t 2=20,则BD =⎠⎛020(24-1.2t )d t =(24t -0.6t 2) ⎪⎪⎪20=240(m).即B ,D 间的距离为240 m .。
高中数学人教A版选修2-2(课时训练):1.6 微积分基本定理 Word版含答案
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(3)含有字母参数的绝对值问题要注意分类讨论.
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高中数学人教A版选修2-2(课时训练):1.6 微积分基本定理 pdf版含答案
2.函数 f(x)与其一个原函数的关系
(1)若 f(x)=c(c 为常数),则 F(x)=cx;
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ax (5)若 f(x)=ax,则 F(x)=ln a(a>0 且 a≠1); (6)若 f(x)=sin x,则 F(x)=-cos_x; (7)若 f(x)=cos x,则 F(x)=sin_x.
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[预习导引]
1.微积分基本定理
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函数、正、余弦函数、指数、对数函数与常数的和与差.
(2)确定积分区间,分清积分下限与积分上限.
【配套K12】2018年高中数学课时跟踪检测十一微积分基本定理新人教A版选修2_2
课时跟踪检测(十一) 微积分基本定理层级一 学业水平达标1.下列各式中,正确的是( ) A.⎠⎛a bF ′(x )d x =F ′(b )-F ′(a ) B.⎠⎛a bF ′(x )d x =F ′(a )-F ′(b ) C.⎠⎛a bF ′(x )d x =F (b )-F (a ) D.⎠⎛a bF ′(x )d x =F (a )-F (b )解析:选C 由牛顿-莱布尼茨公式知,C 正确. 2.⎠⎛0π(cos x +1)d x 等于( ) A .1 B .0 C .π+1D .π解析:选D ⎠⎛0π(cos x +1)d x =(sin x +x ) ⎪⎪⎪π=sin π+π-0=π.3.已知积分⎠⎛01(kx +1)d x =k ,则实数k =( ) A .2 B .-2 C .1D .-1解析:选A 因为⎠⎛01(kx +1)d x =k ,所以⎝ ⎛⎭⎪⎫12kx 2+x ⎪⎪⎪1=k .所以12k +1=k ,所以k =2.4. ⎠⎛-a a|56x |d x ≤2 016,则正数a 的最大值为( ) A .6 B .56 C .36D .2 016解析:选A ⎠⎛-a a|56x |d x =2⎠⎛0a56x d x =2×562x 2⎪⎪⎪a=56a 2≤2 016,故a 2≤36,即0<a ≤6.5.⎠⎛03|x 2-4|d x =( )A.213B.223C.233D.253解析:选C ∵|x 2-4|=⎩⎪⎨⎪⎧x 2-4,2≤x ≤3,4-x 2,0≤x ≤2,∴⎠⎛03|x 2-4|d x =⎠⎛23(x 2-4)d x +⎠⎛02(4-x 2)d x=⎝ ⎛⎭⎪⎫13x 3-4x ⎪⎪⎪32+⎝⎛⎭⎪⎫4x -13x 3⎪⎪⎪2=⎣⎢⎡⎦⎥⎤--⎝ ⎛⎭⎪⎫83-8+⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫8-83-0 =-3-83+8+8-83=233.6.⎠⎛02(x 2-x )d x =__________.解析:∵⎝ ⎛⎭⎪⎫x 33-12x 2′=x 2-x ,∴原式=⎝ ⎛⎭⎪⎫x 33-12x 220=⎝ ⎛⎭⎪⎫83-2-0=23.答案:237. 设f (x )=⎩⎪⎨⎪⎧x 2,x ≤0,cos x -1,x >0.则⎠⎛1-1f (x )d x =_________. 解析:⎠⎛-11f (x )d x =⎠⎛-11x 2d x +⎠⎛01(cos x -1)d x=13x 3⎪⎪⎪-1+(sin x -x ) ⎪⎪⎪1=⎣⎢⎡⎦⎥⎤13×03-13-3+[(sin 1-1)-(sin 0-0)] =sin 1-23.答案:sin 1-238.已知等差数列{a n }的前n 项和为S n ,且S 10=⎠⎛03(1+2x )d x ,则a 5+a 6=__________.解析:S 10=⎠⎛03(1+2x )d x =(x +x 2)30=3+9=12.因为{a n }是等差数列, 所以S 10=a 5+a 62=5(a 5+a 6)=12,所以a 5+a 6=125.答案:1259.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:由f (-1)=2得a -b +c =2, ① 又f ′(x )=2ax +b ,∴f ′(0)=b =0, ②而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x=⎝ ⎛⎭⎪⎫13ax 3+12bx 2+cx 10=13a +12b +c ,∴13a +12b +c =-2, ③ 由①②③式得a =6,b =0,c =-4.