第三章一元函数积分学(不定积分) 求下列不定积分 解 d In 2.∫cosx+sinx+11+snx (+cos x)- 1+cosx cosx+sinx+11+ sin x 1+sin x, 1+sinx 1(1+sin x 解」(1 1+cosx 1+ cosx 1+cos x 1+cosx 2(1+cosx x3+1) 解,方法一令,_1 dt x3+1)了1(1 In(1+r)+c In 1+ 方法 x(x3+1)x(x3+1) dx l[d(1+x8)-nIx8x)+c=-In(1+8+c 1+ 求下列不定积分 (x+1)2√x2+2x+2 解 (x+1)2√x2+2x+21(x+1)2√(x+1)2+1 tan"t sec t tdt +c sin t sin t x+1 + x 解.令x=tant, coS t dt ds sint r dsin t
第三章 一元函数积分学(不定积分) 一. 求下列不定积分: 1. Ú - + - dx x x x 1 1 ln 1 1 2 解. = - + - Ú dx x x x 1 1 ln 1 1 2 c x x x x d x x ˜ + ¯ ˆ Á Ë Ê - + = - + - + Ú 2 1 1 ln 4 1 1 1 ln 1 1 ln 2 1 2. Ú + + × + + + dx x x x x x 1 cos 1 sin (1 cos ) cos sin 1 2 解. c x x x x d x x dx x x x x x ˜ + ¯ ˆ Á Ë Ê + + = + + + + = + + × + + + Ú Ú 2 2 1 cos 1 sin 2 1 1 cos 1 sin 1 cos 1 sin 1 cos 1 sin (1 cos ) cos sin 1 3. Ú ( +1 ) 8 x x dx 解. 方法一: 令 t x 1 = , t c t t dt dt t t t x x dx = - + + + = - ˜ ¯ ˆ Á Ë Ê + - = + Ú Ú Ú ln(1 ) 8 1 1 1 1 1 1 ( 1 ) 8 8 7 8 2 8 = c x ˜ + ¯ ˆ Á Ë Ê - + 8 1 ln 1 8 1 方法二: Ú Ú Ú + = - - + = + dx x x x x x x dx x x dx ) 1 1 1 ( ( 1 ) ( 1 ) 8 8 7 8 8 7 8 = x x c x d x x dx = - + + + + - Ú Ú ln(1 ) 8 1 ln | | 1 (1 ) 8 1 8 8 8 = c x ˜ + ¯ ˆ Á Ë Ê - + 8 1 ln 1 8 1 二. 求下列不定积分: 1. Ú ( + 1 ) + 2 + 2 2 2 x x x dx 解. Ú Ú + + + + = + + + ( 1) ( 1) 1 ( 1) ( 1) 2 2 2 2 2 2 x x d x x x x dx 令x + 1 = tan t Ú t t t dt tan sec cos 2 2 = Ú + + + + = - + = - c x x x c t t tdt 1 2 2 sin 1 sin cos 2 2 2. Ú + 4 2 x 1 x dx 解. 令 x = tan t, Ú Ú Ú Ú Ú = = = - = - + + + c t t t d t t d t dt t t t t t dt x x dx sin 1 3 sin 1 sin sin sin sin sin cos tan sec cos 1 4 4 2 3 3 4 2 4 2
l(√1+x dx +x 解.令x=tant ec t coS t d sin t dt dt 2x2+1)√1+ 2 tant+1)sect J2sin2t+cost arctan sin t+c= arctan +c 解.令x= a sin t dx ,2 sin 2t a cos tdt 2( 1-cos2t dt2q't--a sin 2t+c a cost arcsin 解.令x=sint ∫√-x)dk=」osth=∫ (1+ CoS20).[1+2 cos 2t+cos" 2t I+sin 2t+l(1+cos 4n)dt=t+sin 2t+asin 4t+c --arcsin x+sin 2t(1+cos 20)+c 4+1 arcsin x +-2 sin t cost( )+C =Arcsin x+-xV1-x2(5-2x2)+c x dx =-m全=sm-mosm cos u+c 求下列不定积分
