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1、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.A.卜列各式中，正确的是 ( bF (x)dx=F (b)F a(a)B.bF (x)dx=F (a)F a(b)C.bF (x)dx=F(b) -F(a) aD.bF (x)dx=F(a)F(b) a解析：选C由牛顿莱布尼茨公式知,C正确.2.兀o (cos x+1)dx 等于()A.B. 0C.解析：选兀 (cos x+1)dx=(sinx+x)=sin 兀+兀- 0 =兀.03.已知1(kx+1)dx = k,则实数 k=(A.C.解析:因为1o( kx+ 1)d x = k所以1kx2+x20=k. . 1 ，所以2k+1 = k
2、,所以k=2.0x 1f (x)=1,1x 2,则定积分20f (x)dx 等于(A.34C.； D.3解析：选C2f(x)dx=x dx+1dx=5334-.故选C.35.设f (x) =x2, xC 0 ,2-x, xC1,2,则/0f(x)dx 等于()3A.4B.5C.6D.不存在解析:0 of (x)d x=1 2 .0 ox dx +2 _1(2 -x)dx1 33x1 2 2xx256.26.1解析:11-+ f dx = x xInIn 2112 一(ln 1 1) = ln 2 +2.1.十 x答案:1 In 2 + 2x2,7.设 f(x) = cosx0.解析:11f(x
3、)dx02.x dx+1(cosx 1)d x1 3 3x-1十 (sinx x)3X0-3X3十 (sin 11)(sin 0 0)= sin 1 -2.8.已知f (x) = 3x2+ 2x+ 1,3解析：:111f (x)dx=1 (3x2+2x+ 1)dx= (x3+ x2 + x)1=(1 +1 + 1) -( -1+ 1 1) = 4,2f(a)=4, f(a)=2, 即 3a2+2a+1 = 2,3a2+2a1 = 0.解得a= 1或a=g.3J, 1答案：i或39 .计算下列定积分.5一 2 一(1) 2(3x 2x+5)dx；02(2) J 0 (cos x+ sin x)d
4、 x；、2 x 1 , xdx；22|x2-x|dx.解：(1)X3x22x+5)dx5552=3x dx2xdx+5dx222=x3x2+ 5x=(5 3 23)(5 2 22) + 5(5 2) = 111.222，八、。2兀，-、,，-、 2兀(3) J o (cos x+sin x)dx=(sin x cos x)|o2x)1=(sin 2 兀一cos 2 兀)一(sin 0 cos 0) = 0.2 X 1, x ,1 e x dx= (e In =(e 2 In 2) (e 1 In 1) =e2e In 2.(4)2 I x2-x|d x0=2(x2-x)dx+12o( x x2
5、)d x +(x2 x)d x131201 21 3 1 31 22铲一2x-2+ ?-3x 0+3x-2x 117-3.1117.2fx ,一10.若f (x)是一次函数，且0f(x)dx=5，加(x)dx =万，求 1dx 的值.解：设 f (x) = kx+b( kw 0),k 2 一则 o(kx+b)dx= 2x +bxk kx3bx2十5k b_17 3+2=6,=2+ b= 5,k= 4, 联立可得b=3.24x+ 31 xdx所以 f (x) =4x+ 3.xdx =x34 + x dx= (4 x+ 3lnx)=(8 +3ln 2) (4 +3ln 1)= 4+3ln 2.层级
6、二应试能力达标21B.-21 .若函数 f(x) = xm+ nx 的导函数是 f (x) = 2x+1,则 f ( -x)dx =()5 A.-2C.361D.6解析：选A.1 f (x) = xm+ nx的导函数是f (x)=2x+1,，f(x)=x2+x21f (一x)dx21(x2x)d x2.已知函数f(a)=a兀osin xdx,则f f万 等于(A. 1C. 0兀兀解析：选B f万=1.。兀. 一，一92 msin xdx = cos x 20兀f f - =f(1) = sin xdx=cos x =1 cos 1.3 .若2x+1dx=3+ ln 2 ,则 a 的值是()1
7、xA. 6B . 4C. 3解析：选D1 2x + x dx= (x2+ ln x)=(a2+ ln a) (1 +ln 1) = (a2- 1) + ln a= 3+ln 2.a21 = 3,a= 2.a1,a=2,4.若 f(x)=x2+1f (x)dx,则10f (x)dx=(A. 11C.3解析：选1f (x)d x= c,则01 /c=(x02+ 2c)d x=1x3+2cx 31=3+ 2c,03解得c5.已知0,a2 ,则当 0 (cos x- sin x)dx取最大值时,解析:(cosx sin x)d x = (sin x + cos x)0=sincos1=2i兀当2 si
8、na +-4 - 1取最大值时，sin=1,7t6.函数丫 = /与丫=4（卜0）的图象所围成的阴影部分的面积为92,则y= kx, 解析：由y x2解得x = k, 或 y = k2.由题意得,k 2121 3131313g 40(kx-x )d x= 2kx -3x=2k _3k =” =3，故 k=3.答案：37.已知 _(x3+ax+3ab)dx= 2a+ 6,且 f(t)= (x3 + ax+3ab)dx 为偶函数，求a, b.解：= f (x) = x3+ax为奇函数,1-(x3+ ax)d x= 0,1-(x3+ ax+ 3a b)d xi-(x3+ ax)d x +1T (3
9、a b)dx= 0+(3ab)1 -(-1)=6a 2b,6a-2b=2a+6,即 2ab=3.x4 a 2又 f(t)=疝+2x +t4 at2，而十 + (3 a-b)t为偶函数,，3ab = 0.由得a= 3, b= 一 9.8.已知S为直线x=0, y=4 t2及y = 4-x2所围成图形的面积，S2为直线x=2, y = 4t2及y=4x2所围成图形的面积(t为常数).(i)若 t=q2 时，求 s.-H(2)若t e (0,2),求S+S2的最小值.解：(1)当t =/时，区=/t2 f支)也(历一1).t(2) t e (0,2) , S= 0(4 -x2)-(4-t2)d xft2x 1 323x1=3t2S2=t(4 t2) (4x2)d x3x3-t2x 33-2t2+2-3令 S= S1 + S24t3-S = 4t2-4t = 4t(t -1),令S = 0得t =0(舍去)或t = 1,当 0t1 时，S 0,当 1t0, 所以当t =1时，Smin=2.答案：sin 11f(x)dx=2f(a)成立,则 a=1