法二:设f (x )=|2x +3|+|3-2x |=⎩⎪⎨⎪⎧-4x ,-3≤x <-32,6,-32≤x ≤32,4x ,32<x ≤3.如图,所求积分等于阴影部分面积,即⎠⎛3-3(|2x +3|+|3-2x |)d x =S =2×12×(6+12)×32+3×6=45.层级二 应试能力达标1.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .F ′(x )=cos xB .F ′(x )=sin xC .F ′(x )=-c o s xD .F ′(x )=-sin x解析:选A F (x )=⎠⎛0xcos t d t =sin t ⎪⎪⎪x=sin x -sin 0=sin x .所以F ′(x )=cos x ,故应选A.2.若函数f (x )=x m+nx 的导函数是f ′(x )=2x +1,则⎠⎛12f (-x )d x =( )A.56B.12C.23D.16解析:选A ∵f (x )=x m+nx 的导函数是f ′(x )=2x +1,∴f (x )=x 2+x ,∴⎠⎛12f (-x )d x=⎠⎛12(x 2-x )d x =⎝ ⎛⎭⎪⎫13x 3-12x 2⎪⎪⎪21=56. 3.若⎠⎛1a⎝ ⎛⎭⎪⎫2x +1x d x =3+ln 2,则a 的值是( )A .6B .4C .3D .2解析:选 D ⎠⎛1a⎝ ⎛⎭⎪⎫2x +1x d x =(x 2+ln x )a 1=(a 2+ln a )-(1+ln 1)=(a 2-1)+ln a =3+ln 2.∴⎩⎪⎨⎪⎧a 2-1=3,a >1,a =2,∴a =2.4.若f (x )=x 2+2⎠⎛01f (x )d x ,则⎠⎛01f (x )d x =( )A .-1B .-13C.13D .1解析:选B 设⎠⎛01f (x )d x =c ,则c =⎠⎛01(x 2+2c )d x =⎝ ⎛⎭⎪⎫13x 3+2cx ⎪⎪⎪10=13+2c ,解得c =-13.5.函数y =x 2与y =kx (k >0)的图象所围成的阴影部分的面积为92,则k =________________.解析:由⎩⎪⎨⎪⎧y =kx ,y =x 2,解得⎩⎪⎨⎪⎧x =0,y =0或⎩⎪⎨⎪⎧x =k ,y =k 2.由题意得,⎠⎛0k (kx -x 2)d x =⎝ ⎛⎭⎪⎫12kx 2-13x 3⎪⎪⎪k=12k 3-13k 3=16k 3=92,∴k =3. 答案:36.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________解析:长方形的面积为S 1=3,S 阴=⎠⎛013x 2d x =x 3⎪⎪⎪1=1,则P =S 阴S 1=13. 答案:137. 已知S 1为直线x =0,y =4-t 2及y =4-x 2所围成图形的面积,S 2为直线x =2,y =4-t 2及y =4-x 2所围成图形的面积(t 为常数).(1)若t =2时,求S 2.(2)若t ∈(0,2),求S 1+S 2的最小值. 解:(1)当t =2时,S 2=([2-(4-x 2)]d x =⎝ ⎛⎭⎪⎫13x 3-2x =43(2-1). (2)t ∈(0,2),S 1=⎠⎛0t[(4-x 2)-(4-t 2)]d x=⎝⎛⎭⎪⎫t 2x -13x 3⎪⎪⎪t0=23t 3, S 2=⎠⎛t2[(4-t 2)-(4-x 2)]d x =⎝ ⎛⎭⎪⎫13x 3-t 2x ⎪⎪⎪2t=83-2t 2+23t 3, 所以S =S 1+S 2=43t 3-2t 2+83,S ′=4t 2-4t =4t (t -1),令S ′=0得t =0(舍去)或t =1, 当0<t <1时,S ′<0,S 单调递减, 当1<t <2时,S ′>0,S 单调递增,所以当t =1时,S min =2.8.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.解:抛物线y =x -x 2与x 轴两交点的横坐标x 1=0,x 2=1,所以,抛物线与x 轴所围图形的面积S =⎠⎛01(x -x 2)d x =⎝ ⎛⎭⎪⎫x 22-x 33⎪⎪⎪1=12-13=16. 抛物线y =x -x 2与直线y =kx 两交点的横坐标为x ′1=0,x ′2=1-k ,所以S2=(x -x 2-kx )d x =⎝ ⎛⎭⎪⎫1-k 2x 2-x 33=16(1-k )3,又知S =16,所以(1-k )3=12. 于是k =1-312=1-342.。
2018年秋高中数学 课时分层作业10 微积分基本定理 新人教A版选修2-2