= c x x x x + + + ˜ ˜ ¯ ˆ Á Á Ë Ê + - 2 3 2 1 1 3 1 3. Ú + + 2 2 (2 x 1 ) 1 x dx 解. 令 x = tan t Ú Ú Ú Ú + = + = + = + + t d t dt t t t dt t t t x x dx 2 2 2 2 2 2 2 1 sin sin 2 sin cos cos (2 tan 1)sec sec (2 1) 1 = c x x t c + + + = 2 1 arctan sin arctan 4. Ú - 2 2 2 a x x dx (a > 0) 解. 令 x = a sin t Ú Ú Ú = - + - = × = - dt a t a t c t a a t a t a tdt a x x dx sin 2 4 1 2 1 2 1 cos2 cos sin cos 2 2 2 2 2 2 2 2 = a x c a x a a x ˜ + ¯ ˆ Á Ë Ê - - 2 2 2 2 arcsin 2 5. Ú - x dx 2 3 (1 ) 解. 令 x = sin t Ú Ú Ú Ú + + = + - = = dt t t dt t x dx tdt 4 1 2 cos2 cos 2 4 (1 cos2 ) (1 ) cos 2 2 2 3 4 = Ú t + t + + t dt = t + t + sin 4t + c 32 1 sin 2 4 1 8 3 (1 cos 4 ) 8 1 sin 2 4 1 4 1 = x + t + cos 2t) + c 4 1 sin 2 (1 4 1 arcsin 8 3 = c t x t t + + - + ) 4 4 1 2 sin 2 sin cos ( 4 1 arcsin 8 3 2 = x + x 1- x (5 - 2x ) + c 8 1 arcsin 8 3 2 2 6. Ú - dx x x 4 2 1 解. 令 t x 1 = Ú Ú Ú ˜ = - - ¯ ˆ Á Ë Ê - - = - dt t t dt t t t t dx x x 2 2 4 2 2 4 2 1 1 1 1 1 令t = sin u Ú - u udu 2 sin cos = c x x u c + - + = 3 2 3 3 3 ( 1 ) cos 3 1 三. 求下列不定积分:
e+e 解 e +e dx ts∫d (er-e-r)211=arctan(e-e )+c 2(1+4) 解令1=23,a=d t In 2 +4)=](+t)h2in2(-1+r2 tIn 2 In 2 (2+ arctan 2) 四.求下列不定积分 解 99 xd(x-2)-9 x(x-2) 99(x-2 5.4 99xX-2)999×98(x-2)99.9」x(x-2)ax 5.4.3x2 99x-2)998(x-2)99.98.97x-2y99.9897.96(x-2) 5·4·3·2x 5·4·3.2 9098996.9x-2)-9.9.97.96.95(x-2y+c x√1+x tdt 解 令x=1/t 1⊥,4 tan u I rsec udu=- In tanu+ secu+ 五.求下列不定积分: x cos xdx 解.| x cos xdr x(+cos 2x)dx=x d sin 2x x +-xsin 2 in 2xdx
1. Ú - + + dx e e e e x x x x 1 4 2 3 解. Ú Ú Ú = - + - + - = - + + = - + + - - - - - e e c e e d e e dx e e e e dx e e e e x x x x x x x x x x x x x x arctan( ) ( ) 1 ( ) 1 1 4 2 2 2 2 3 2. Ú 2 (1 + 4 ) x x dx 解. 令 x t = 2 , t ln 2 dt dx = c t t dt t t t t dx dt x x ˜ = - - + ¯ ˆ Á Ë Ê + = - + = + Ú Ú Ú ln 2 arctan ln 2 1 1 1 1 ln 2 1 2 (1 4 ) (1 )ln 2 2 2 2 2 = c x x - + + - (2 arctan 2 ) ln 2 1 四. 