高等数学经典题型整理.pdf

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1、1y²KŠ9ϼêf(b)f(a)g(x)=f(a)+(xa)baKg(x)•/XAx+B¼ê§Ï•f(x)Ø•/XAx+B¼ê§¤±–•3˜:c2[a;b]§¦f(c)6=g(c)§´f(a)=g(a);f(b)=g(b)(1)f(c)>g(c)§džf(c)f(a)g(c)g(a)>cacaf(b)f(a)f(a)+(ca)f(a)ba=caf(b)f(a)=baf(c)f(a)f(b)f(a)=>cabaÏ•[a;c][a;b]§¤±d¥Š½n•912(a;c)§¦0f(c)f(a)f

2、(1)=calk0f(b)f(a)f(1)>ba(2)f(c)bcbcf(b)f(a)f(b)f(a)+(ca)ba=bcf(b)f(a)=baf(b)f(c)f(b)f(a)=>bcbaÏ•[c;b][a;b]§¤±d¥Š½n•922(c;b)§¦0f(b)f(c)f(2)=bc1lk0f(b)f(a)f(2)>banþ¤ã§3(a;b)S–k˜:¦0f(b)f(a)f()>ba2¦:‡ê¦e¡ü‡•‚:(y=4lnx

3、+ky=4x+ln4x=4lnx+k=4x+ln4xþªŒ±=z•ln4x4lnx+4x=k-f(x)=ln4x4lnx+4xK•§=z•f(x)=k(1)-4(ln3x1+x)f0(x)==0x)Ñx=1(f0(x)=0•kùo˜‡)§Œ±ùog•µf0(x)=0,ln3x1+x=0,´•g(x)=ln3x+x1´üNO¼ê§§†x¶•õƒ˜g§¤±•kx=1ù˜‡))x(0;1)1(1;+1)f0(x)0+f(x)&fmin(x)=f(1)=4%Table1:f(x)Úf0(x)Š2·‚ŠâTable.1Œ±‰Ñ¼êf(x)

4、úãXFig.1§l㥌±wÑ•§(1))‡ê‘kŠCzFigure1:f(x)úã(y=4lnx+kik<4ž§•§(1)k0‡)§=ü•‚vk:¶y=4x+ln4x(y=4lnx+kiik=4ž§•§(1)k1‡)§=ü•‚k˜‡y=4x+ln4x:¶(y=4lnx+kiiik>4ž§•§(1)k2‡)§=ü•‚kü‡y=4x+ln4x:"3pdÈ©½È©Z+1x2I=edx(2)1duȼêex2>0,¤±½È©I>0.duUCÈ©CþØUC½È©Š§¤±(2)•Œ±¤Z+1y2I=edy(3)1(2)

5、(3)Z+1Z+1Z+1Z+1I2=ex2dx
ey2dy=dxdye(x2+y2)(4)11113(4)´†‹IXe鲡•È©§Œ±=z¤4‹Ie鲡•È©Z2Z+1I2=drdrer2(5)00(5)¥•Ü©È©é{ü§´2,,(5)ÒC¤Z+1I2=2rdrer2(6)0Z+1r2=2rdre(7)0Z+1=er2dr2(8)0+1r2=(e)(9)0=(10)duÈ©I>0,¤±Œ±ò(10)†m•§pI=(11)duȼêex2´‡ó¼ê§¤±·‚„U

6、Z+1x21K=edx=I(12)02¤±½È©Zp+1x2K=edx=(13)02Z+12pI=exdx=(14)14VÝ‚¡È3†‹I¥VÝ‚•§´(x2+y2)2=x2y2(15)ª(15)Œ±z¤4‹Ieëꕧ§Šâ(x=rcos(16)y=rsinòª(16)‘ª(15),r4=r2cos2r2sin2(17)41.r6=0ž§ª(17)ü>Ӟرr2§r2=cos2sin2=cos2(18)(r=02.r=0ž§ª(17)L«´§=ª(17)L«´:"•?¿Š´§·‚uyª

7、(18)¥•¹:(r=0;==4),T:•L«:"¤±§r=0œ¹Œ±8r6=0œ¹"nܱþ?ا·‚•VÝ‚34‹Ieëꕧ´r2=cos2(19)e¡¦1˜–•4‰Œ"Äk§Ï•·‚?n´1˜–•œ¹§¤±0<<(20)4,§Ï•r2>0,¤±ª(19)7L÷vr2=cos2>0(21)dت(21)hi32n;+n[+n;+n(22)44d(20)(22)31˜–•S4Cz‰Œ´hi20;(23)4Figure2:ã¥ÒKÜ©´È©¡È§Ù¡ÈŒ±Cq•˜‡n

8、/¡È5ŠâVÝ‚A:§·‚•I‡¦ÑÙ31˜–•S¡È§,¦±o=Œ"3Fig.2¥§·‚Œ±w§ÒKÜ©“L