课时分层作业(十) 微积分基本定理(建议用时:40分钟)[基础达标练]一、选择题 1.(e x +2x )d x 等于( )A .1B .e -1C .eD .e +1C [∵(e x+2x )d x =()e x +x 2=e +1-1=e ,故选C.]2.已知积分 (kx +1)d x =k ,则实数k =( )A .2B .-2C .1D .-1A∴k =2.]3.设f (x )=⎩⎪⎨⎪⎧x 2,0≤x <1,2-x ,1<x ≤2,则f (x )d x =( )【导学号:31062095】A.23B.34C.45D.56D [f (x )d x =x 2d x + (2-x )d x==13+12=56.] 4.若函数f (x )=x m +nx 的导函数是f ′(x )=2x +1,则f (-x )d x =( )A.56B.12C.23D.16A [∵f (x )=x m+nx 的导函数是f ′(x )=2x +1, ∴f (x )=x 2+x , ∴f (-x )d x = (x 2-x )d x=⎝ ⎛⎭⎪⎫13x 3-12x 2=56.] 5.设a =d x ,b =x 2d x ,c =x 3d x ,则a ,b ,c 的大小关系是( )A .a >b >cB .c >a >bC .a >c >bD .c >b >a∴a >b >c .] 二、填空题6.⎝ ⎛⎭⎪⎫1-2sin 2θ2d θ=________. 【导学号:31062096】[解析]=32. [答案] 327.(2-|x |)d x =________.[解析] 因为f (x )=2-|x |=⎩⎪⎨⎪⎧2+x ,x ≤0,2-x ,x ≥0,所以[答案] 728.已知x ∈(0,1],f (x )=(1-2x +2t )d t ,则f (x )的值域是________.[解析] f (x )= (1-2x +2t )d t =(t -2xt +t 2)=-2x +2(x ∈(0,1]).∴f (x )的值域为[0,2). [答案] [0,2) 三、解答题 9.计算定积分:(|2x +3|+|3-2x |)d x .[解] 设f (x )=|2x +3|+|3-2x |,x ∈[-3,3],则f (x )=⎩⎪⎨⎪⎧-4x ,-3≤x <-32,6,-32≤x ≤32,4x ,32<x ≤3.=-2×⎝ ⎛⎭⎪⎫94-9+6×⎝ ⎛⎭⎪⎫32+32+2×⎝ ⎛⎭⎪⎫9-94=45. 10.设函数f (x )=ax 2+c (a ≠0),若f (x )d x =f (x 0),0≤x 0≤1,求x 0的值.【导学号:31062097】[解] 因为f (x )=ax 2+c (a ≠0),且⎝ ⎛⎭⎪⎫a3x 3+cx ′=ax 2+c ,所以f (x )d x =(ax 2+c )d x =⎝ ⎛⎭⎪⎫a 3x 3+cx=a 3+c =ax 20+c ,解得x 0=33或x 0=-33(舍去).即x 0的值为33. [能力提升练]1.若y = (sin t +cos t ·sin t )d t ,则y 的最大值是( )A .1B .2C .-1D .0B [y =(sin t +cos t ·sin t )d t==-cos x +1-14(cos 2x -1)=-14cos 2x -cos x +54=-12cos 2x -cos x +32=-12(cos x +1)2+2≤2.]2.若f (x )=x 2+2f (x )d x ,则f (x )d x 等于( )A .-1B .-13C .13D .1B [∵f (x )d x 是常数,所以可设f (x )=x 2+c (c 为常数), 所以c =2f (x )d x =2(x 2+c )d x =2⎝ ⎛⎭⎪⎫13x 3+cx ,解得c =-23,f (x )d x = (x 2+c )d x =3.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 =.[答案] 434.已知f (x )=若f (f (1))=1,则a =__________.[解析] 因为f (1)=lg 1=0, 且3t 2d t =t 3|a 0=a 3-03=a 3,所以f (0)=0+a 3=1,所以a =1. [答案] 1 5.已知f (x )=(12t +4a )d t ,F (a )=[f (x )+3a 2]d x ,求函数F (a )的最小值.【导学号:31062098】[解] 因为f (x )= (12t +4a )d t =(6t 2+4at )=6x 2+4ax -(6a 2-4a 2)=6x2+4ax -2a 2,因为F (a )=[f (x )+3a 2]=(6x 2+4ax +a 2)d x =(2x 3+2ax 2+a 2x )| 10=2·13+2a ·12+a 2·1=(a +1)2+1≥1.所以当a =-1时,F (a )的最小值为1.。
高中数学课时跟踪检测(十)微积分基本定理(含解析)新人教A版选修22
