微积分试卷-英文版3.doc

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1、Four short words sum up what has lifted most successful individuals above the crowd: a little bit more.-author-date微积分试卷-英文版31、 (1) _x0001_ 0 .1、 (1) 0 . (2) 1/(1+x2) .(3) -cotx .(4). 泰勒展开式（书上有。） .(5) 26/3 .2、 (6) The right proposition in the following propositions is _A_.A. Ifexists anddoes not exi

2、st thendoes not exist.B. If,do both not exist thendoes not exist. C. Ifexists anddoes not exist thendoes not exist. D. Ifexists anddoes not exist thendoes not exist.(7) The right proposition in the following propositions is _A_.A. Ifthenexists. B. If thendoes not exist. C. Ifdoes not exist then.D. I

3、fdoes not exist then the curedoes not have tangent at.(8) The right statement in the following statements is _D_. A. B. C. D. (9) For continuous function, the erroneous expression in the following expressions is _C_.A. B. C. D. (10) The right proposition in the following propositions is _D_.A. Ifis

4、discontinuous onthenis unbounded on.B. Ifis unbounded onthenis discontinuous on. C. Ifis bounded onthenis continuous on.D. Ifhas absolute extreme values onthenis continuous on.3、Evaluate 1/24Find andif. 隐函数求导。注意右边第一项求导为2*x*y(x) (y(x)就是y)5、 Find. arctanx/x2-arctanx/(1+x2)=-(arctanx)/x+ dx/x(1+x2)=-(a

5、rctanx)/x+(-x) dx/(1+x2)+ dx/x=-(arctanx)/x-(1/2)d(1+x2)/(1+x2)+ dx/x=-(arctanx)/x-(1/2)ln(1+x2)+ln|x|+C6、 Given that. (1) Find the intervals on whichis increasing or decreasing. F(x)=0，x=0，0为分界点(2) Find the local maximum and minimum values of 0、1（逐渐接近但不能等于）(3)Find the intervals of concavity and the

6、 inflection points. F(x)=0,x=根号2-1或-根号2-1，讨论(4) Find the asymptote lines of the cure y=17、Let R be the region bounded by the curve , and the line and.(a)Evaluate the area of the region R. 3/2-In2(b)Find the volume of the solid generated by revolving the R about the y-axis. 8/38、 Determine the production level that will maximize the profit for a company with cost and demand functionsand. 4009、State the second derivative test theorem testing maximum and prove it.书上有10、Show that the equationhas at most one root in the interval.设有2根，f(x)在-2到2上连续,f(x)=0得出2根为-根号5和根号5，即在-2到2上单调。所以最多一根-