求下列不定积分: 1. Ú - dx x x 100 5 ( 2 ) 解. Ú Ú Ú - - + - - = - - = - - x x dx x x dx x d x x x 4 99 99 5 5 99 100 5 ( 2 ) 99 5 99 ( 2 ) ( 2 ) 99 1 ( 2 ) = Ú - - × × + ¥ - - - - x x dx x x x x 3 98 98 4 99 5 ( 2 ) 99 98 5 4 99 98 ( 2 ) 5 99 ( 2 ) = 96 2 97 3 98 4 99 5 99 98 97 96 ( 2 ) 5 4 3 99 98 97 ( 2 ) 5 4 99 98 ( 2 ) 5 99( 2 ) × × × - × × - × × - × - × - - - - x x x x x x x x c x x x + × × × × - × × × × - × × × × - × × × - 95 94 99 98 97 96 95 ( 2 ) 5 4 3 2 99 98 97 96 95 ( 2 ) 5 4 3 2 2. Ú + 4 x 1 x dx 解. Ú Ú Ú Ú + = - + = - + - = + 2 2 2 4 4 4 2 4 2 1 ( ) 1 1 1 1 1 1 / 1 t dt t tdt t t t dt t x t x x dx 令 c x x du u u c u u t u + + = - = - + + = - Ú 2 2 4 2 1 ln 2 1 ln | tan sec | 2 1 sec sec 2 1 令 tan 五. 求下列不定积分: 1. Ú x xdx 2 cos 解. Ú Ú Ú x xdx = x + x dx = x + xd sin 2x 4 1 4 1 (1 cos 2 ) 2 1 cos 2 2 Ú = x + x x - sin 2xdx 4 1 sin 2 4 1 4 1 2
x+-xsin 2x +-cOS 2 dx 解∫ sec'xdr=」 j sec xd tan.x= sextans- j tan x sec x tan x =sec x tanx-(sec x-1)sec xdx=sec x tan x+ In sec x+ tan x -sec'xdx 1 xdx=sec x tan x +In sec x tan x +c ∫(mx)d In x) 3(nx)2 (n x)_3(n x)+[ 6Inx dx=_(nx)_3(n-x)_6n x +[6 x x (In x) 3(In x)2 x 6 +c 4. cos(In x ) dx 解∫ cos(In x)dx= x cos(In x)+smmx)= x[cos(In x)+mmx-∫omx)d cos(In x)dx=-[cos(In x)+sin(In x)+c 六.求下列不定积分 x+√1+ tR(x inx+i+xdx= In(x+1+x2yd-I =ln(x+√1+x2) dx √1+ 令x= tant In(r+Ⅵ+x2)1r1 2(1-x2)21-tan2t 2)1 2(1-x2)21-2sin X+ d√2sint 2(1-x2)2√21-2sin2t n(x+√1+x2) 2)4√2 ′oh +c +c 2(1-x2) 1+x2-√2
= x + x x + cos 2x + c 8 1 sin 2 4 1 4 1 2 2. Ú xdx 3 sec 解. Ú Ú Ú sec xdx = sec xd tan x = sec x tan x - tan x sec x tan xdx 3 = Ú Ú x x - x - xdx = x x + x + x - xdx 2 3 sec tan (sec 1 )sec sec tan ln | sec tan | sec xdx = x x + x + x + c Ú ln | sec tan | 2 1 sec tan 2 1 sec3 3. Ú dx x x 2 3 (ln ) 解. Ú Ú Ú = - = - + dx x x x x x dx x d x x 2 2 3 3 2 3 3 (ln ) (ln ) 1 1 (ln ) (ln ) Ú = - - + dx x x x x x x 2 3 2 (ln ) 3 (ln ) 6 ln Ú = - - - + dx x x x x x x x 2 3 2 (ln ) 3 (ln ) 6 ln 6 c x x x x x x x = - - - - + (ln ) 3 (ln ) 6 ln 6 3 2 4. Ú cos(ln x)dx 解. Ú Ú Ú cos(ln x)dx = x cos(ln x ) + sin(ln x )dx = x [cos(ln x ) + sin(ln x )] - cos(ln x )dx \ x x c x x dx = + + Ú [cos(ln ) sin(ln )] 2 cos(ln ) 六. 