高中数学课时跟踪检测(十)微积分基本定理(含解析)新人教A版选修22课时跟踪检测(十) 微积分基本定理一、题组对点训练对点练一 求简单函数的定积分 1.⎠⎛02(x -1)d x 等于( ) A .-1 B .1 C .0D .2解析:选C ⎠⎛02(x -1)d x =⎝ ⎛⎭⎪⎫12x 2-x ⎪⎪⎪2=12×22-2=0. 2.⎠⎛01(e x+2x )d x 等于( )A .1B .e -1C .eD .e +1解析:选C ⎠⎛01(e x +2x )d x =(e x +x 2) ⎪⎪⎪10=(e 1+1)-e 0=e.3.⎠⎜⎛-π2π2 (1+cos x )d x =( ) A .π B .2 C .π-2D .π+2解析:选D ∵(x +sin x )′=1+cos x ,∴⎠⎜⎛-π2π2 (1+cos x )d x =(x +sin x ) ⎪⎪⎪π2-π2=π+2. 4.计算定积分⎠⎛-11 (x 2+sin x )d x =________.解析:⎠⎛-11 (x 2+sin x )d x =⎝ ⎛⎭⎪⎫x33-cos x ⎪⎪⎪1-1=23. 答案:23对点练二 求分段函数的定积分5.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 .不存在解析:选C ⎠⎛02f (x )d x =⎠⎛01x 2d x +⎠⎛12(2-x )d x =13x 3⎪⎪⎪1+⎝⎛⎭⎪⎫2x -12x 221=13+⎝ ⎛⎭⎪⎫4-2-2+12=56. 6.计算下列定积分: (1)⎠⎛25|x -3|d x ;(2)若f (x )=⎩⎪⎨⎪⎧x 2,x ≤0,cos x -1,x >0,求⎠⎛-12πf (x )d x .解:(1)∵|x -3|=⎩⎪⎨⎪⎧3-x ,x ∈[2,3),x -3,x ∈[3,5],∴⎠⎛25|x -3|d x =⎠⎛23|x -3|d x +⎠⎛35|x -3|d x=⎠⎛23(3-x )d x +⎠⎛35(x -3)d x =⎝⎛⎭⎪⎫3x -12x 2⎪⎪⎪32+⎝ ⎛⎭⎪⎫12x 2-3x ⎪⎪⎪53=⎝ ⎛⎭⎪⎫9-12×9-6+2+⎝ ⎛⎭⎪⎫252-15-92+9=52.(2)由已知⎠⎛-12πf (x )d x =⎠⎛-10x 2d x +⎠⎛02π(cos x -1)d x=13x 3⎪⎪⎪-1+(sin x -x ) ⎪⎪⎪⎪π2=13+⎝⎛⎭⎪⎫1-π2=43-π2.对点练三 根据定积分求参数7.若⎠⎛1a ⎝⎛⎭⎪⎫2x +1x d x =3+ln 2,则a 的值是( )A .6B .4C .3D .2解析:选D ⎠⎛1a ⎝⎛⎭⎪⎫2x +1x d x =(x 2+ln x ) ⎪⎪⎪a1=(a 2+ln a )-(1+ln 1)=(a 2-1)+ln a =3+ln 2.∴⎩⎪⎨⎪⎧a 2-1=3,a >1,a =2,∴a =2.8.设f (x )=⎩⎪⎨⎪⎧lg x ,x >0,x +⎠⎛0a 3t 2d t ,x ≤0,若f (f (1))=1,则a =________.解析:显然f (1)=lg 1=0,f (0)=0+⎠⎛0a3t 2d t =t 3⎪⎪⎪a0=a 3,得a 3=1,a =1.答案:19.已知2≤⎠⎛12(kx +1)d x ≤4,则实数k 的取值范围为________.解析:⎠⎛12(kx +1)d x =⎝ ⎛⎭⎪⎫12kx 2+x ⎪⎪⎪21=(2k +2)-⎝ ⎛⎭⎪⎫12k +1=32k +1,所以2≤32k +1≤4,解得23≤k ≤2. 答案:⎣⎢⎡⎦⎥⎤23,2 10.已知f (x )是二次函数,其图象过点(1,0),且f ′(0)=2,⎠⎛01f (x )d x =0,求f (x )的解析式.解:设f (x )=ax 2+bx +c (a ≠0),∴a +b +c =0. ∵f ′(x )=2ax +b ,① ∴f ′(0)=b =2.②⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x =⎝ ⎛⎭⎪⎫13ax 3+12bx 2+cx ⎪⎪⎪1=13a +12b +c =0.③ 由①②③得⎩⎪⎨⎪⎧a =-32,b =2,c =-12,∴f (x )=-32x 2+2x -12.二、综合过关训练1.已知⎠⎛02f (x )d x =3,则⎠⎛02[f (x )+6]d x =( )A .9B .12C .15D .18解析:选C ⎠⎛02[f (x )+6]d x =⎠⎛02f (x )d x +⎠⎛026d x =3+6x ⎪⎪⎪20=3+12=15.2.若函数f (x )=x m+nx 的导函数是f ′(x )=2x +1,则⎠⎛12f (-x )d x =( ) A.56 B.12 C.23D.16解析:选A ∵f (x )=x m+nx 的导函数是f ′(x )=2x +1,∴f (x )=x 2+x ,∴⎠⎛12f (-x )d x=⎠⎛12(x 2-x )d x =⎝ ⎛⎭⎪⎫13x 3-12x 2⎪⎪⎪21=56. 3.若y =⎠⎛0x (sin t +cos t ·sin t )d t ,则y 的最大值是( )A .1 `B .2C .-1D .0解析:选B y =⎠⎛0x (sin t +cos t ·sin t )d t =⎠⎛0x sin t d t +⎠⎛0x sin 2t 2d t =-cos t ⎪⎪⎪x0-14cos2t ⎪⎪⎪x=-cos x +1-14(cos 2x -1)=-14cos 2x -cos x +54=-12cos 2x -cos x +32=-12(cos x +1)2+2≤2.4.