英文版微积分试卷答案

1、1、limxsin 2x21.(2) d(arctan x)2 dx1+x一 1(3) dxx-cot x+C2 sin x(4) . (e2x) 2ne2x，、4 ,1 2xdx 26/30 2、(6) The right proposition in the following propositions isAA. If lim f (x)exists and lim g(x)does not exist then lim( f(x) g(x) does not x ax ax aexist.B. If lim f(x), lim g(x)do both not exist then li

2、m( f(x) g(x) does not exist. x ax ax aC. If lim f (x) exists and lim g(x) does not exist then lim f (x)g(x) does not x ax ax aexist.D. If lim f (x)existsx af (x)andlim g(x)does not exist then lim does not exist. x ax a g(x)(7) The right proposition in the following propositions is_BA. If lim f (x) f

3、 (a)then f (a) exists. x aB. If lim f (x) f (a) then f (a) does not exist. x aC. If f (a) does not exist then lim f (x) f(a). x aD. If f (a) does not exist then the cure y f (x) does not have tangent at (a, f (a).(8) The right statement in the following statements is Dsin xA. lim 1B.x x11C. x dx x C

4、 D.11lim(1 x)x exb 1a 1 x5dxa1 y5dy(9) For continuous function expressions is D_.d b 一一A. ( f(x)dx)f (b)db af (x), the erroneous expression in the followingB.d b .rd；(af(x)dx)f(a)dbdbC. Fjf(x)dx)0D.Fjf(x)dx)f(b)f(a)dx adx a(10) The right proposition in the following propositions is_BA. If f (x) is d

5、iscontinuous on a,b then f (x) is unbounded on a,b.B. If f (x) is unbounded on a,b then f (x) is discontinuous ona,b.a,b.C. If f (x) is bounded on a, b then f (x) is continuous onD. If f (x) has absoluteextreme values on a,b then f(x) is continuousona, b.ex 1 13、Evaluate lim(-) x 0 x xex 二1 xex 1丁)=

6、典(云x、 e 1)=lim =x 0 22y 0 y(、t)dty(x) ex 10dxarctan x ,5、 Find dx =x2(1 x2)(22.(x 1 x ) arctan x ,22dxx2(1 x2)（考点课本节洛比达法则，每年都会有一道求极限的解答题，大多数都是用洛比 达法则去求解，所以大家要注意节的内容。注意洛比达法则的适用范围。）24. Find dy |x °andy (0) if x2(x y) ( : y( .t)dt ex) 1 y' 2x y(x) exy' 2xdyX0 (2 0 y(0) e0 1)dxy'' (

7、2x y(x) ex 1)' 2y(x) 2xy'(x) ex y:y(,t)dtex-x y(0) 0e0-01y''(0) 2y(0)2 0y'(0) e°=3（考察微积分基本定理与微分，书上节）arctan x , arctan x ,2 dx dxx(1 x )-1 _="x arctan x+ dx 1arctan2x x+x32-1 _="x arctan x+1+x +x2r dxx(1+x-arctan2 x 2-1 _="x arctan x+1dx x(1+x2),1,2dx - arctan

8、x-1, ,=-x arctanx+In x -In 1+x1 arctan2 x2=-x-1 arctanx+In.1+x212-arctan x+C2（凑微分求不定积分，积分是微积分的重点及难点，大家一定要掌握透彻。2x6、Given that f (x) x(1) Find the intervals on whichf (x) is increasing or decreasing.f (x)2x(x2 1) x2 2x(x21)22x When f/ 221(x 1)(x) 0 x 0 f(x) 0Therefore, the increasinginterval is 0,thed

9、ecreasingintervalis,0(2) Find the local maximum and minimum values off(x)f (x) 0 x 0 The function is increasing in interval0,decreasing ininterval,0 , therefore, the function existthelocalminimum value, itis f (x) 0(3)Find the intervals of concavity and the inflection points.''(x)2x(/ 1)22(

10、x2 1)2 8x 2(x2 1)(x2 1)46x4(x24x2''(x)6x4 4x2 2-/2T-4(x 1)、.3x or3''(x)6x4 4x2 2 (x2 1)40 or''(f''()=4Therefore, the concave upward interval are,the concavedownwardinterval areU,03andtheinflectionpoints are、3 1一，一3 4(4) Find the asymptote lines of the curef(x)limxx2 1

11、 1+工 xTherefore, the lineis a horizontal asymptote11y 一 , and the line y x（考点：节，、节。近几年经常会考一道作图题。这种题目应该在注意的点主要包括函数的定义 域，对称性，增减区间，极值点，凹凸性，拐点，以及渐近线等。大家参照课本的节进行作图）7、Let R be the region bounded by the curveandx 2.(a)Evaluate the area of the region R.21R= x dx =2x2In x=1 222In212In1 =- In22(b)Find the y-a

12、xis .volume of thesolidgenerated by revolving the R about the2V=1（42 -y )dy11 ?(424x1 3y4y2314-22求面积以及体积，课本、节。这类题目是常考题，较简单。望同学一2 _ 20.01x(1450 36x 0.58x24x 14502x 20 since x>0, then x=40030.001x ) _.When x=400 P (x)0.006 400 1.141.26 0定要做相应的题目加以巩固。）8、 Determine the production level that will maxi

13、mize the profit for acompany with cost and demand functionsC(x) 1450 36x 0.58x2 0.001x3and p(x) 60 0.01x.2Solution R(x) (60 0.01x)x 60x 0.01xP(x) R(x) C(x) 60x320.001x0.57x_'_ 2P(x) 0.003x1.14x._ Let P (x) 0 x 400orP (x)0.006x 1.1432P(400)0.001 4000.57 40024 400 1450 35350Therefore, when the pro

14、duction level is 400 that will maximize the profit 35350（考点：经济函数，课本节。此题型为常考题，属于送分题，大家可以做相应的节的练习 加以巩固）9、 State the second derivative test theorem testing maximumand prove it.Suppose f '' is continuous near cIf f( ' c)=0 and f('' c)< 0 , then has a local maximum at c .Proof: Bec

15、ause near f('' c)< 0 c and so f is concave downward near c. This meansthat the graph of lies below its horizontal tangent at c and so has a local maximum at c.10、 Showthat the equation x3 15x c 0 has at most one root in the interval 2,2 .3Suppose the equation x 15x c 0 has two root x1, x2 and x1 x2 in the intervalx1 ,x2 2, 2 .3Let F(x) x 15x c then there exist F(x1) F(x2)=0Using the Rolle s Theorem we can know that there exist one number c can satisfy