求下列不定积分: 1. Ú - + + dx x x x x 2 2 2 (1 ) ln( 1 ) 解. Ú Ú - = + + - + + 2 2 2 2 2 1 1 ln( 1 ) 2 1 (1 ) ln( 1 ) x dx x x d x x x x = Ú + × - - - + + dx x x x x x 2 2 2 2 1 1 1 1 2 1 1 1 ln( 1 ) 2 1 令x = tan t tdt x t t x x 2 2 2 2 sec sec 1 1 tan 1 2 1 2(1 ) ln( 1 ) × × - - - + + Ú = dt t t x x x Ú - - - + + 2 2 2 1 2 sin cos 2 1 2(1 ) ln( 1 ) = Ú - - - + + t d t x x x 2 2 2 1 2 sin 2 sin 2 2 1 2(1 ) ln( 1 ) = c t t x x x + - + - - + + 1 2 sin 1 2 sin ln 4 2 1 2(1 ) ln( 1 ) 2 2 = c x x x x x x x + + - + + - - + + 1 2 1 2 ln 4 2 1 2(1 ) ln( 1 ) 2 2 2 2
r arctan x dx arctan 1+x arctan x ∫n-d=Ⅵ+ x?arctan x-m(x+1+x)+c arctan e 解 2/arctane' de= e arctan e"+ -2x e e arctan e"+ - dx dx 1+e 2e(1+e2) an x)+c ln(1+x2)-3 ≥0 七.设f(x)= 求|f(x)dtx (x2+2x-3)ex< ∫(xm1+x2)-3dt 解.」f(x)d ln(1+x2)-[x2-hn(1+x2)]-3 e-2+c1 考虑连续性,所以 x2ln(1+x2)-[x2-ln(1+x2)]-3x+c f(x)x=12 x2+4x+1 0 八.设f(ex)= a sinx+ bcos x,(a,b为不同时为零的常数,求fx) 解.令t=e',x=lnt,f(1) n)+bcos(lnt),所以 f(x)=[asin(In x)+ b cos(In x)]dx la+b)sin(In x)+(b-a)cos(In x)+c 九.求下列不定积分 x=2sin x'v4-x2dx=32 sin t cos'tdt =-32 (1-cos ()cos ld cost
2. Ú + dx x x x 2 1 arctan 解. Ú Ú Ú + + = + = + - + dx x x dx xd x x x x x x 2 2 2 2 2 1 1 arctan 1 1 arctan 1 arctan = dx x x x x c x x x = + - + + + + + - Ú 1 arctan ln( 1 ) 1 1 1 arctan 2 2 2 2 3. Ú dx e e x x 2 arctan 解. dx e e dx e de e e e e e x x x x x x x x x Ú Ú Ú + = - = - + - - - 2 2 2 2 2 2 1 1 arctan 2 1 arctan 2 arctan 1 dx e e e e x x x x Ú + = - + - - 2 2 2 1 1 arctan 2 1 Ú + = - + - dx e e e e x x x x (1 ) 1 2 1 arctan 2 1 2 2 dx e e e x c e e e e e x x x x x x x x = - + + + + = - + - - - - Ú ( arctan arctan ) 2 1 ) 1 1 ( 2 1 arctan 2 1 2 2 2 七. 设 Ó Ì Ï + - + - = -x x x e x x f x ( 2 3 ) ln(1 ) 3 ( ) 2 2 0 0 < ³ x x , 求 Ú f (x )dx . 解. Ô Ó Ô Ì Ï + - + - = - Ú Ú Ú x x e dx x x dx f x dx x ( 2 3) ( ln(1 ) 3) ( ) 2 2 Ô Ó Ô Ì Ï - + + + + - - + - + = - 1 2 2 2 2 2 ( 4 1 ) [ ln(1 )] 3 2 1 ln(1 ) 2 1 x x e c x x x x x c x 0 0 < ³ x x 考虑连续性, 所以 c =－1+ c1, c1 = 1 + c Ú f (x )dx Ô Ó Ô Ì Ï - + + + + + - - + - + = - x x e c x x x x x c x ( 4 1 ) 1 [ ln(1 )] 3 2 1 ln(1 ) 2 1 2 2 2 2 2 0 0 < ³ x x 八. 设 f e a x b x x '( ) = sin + cos , (a, b 为不同时为零的常数), 求 f(x). 解. 令t e x t x = ， = ln , f '(t ) = a sin(ln t ) + b cos(ln t ) , 所以 Ú f (x ) = [a sin(ln x ) + b cos(ln x )]dx = a b x b a x c x [( + )sin(ln ) + ( - ) cos(ln )] + 2 九. 求下列不定积分: 1. Ú x - x dx 3 2 4 解. 令 x = 2sin t Ú Ú Ú x 4 - x dx = 32 sin t cos tdt = - 32 (1 - cos t) cos td cos t 3 2 3 2 2 2