若f (x )=x 2+2⎠⎛01f (x )d x ,则⎠⎛01f (x )d x 等于( )A .-1B .-13C.13D .1解析:选B 因为⎠⎛01f (x )d x 是常数,所以f ′(x )=2x ,所以可设f (x )=x 2+c (c 为常数),所以c =2⎠⎛01f (x )d x =2⎠⎛01(x 2+c )d x =2⎝ ⎛⎭⎪⎫13x 3+cx ⎪⎪⎪10,解得c =-23,⎠⎛01f (x )d x =⎠⎛01(x 2+c )d x=⎠⎛01⎝⎛⎭⎪⎫x 2-23d x =⎝ ⎛⎭⎪⎫13x 3-23x ⎪⎪⎪10=-13.5.⎠⎛02(4-2x )(4-3x 2)d x =________.解析:⎠⎛02(4-2x )(4-3x 2)d x =⎠⎛02(16-12x 2-8x +6x 3)d x =⎝ ⎛⎭⎪⎫16x -4x 3-4x 2+32x 4⎪⎪⎪2=8.答案:86.若f (x )=⎩⎪⎨⎪⎧x 2,x ≤0,sin x -1,x >0,则⎠⎛-11f (x )d x =________.解析:⎠⎛-11f (x )d x =⎠⎛-10x 2d x +⎠⎛01(sin x -1)d x=13x 3⎪⎪⎪-1+(-cos x -x ) ⎪⎪⎪1=13-cos 1. 答案:13-cos 17.计算下列定积分.(1) ⎠⎛-33(|2x +3|+|3-2x |)d x ;(2)⎠⎛14⎝⎛⎭⎪⎫2x -1x d x .解:(1)∵|2x +3|+|3-2x |=⎩⎪⎨⎪⎧-4x ,x ≤-32,6,-32<x <32,4x ,x ≥32.∴⎠⎛-33 (|2x +3|+|3-2x |)d x=⎠⎜⎛-3-32 (-4x )d x +⎠⎜⎜⎛-32326d x +⎠⎜⎛3234x d x=-2x 2⎪⎪⎪⎪-32-3+6x ⎪⎪⎪32-32+2x 2⎪⎪⎪⎪332=(-2)×⎝ ⎛⎭⎪⎫-322-(-2)×(-3)2+6×32-6×⎝ ⎛⎭⎪⎫-32+2×32-2×⎝ ⎛⎭⎪⎫322=45.(2)⎠⎛14⎝⎛⎭⎪⎫2x-1x d x =⎠⎛142xd x -⎠⎛141xd x =2xln 2⎪⎪⎪41-2x ⎪⎪⎪41=⎝ ⎛⎭⎪⎫16ln 2-2ln 2-(24-2)=14ln 2-2. 8.已知f (x )=⎠⎛-ax (12t +4a )d t ,F (a )=⎠⎛01[f (x )+3a 2]d x ,求函数F (a )的最小值.解:∵f (x )=⎠⎛-ax (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 ) ⎪⎪⎪1=a 2+2a +2=(a +1)2+1≥1,∴当a =-1时,F (a )最小值=1.。
2017-2018学年人教A版数学选修2-2课时跟踪检测(十一) 微积分基本定理含解析
课时跟踪检测(十一)微积分基本定理一、选择题1。
错误!(x3+x2-30)d x等于( )A.56 B.28C.14 D。
错误!解析:选D 错误!(x3+x2-30)d x=错误!x4+错误!x3-30x错误!=错误!(44-24)+错误!(43-23)-30×(4-2)=错误!。
2.错误!-2错误!d x等于()A.错误!B.错误!C。
错误!D。
错误!解析:选A 错误!-2错误!d x=错误!-2x2d x+错误!-2错误!d x=错误!x3错误!+错误!错误!=错误!(x3-x-3)错误!=错误!错误!-错误!错误!=错误!。
3.设f(x)=错误!则错误!f(x)d x等于( )A。
错误! B.错误!C。
错误!D.不存在解析:选C 错误!f(x)d x=错误!x2d x+错误!(2-x)d x=错误!x3错误!+错误!错误!=错误!+错误!=错误!.4.计算错误!(1+错误!)d x的结果为()A.1 B。
错误!C.1+错误!D.1+错误!解析:选C∵错误!错误!d x=错误!,∴错误!(1+错误!)d x=错误!1d x+错误!错误!d x=1+错误!。
5.(江西高考)若f(x)=x2+2错误!f(x)d x,则错误!()A.-1 B.-错误!C.13D.1解析:选B∵f(x)=x2+2错误!f(x)d x,∴错误!f(x)d x=错误!错误!=错误!+错误!∴错误!f(x)d x=-错误!。
二、填空题6.若错误!(2x-3x2)d x=0,则k=________.解析:错误!(2x-3x2)d x=(x2-x3)错误!=k2-k3=0,解得k=0(舍去)或k=1。
答案:17.计算定积分错误!-1(x2+sin x)d x=________.解析:错误!-1(x2+sin x)d x=错误!错误!=错误!。
答案:错误!8.设f(x)=错误!若f(f(1))=1,则a=________.解析:显然f(1)=lg1=0,f(0)=0+错误!3t2d t=t3错误!=a3,得a3=1,a=1.答案:1三、解答题9.计算下列定积分.(1)∫错误!0(sin x-sin 2x)d x;(2)错误!-3(|2x+3|+|3-2x|)d x。
