電大?？啤段⒎e分初步》復(fù)習(xí)題及答案

1、電大微積分初步考試小抄電大微積分初步考試小抄一、填空題一、填空題⒈函數(shù)的定義域是xxf??51)(（－（－∞，5）5－＞0→＜5xx⒉1???xxx1sinlim，1sinlim???xxx01???xx時(shí)，⒊已知，則=xxf2)(?)(xf??2ln22）（x⒋若，則???cxFxxf)(d)(???xxfd)32(CxF??)32(21⒌微分方程的階數(shù)是三階三階yxxyyx???????esin)(4∵y???6.函數(shù)的定義域是（2

2、2，11）U（)2ln(1)(??xxf1，∞)∞)??????1221ln)2(ln2x02ln02?????????xxxxx，＞，＞，＞∴??12x|?且＞x7.2??xxx2sinlim02112122sinlim2sinlim00?????xxxxxx21:222sinlim0???xxx8.若y=x(x–1)(x–2)(x–3)，則(0)=y?66y=x(x1)(x2)(x3)=(x2x)(x25x6)=x45x36x2x3

3、5x26x=x46x311x26x622184y23x?????xx（把0帶入X）?6)0(????y9.???xxded2dxxe?2或)()(xfdxxf???）（dxxfdxxfd)())((??10.微分方程的特解為y=ey=ex.1)0(???yyyyy??ydxdy??????dxdydxydyy1兩邊積分又y(0)=1(x=0y=1)ecxy???cxy???ln010????cec，11.函數(shù)的定義域是24)2ln(1)

4、(xxxf????????2112?，??????????????????????????????????????122122x21ln)2ln(22x20)2(ln02042xxxxxxxx＜＜＞＞12.若函數(shù)在處連續(xù)，?????????0013sin)(xkxxxxf0?x則1?k(在處連續(xù))∵)()(lim00xxfxfx??)(xfx0kf?)0(（無窮小113sin0lim)13sin(0lim????????xxxxxx量x

5、有界函數(shù)）13.曲線在點(diǎn)處的切線方程是xy?)11(2121y??x，xxy21??xy2121???切ky?????211x|2121y)1(211y????????xx方程14.sinsinxcxc???xxsd)in(15.微分方程的階數(shù)為三階三階xyyxysin4)(53???????16.函數(shù)的定義域是（2323）)2ln()(??xxxfU（3，∞）??3x2x|122)2ln(20)2ln(02??????????????

6、??????且＞＞＞＞xxxnxxxx17.1212???xxx2sinlim18.已知，則=2727ln32727ln3xxxf3)(3??)3(f?3ln3)(32xxxf???3ln2727)3(????f19.=ex2x2cc?2dex20.微分方程的階數(shù)為四階四階xyxyysin4)(7)4(3????二、單項(xiàng)選擇題二、單項(xiàng)選擇題⒈設(shè)函數(shù)，則該函數(shù)是（偶函數(shù)）（偶函數(shù)）∵2eexxy???⒉函數(shù)所以是偶函數(shù))(2ee)(xfx

7、fxx?????的間斷點(diǎn)是（）分母233)(2????xxxxf21??xx無意義的點(diǎn)是間斷點(diǎn)∴210232?????xxxx⒊下列結(jié)論中（在處不連續(xù)，則一定在處不連續(xù)，則一定在)(xf0xx?y1=xxxcossin2321?∴dy=(∴dy=()dx)dxxxxcossin2321?7.7.計(jì)算不定積分計(jì)算不定積分xxd)21(9??解：解：dxx??)21(9令u=12xu=12xu′=u′=22∴dudxxdu212?????c

8、cduduxuuu???????????????20192121)21()21(101099?8.8.計(jì)算定積分計(jì)算定積分xxxde10??解：解：u=xu=xeexxvv??????，)()(101101010|xddxxdxxeeeeexxxx?????????????????=1)11(1|101???????eeeex9.9.計(jì)算極限計(jì)算極限4586lim224?????xxxxx3212lim)4)(1()4)(2(lim44

9、??????????xxxxxxxx10.10.設(shè)，求，求xyx3sin2??ydy1=sin3x=sin3xy1=1=sinusinuu=3xu=3xxy3cos3x3sinu1??????）（）（∴y′=2∴y′=2xln23cos3xln23cos3x∴dy=(2∴dy=(2xln23cos3x)dxln23cos3x)dx11.11.計(jì)算不定積分計(jì)算不定積分xxxdcos?u=xu=xv′=cosxv′=cosxv=sinxv=

10、sinx?xdxxcos?????????cxxxxdxxxxdxx)cos(sinsinsincos12.12.計(jì)算定積分計(jì)算定積分xxxdln51e1??令u=lnxu=lnxu′=u′=?????????eeeedxxxdxxxxdxxxdxx11e111ln51ln5lnln51|x1du=du=dxdx1≤x≤e1≤x≤e0≤lnx≤10≤lnx≤1x1∴2121ln|102101?????uududxxxe∴原式原式=15=

11、15=212713.13.計(jì)算極限計(jì)算極限623lim222?????xxxxx解：解：5131lim)2)(3x()1)(2(lim22??????????xxxxxxx14.14.設(shè)，求，求xxy12e?y?解：解：exxy12??()eyx11?eyu?1xu1?)xexeeyxuux21211)1()1()(?????????eexexeexexx1x12x12x1x12x122)(2)()(y??????????????xx1

12、5.15.計(jì)算不定積分計(jì)算不定積分xxd)12(10??解：解：u=2x1u=2x1=2=2du=2dxdu=2dxdxx??)12(10d?∴cdududxuuux???????????1121212111101010)12(cx???）（12112116.16.計(jì)算定積分計(jì)算定積分?10dexxx解：解：u=xu=xdxxex??10exv??exv?1)1(101010|?????????eedxxdxxeeexxx四、應(yīng)用題（本

13、題四、應(yīng)用題（本題1616分）分）用鋼板焊接一個(gè)容積為用鋼板焊接一個(gè)容積為4的底為正方形的無蓋的底為正方形的無蓋3m水箱，已知鋼板每平方米水箱，已知鋼板每平方米1010元，焊接費(fèi)元，焊接費(fèi)4040元，問水箱元，問水箱的尺寸如何選擇，可使總費(fèi)最低？最低總費(fèi)是多少？的尺寸如何選擇，可使總費(fèi)最低？最低總費(fèi)是多少？解：設(shè)水箱的底邊長為x，高為h表面積為s，且有h=x24所以S(x)=x24xh=x2x16?xxS2162???令（x）=0，得x