32 cost+—cos +(4-x2)-4-x2)2 dx(a>0 解.令x=asec jrra-dra>o)=atan a sec tant=a tan d =a tani-at- (1+e e(1+e)d= 1rd(1 = arcsin e-√1- d(a>0) v2a=x d全a=x2「n=hn全m=y2nsmt8 Ba tdt 2a 8a2〔-c9202 dt= 2a(1-2 cos 2t+cos 2t)dt =2at-2a- sin 2t+ 2a 1+cos 4t dt= 3a t-2a sin 2t +-sin 4t+c =3a2t-4a sin t cost+a2 sin t cost (1-2sin2t)+c =3att-3a sin tcost-2a' sint cost+c 2a2 Vav 2 2a v 2av 2 3a arcsin 3a+x a)√x(2a-x)+c 十.求下列不定积分 2-sin x 2 +coSx 2-sin x 1dr+rd(2 cos cosx 2+cos x 2+cosx 2dt 令m2=:2-1+2-+ln12 2dt cOSx 2「 +In 2+cosx
= - t + t + c = - x - - x + c 2 3 2 2 5 3 5 2 (4 ) 3 4 (4 ) 5 1 cos 5 32 cos 3 32 2. Ú > - ( 0 ) 2 2 dx a x x a 解. 令 x = a sec t Ú Ú Ú > = = = - + - a t t a tdt a t at c a t a t dx a x x a sec tan tan tan sec tan ( 0 ) 2 2 2 = c x a x - a - a arccos + 2 2 3. dx e e e x x x Ú - + 2 1 (1 ) 解. = - + Ú d e e e x x x 2 1 (1 ) Ú - dx e e x x 2 1 + dx e e x x Ú - 2 2 1 = Ú - x x e de 2 1 － dx e d e x x Ú - - 2 2 1 (1 ) 2 1 = e e c x x - - + 2 arcsin 1 4. Ú - dx a x x x 2 (a > 0) 解. Ú - dx a x x x 2 令u = x Ú - du a u u 2 4 2 2 令u = 2a sin t Ú a tdt 2 4 8 sin = Ú Ú = - + - dt a t t dt t a 2 (1 2 cos2 cos 2 ) 4 (1 cos 2 ) 8 2 2 2 2 = t c a dt a t a t t a t a t a = - + + + - + Ú sin 4 4 3 2 sin 2 2 1 cos 4 2 2 sin 2 2 4 2 2 2 2 2 =3 a t - 4 a sin t cost + a sin t cost (1 - 2 sin t ) + c 2 2 2 2 =3 a t - 3 a sin t cost - 2 a sin t cost + c 2 2 2 3 = c a a x a x a x a a a x a x a a x a + - - - - 2 2 2 2 2 2 2 2 3 2 3 arcsin 2 2 2 = x a x c a x a x a - + + - (2 ) 2 3 2 3 arcsin 2 十. 求下列不定积分: 1. Ú + - dx x x 2 cos 2 sin 解. Ú Ú Ú + + + + = + - x d x dx x dx x x 2 cos (2 cos ) 2 cos 1 2 2 cos 2 sin t x = 2 令tan Ú Ú + + + + + = + - + + ln | 2 cos | 3 2 ln | 2 cos | 2 1 1 2 1 2 2 2 2 2 2 x t dt x t t t dt