2017-2018学年高中数学 课时跟踪训练(十五)微积分基本定理 北师大版选修2-2
课时跟踪训练(十五) 微积分基本定理1.下列积分值等于1的是( ) A.∫10x d x B.∫10(x +1)d xC.∫101d xD.∫1012d x2.(福建高考)⎠⎛01(e x+2x )d x =( )A .1B .e -1C .eD .e +13.∫30|x 2-4|d x =( ) A.213 B.223 C.233D.2534.函数F (x )=∫x0t (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5.若∫a -a x 2d x =18(a >0),则a =________. 6.(陕西高考)设f (x )=⎩⎪⎨⎪⎧lg x , x >0,x +⎠⎛0a 3t 2d t ,x ≤0,若f (f (1))=1,则a =________.7.求下列定积分: (1)∫212x 2+x +1xd x ;(2)∫π02sin ⎝⎛⎭⎪⎫x +π4d x .8.A ,B 两站相距7.2 km ,一辆电车从A 站开往B 站,电车开出t s 后到达途中C 点,这一段的速度为1.2t m/s ,到C 点的速度为24 m/s ,从C 点到B 站前的D 点这段路程做匀速行驶,从D 点开始刹车,经t s 后,速度为(24-1.2t ) m/s ,在B 站恰好停车,试求:(1)A ,C 间的距离; (2)B ,D 间的距离.答 案1.选C ∫101d x =x ⎪⎪ 1=1.2.选C ⎠⎛01(e x +2x )d x =(e x +x 2)|1=(e 1+1)-e 0=e.3.选C ∫3|x 2-4|d x =∫20(4-x 2)d x +∫32(x 2-4)d x =⎝⎛⎭⎪⎫4x -13x 3⎪⎪⎪20+⎝ ⎛⎭⎪⎫13x 3-4x ⎪⎪⎪32=233,故选C. 4.选 B F (x )=∫x 0(t 2-4t )d t =⎝ ⎛⎭⎪⎫13t 3-2t 2⎪⎪⎪x=13x 3-2x 2(-1≤x ≤5).F ′(x )=x 2-4x ,由F ′(x )=0,得x =0或4,列表如下:可见极大值F (0)=0,极小值F (4)=-3.又F (-1)=-3,F (5)=-3,所以最大值为0,最小值为-323.5.解析:∫a-a x 2d x =x 33| a-a =a 33--a33=18⇒a =3.答案:36.解析:显然f (1)=lg 1=0,f (0)=0+∫a03t 2d t =t 3⎪⎪⎪a=1,得a =1.答案:17.解:(1)∫212x 2+x +1xd x=∫21(2x +1x+1)d x=∫212x d x +∫211xd x +∫211d x=x 2|21+ln x |21+x |21=(4-1)+ln 2-ln 1+2-1 =4+ln 2.(2)∵2sin(x +π4)=2⎝ ⎛⎭⎪⎫sin x ·22+cos x ·22=sin x +cos x ,(-cos x +sin x )′=sin x +cos x , ∴∫π2sin(x +π4)d x =∫π0(sin x +cos x )d x=(-cos x +sin x ) |π=(-cos π+sin π)-(-cos 0+sin 0)=2.8.解:(1)设从A 到C 的时间为t 1 s ,则1.2t 1=24,解得t 1=20, 则AC =∫2001.2t d t =0.6t 2|200=240(m).即A ,C 间的距离为240 m.(2)设从D 到B 的时间为t 2 s ,则24-1.2t 2=0, 解得t 2=20,则BD =∫200(24-1.2t )d t =(24t -0.6t 2) |200=240(m).即B ,D 间的距离为240 m.。
20172018学年高中数学人教A版选修22:课时跟踪检测(十一) 微积分基本定理 Word版含解析
(1+ 2x)dx, 则 a5+ a6=
因为{an}是等差数列, 所以 S10=10a52+a6=5(a5+a6)=12,所以 a5+a6=152.
12 答案: 5
1
∫ 9.已知 f(x)=ax2+bx+c(a≠0),且 f(-1)=2,f′(0)=0, 0
b,c 的值.
解:由 f(-1)=2 得 a-b+c=2, ①
D.F′(x)=-sin x
x
∫ 解析:选 A F(x)= cos tdt=sin tError!=sin x-sin 0=sin x. 0
所以 F′(x)=cos x,故应选 A.
[ ] =
1
1
3 × 03-3 × -13
+[(sin 1-1)-(sin 0-0)]
=sin 1-23.
答案:sin 1-23
3
∫ 8. 已 知 等 差 数 列 {an}的 前 n 项 和 为 Sn, 且 S10= 0
__________.
3
∫ 解析:S10= 0
(1+2x)dx=(x+x2)30=3+9=12.
56xdx= 2×526x2Error!= 56a2≤2 016, 故
a2≤36,即 0<a≤6.
3
∫5. |x2-4|dx=( ) 0
A.231
B.232
C.233
D.235
解析:选 C ∵|x2-4|=Error!
