定積分習(xí)題及答案

1/311""01404010定積分0___"20j4000dxx2/312x)+wx2+1n)wn20其它0))0kk0xxet-16-2x)+w(x+a)x)+w(x+a)x>0101003/3130f(0)。a(2)24a22a00x)0x47．設(shè)f(x)在[a,b]上連續(xù)可微，若f(a)=f(b)=0，則 k)0ak04/3140000dx2h)h)0+a00000005/315aaa227．設(shè)f(x)處處二階可導(dǎo)，且f,(x)之0，又u(t)為任一連續(xù)函數(shù)，則a0(a0)a(2)aa定積分0解：原式=-jcos3xdx=-1cos4x=104400爪爪204080=a4"-a41sin4t="a48284160""""44"42=2-33-1511解：令5-4x=u，則x=-u2，dx=-udu44238613346/316422jedlnxjed(1+lnx)100幾02_幾_幾幾_2007/3178/318000000240"=3"+1sin4x2=3"2420解：∵x3sin2x為奇函數(shù)4""4""44""441x19/31902082200=一420000000005000002202064006401x1112""–4"–4"=j0"–4""–04442=–3310/31100(cos2x)0(cos2x)0cos2x00cos2x0cosx0402原式=一j一(2)(2)dt (2)011/311112/3112800x2-11028x2xxx)|2+2(x) (x)==2arctgx=2"x20+0解：原式=jln(|2sinx-cosx)|dx令x=2t2j(ln2+lnsint+lncost)dt0(22)0200213/3113tt200dx解：將兩邊對(duì)x求導(dǎo)得014/31142)2)xarctgxx0012 (2)63 (2)6301xwx2+1ctgxctgx解：lim0w型limx)+wx)+wx)+wx)+w2x)+wxx)+wxx)+w+2+2x4|l0,其它00002200爪02爪15/3115(101+ex-1,當(dāng)x<1時(shí)0n)wn2xdx=3nnnn0kkkk0016/311600222xet_16=2(|"_arctgex_1)|=" (3)61_2 (2) (2)1∴y=e_x2在x=0處取最大值y=1，且y=e_x2在x=土處取最小值e_2211___22217/3117即2e-<j12e-x2dx<2。1-2x)+w(x+a)ax)+w(x+a)aaa1-1-103e0002240118/3118001x00x00x001(2)11121令I(lǐng),=4-=0得a=aa20a00020000003143319/3119000000f(0)。x)0x3x)0x2 3x)0xx)0x2a00 (2)(2)20/3120a(2)240002!00002!000020060002 (2)24a2fxfafxfafbfaxa)aa221/31212axaaxtax0000200(2)202020200x)0x422/3122()()()()()()7．設(shè)f(x)在[a,b]上連續(xù)可微，若f(a)=f(b)=0，則aa2a2a22k)0akakkakaakka+kka故limjbfk(xk)_f(x)1jb+kf(x)dx_lim1ja+kf(x)dxk)0akk)0kbk)0ka23/31230000000000000由f(x)可微，于是解之f(x)=ke_x(k為任意常數(shù))00024/312400000000000dx2解：方程兩邊對(duì)x求導(dǎo)，得25/3125h)h)0+a證：設(shè)F(x)為f(x)的原函數(shù)，則h)0+h0解：等式兩邊對(duì)x求導(dǎo)，得0001212220000026/3126000a0aa0a0a故F(2a)-2F(a)=1b-a27/312700000aa000x00000""0000axaaa飛28/3128a飛aaa0x-10x-1x-1xx00aaaaaaaaaaa29/3129a2aa227．設(shè)f(x)處處二階可導(dǎo)，且f,(x)之0，又u(t)