3
3
2
∫ ∫ ∫ ∴
|x2-4|dx=
(x2-4)dx+
(4-x2)dx
0
2
0
′=x2-x,∴原式=
x3 1 3 -2x2
20=
8 3-2
-0=23.
( 人教A版)2017-2018学年高中数学选修2-2:1.6微积分基本定理课件 (共38张PPT)
=(4x-x2)10 +2x21 =3+2=5.
探究三 微积分基本定理的应用
[典例 3]
已知
1 1
(x3+ax+3a-b)dx=2a+6,且 g(t)=t (x3+ax+3a-b)dx 为偶
0
函数,求 a,b.
[解析] ∵f(x)=x3+ax 为奇函数,
∴
1 1
(x3+ax)dx=0,
2.
0
(x-ex)dx
等于(
)
-1
A.-1-1e
B.-32+1e
C.-1
D.-32
解析:0 -1
(x-ex)dx=12x2-ex0-1 =-1-12-e-1=1e-32.
答案:B
3.若2(2ax+a+1)dx=5,则 a=________. 1
1.1(ex+2x)dx 等于( ) 0
A.1
B.e-1
C.e
D.e+1
解析:1(ex+2x)dx=(ex+x2)10 =(e1+1)-e0=e. 0
答案:C
2.
2
(1+cos x)dx 等于(
)
2
A.π
B.2
C.π-2
D.π+2
解析:∵(x+sin x)′=1+cos x,
1.6 微积分基本定理
考纲定位
重难突破
1.了解并掌握微积分基 重点:1.微积分基本定理.
本定理的含义.
2.利用微积分基本定理求定积分.
2.会利用微积分基本定 难点:用微积分基本定理解决与之相关
理求函数的积分.
的综合问题.
01 课前 自主梳理 02 课堂 合作探究 03 课后 巩固提升
课时作业
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课时跟踪检测(十一) 微积分基本定理层级一 学业水平达标1.下列各式中,正确的是( ) A.⎠⎛a bF ′(x )d x =F ′(b )-F ′(a ) B.⎠⎛a bF ′(x )d x =F ′(a )-F ′(b ) C.⎠⎛a bF ′(x )d x =F (b )-F (a ) D.⎠⎛a bF ′(x )d x =F (a )-F (b )解析:选C 由牛顿-莱布尼茨公式知,C 正确. 2.⎠⎛0π(cos x +1)d x 等于( ) A .1 B .0 C .π+1D .π解析:选D ⎠⎛0π(cos x +1)d x =(sin x +x ) ⎪⎪⎪π=sin π+π-0=π.3.已知积分⎠⎛01(kx +1)d x =k ,则实数k =( ) A .2 B .-2 C .1D .-1解析:选A 因为⎠⎛01(kx +1)d x =k ,所以⎝ ⎛⎭⎪⎫12kx 2+x ⎪⎪⎪1=k .所以12k +1=k ,所以k =2.4. ⎠⎛-a a|56x |d x ≤2 016,则正数a 的最大值为( ) A .6 B .56 C .36D .2 016解析:选A ⎠⎛-a a|56x |d x =2⎠⎛0a56x d x =2×562x 2⎪⎪⎪a=56a 2≤2 016,故a 2≤36,即0<a ≤6.5.⎠⎛03|x 2-4|d x =( )A.213B.223C.233D.253解析:选C ∵|x 2-4|=⎩⎪⎨⎪⎧x 2-4,2≤x ≤3,4-x 2,0≤x ≤2,∴⎠⎛03|x 2-4|d x =⎠⎛23(x 2-4)d x +⎠⎛02(4-x 2)d x=⎝ ⎛⎭⎪⎫13x 3-4x ⎪⎪⎪32+⎝⎛⎭⎪⎫4x -13x 3⎪⎪⎪2=⎣⎢⎡⎦⎥⎤--⎝ ⎛⎭⎪⎫83-8+⎣⎢⎡⎦⎥⎤⎝ ⎛⎭⎪⎫8-83-0 =-3-83+8+8-83=233.6.⎠⎛02(x 2-x )d x =__________.解析:∵⎝ ⎛⎭⎪⎫x 33-12x 2′=x 2-x ,∴原式=⎝ ⎛⎭⎪⎫x 33-12x 220=⎝ ⎛⎭⎪⎫83-2-0=23.答案:237. 设f (x )=⎩⎪⎨⎪⎧x 2,x ≤0,cos x -1,x >0.则⎠⎛1-1f (x )d x =_________. 解析:⎠⎛-11f (x )d x =⎠⎛-11x 2d x +⎠⎛01(cos x -1)d x=13x 3⎪⎪⎪-1+(sin x -x ) ⎪⎪⎪1=⎣⎢⎡⎦⎥⎤13×03-13-3+[(sin 1-1)-(sin 0-0)] =sin 1-23.答案:sin 1-238.已知等差数列{a n }的前n 项和为S n ,且S 10=⎠⎛03(1+2x )d x ,则a 5+a 6=__________.解析:S 10=⎠⎛03(1+2x )d x =(x +x 2)30=3+9=12.因为{a n }是等差数列, 所以S 10=a 5+a 62=5(a 5+a 6)=12,所以a 5+a 6=125.答案:1259.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:由f (-1)=2得a -b +c =2, ① 又f ′(x )=2ax +b ,∴f ′(0)=b =0, ②而⎠⎛01f (x )d x =⎠⎛01(ax 2+bx +c )d x=⎝ ⎛⎭⎪⎫13ax 3+12bx 2+cx 10=13a +12b +c ,∴13a +12b +c =-2, ③ 由①②③式得a =6,b =0,c =-4.法二:设f (x )=|2x +3|+|3-2x |=⎩⎪⎨⎪⎧-4x ,-3≤x <-32,6,-32≤x ≤32,4x ,32<x ≤3.如图,所求积分等于阴影部分面积,即⎠⎛3-3(|2x +3|+|3-2x |)d x =S =2×12×(6+12)×32+3×6=45.层级二 应试能力达标1.函数F (x )=⎠⎛0x cos t d t 的导数是( )A .F ′(x )=cos xB .F ′(x )=sin xC .F ′(x )=-c o s xD .F ′(x )=-sin x解析:选A F (x )=⎠⎛0xcos t d t =sin t ⎪⎪⎪x=sin x -sin 0=sin x .所以F ′(x )=cos x ,故应选A.2.若函数f (x )=x m+nx 的导函数是f ′(x )=2x +1,则⎠⎛12f (-x )d x =( )A.56B.12C.23D.16解析:选A ∵f (x )=x m+nx 的导函数是f ′(x )=2x +1,∴f (x )=x 2+x ,∴⎠⎛12f (-x )d x=⎠⎛12(x 2-x )d x =⎝ ⎛⎭⎪⎫13x 3-12x 2⎪⎪⎪21=56. 3.若⎠⎛1a⎝ ⎛⎭⎪⎫2x +1x d x =3+ln 2,则a 的值是( )A .6B .4C .3D .2解析:选 D ⎠⎛1a⎝ ⎛⎭⎪⎫2x +1x d x =(x 2+ln x )a 1=(a 2+ln a )-(1+ln 1)=(a 2-1)+ln a =3+ln 2.∴⎩⎪⎨⎪⎧a 2-1=3,a >1,a =2,∴a =2.4.若f (x )=x 2+2⎠⎛01f (x )d x ,则⎠⎛01f (x )d x =( )A .-1B .-13C.13D .1解析:选B 设⎠⎛01f (x )d x =c ,则c =⎠⎛01(x 2+2c )d x =⎝ ⎛⎭⎪⎫13x 3+2cx ⎪⎪⎪10=13+2c ,解得c =-13. 5.函数y =x 2与y =kx (k >0)的图象所围成的阴影部分的面积为92,则k =________________.解析:由⎩⎪⎨⎪⎧y =kx ,y =x 2,解得⎩⎪⎨⎪⎧x =0,y =0或⎩⎪⎨⎪⎧x =k ,y =k 2.由题意得,⎠⎛0k(kx -x 2)d x =⎝ ⎛⎭⎪⎫12kx 2-13x 3⎪⎪⎪k=12k 3-13k 3=16k 3=92,∴k =3. 答案:36.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________解析:长方形的面积为S 1=3,S 阴=⎠⎛013x 2d x =x 3⎪⎪⎪1=1,则P =S 阴S 1=13. 答案:137. 已知S 1为直线x =0,y =4-t 2及y =4-x 2所围成图形的面积,S 2为直线x =2,y =4-t 2及y =4-x 2所围成图形的面积(t 为常数).(1)若t =2时,求S 2.(2)若t ∈(0,2),求S 1+S 2的最小值. 解:(1)当t =2时,S 2=([2-(4-x 2)]d x =⎝ ⎛⎭⎪⎫13x 3-2x =43(2-1). (2)t ∈(0,2),S 1=⎠⎛0t[(4-x 2)-(4-t 2)]d x=⎝⎛⎭⎪⎫t 2x -13x 3⎪⎪⎪t0=23t 3, S 2=⎠⎛t2[(4-t 2)-(4-x 2)]d x =⎝ ⎛⎭⎪⎫13x 3-t 2x ⎪⎪⎪2t=83-2t 2+23t 3, 所以S =S 1+S 2=43t 3-2t 2+83,S ′=4t 2-4t =4t (t -1),令S ′=0得t =0(舍去)或t =1, 当0<t <1时,S ′<0,S 单调递减, 当1<t <2时,S ′>0,S 单调递增, 所以当t =1时,S min =2.8.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.解:抛物线y =x -x 2与x 轴两交点的横坐标x 1=0,x 2=1,所以,抛物线与x 轴所围图形的面积S =⎠⎛01(x -x 2)d x =⎝⎛⎭⎪⎫x 22-x 33⎪⎪⎪1=12-13=16. 抛物线y =x -x 2与直线y =kx 两交点的横坐标为x ′1=0,x ′2=1-k ,所以S2=(x -x 2-kx )d x =⎝ ⎛⎭⎪⎫1-k 2x 2-x 33=16(1-k )3,又知S =16,所以(1-k )3=12. 于是k =1-312=1-342.。