arctan+In 2+ cosx +=+arctan(tan)+In 2+cosx +c sIn x cosx sin x+ cosx sin x cos x dx=一 Ir1+2sin x cos x-1 sin x+ cosx sin x+ cos x lr(sin x++ cos)-dx= (sin x+ cos x)dx-i sin x+ cos x 2sin x +cosx d(x os x) sin(x+ =-(sin x-coS x In tan(+I 第三章一元函数积分学(定积分) 若在a,b上连续证明:对于任意选定的连续函数0N均有/(x)p(xk=0 则fx)=0. 证明:假设)≠0,a0.因为fx)在a,b上连续,所以存在δ>0,使 得在-6.5+6上y>0令m=mm。f(x),按以下方法定义,上p在一8 5+8上0(x)=√62-(x-5)2,其它地方④()=0.所以 ∫/x=/x)tm2>0 和[f(xyd(x)dx=0矛盾.所以(x)=0 二设为任意实数,证明:= dx I+(tanx) 1+cot x) 证明:先证 f(sin x) f(cos x) f(sin x)+f(cos x) f(sin x)+f(cos x) 所以 f∫(snx) dx f(cost) d(-1) f(sin x)+f(cos x) f(cost)+f(sint) dt f(cosx) f(cost)+f(sin n) f(cos x)+f(sin x) 于是 f∫(snx) f(sin x)+f(cos x +( COSX du
= x c x x c t + + + = ) + ln | 2 + cos | + 2 (tan 3 1 arctan 3 4 ln | 2 cos | 3 arctan 3 4 2. Ú + dx x x x x sin cos sin cos 解. Ú Ú + + - = + dx x x x x dx x x x x sin cos 1 2 sin cos 1 2 1 sin cos sin cos = Ú Ú Ú + = + - + + - dx x x dx x x dx x x x sin cos 1 2 1 (sin cos ) 2 1 sin cos (sin cos) 1 2 1 2 = Ú + + - - ) 4 sin( ) 4 ( 4 2 (sin cos ) 2 1 p p x d x x x = c x x - x - + ) | + 2 8 ln | tan( 4 2 (sin cos ) 2 1 p 第三章 一元函数积分学(定积分) 一．若 f(x)在[a，b]上连续, 证明: 对于任意选定的连续函数F(x), 均有 ( )F( ) = 0 Ú b a f x x dx , 则 f(x) º 0. 证明: 假设 f(x)¹ 0, a 0. 因为 f(x)在[a，b]上连续, 所以存在d > 0, 使 得在[x－d, x + d]上 f(x) > 0. 令 m = min f (x ) x -d £ x£ x + d . 按以下方法定义[a，b]上F(x): 在[x－d, x + d]上F(x) = 2 2 d - (x -x ) , 其它地方F(x) = 0. 所以 0 2 ( ) ( ) ( ) ( ) 2 F = F ³ > Ú Ú + - x d pd x d f x x dx f x x dx m b a . 和 ( )F( ) = 0 Ú b a f x x dx 矛盾. 所以 f(x) º 0. 二. 设l为任意实数, 证明: Ú + = 2 0 1 (tan ) 1 p l dx x I = 1 (cot ) 4 1 2 0 p p l = + Ú dx x . 证明: 先证: (sin ) (cos ) 4 (sin ) 2 0 p p = + Ú dx f x f x f x = Ú + 2 0 (sin ) (cos ) (cos ) p dx f x f x f x 令 t = - x 2 p , 所以 = + Ú 2 0 (sin ) (cos ) (sin ) p dx f x f x f x Ú - + 0 2 ( ) (cos ) (sin ) (cos ) p d t f t f t f t = = + Ú 2 0 (cos ) (sin ) (cos ) p dt f t f t f t Ú + 2 0 (cos ) (sin ) (cos ) p dx f x f x f x 于是 = + Ú 2 0 (sin ) (cos ) (sin ) 2 p dx f x f x f x + + Ú 2 0 (sin ) (cos ) (sin ) p dx f x f x f x Ú + 2 0 (cos ) (sin ) (cos ) p dx f x f x f x
f∫(sinx)+∫(cosx) 所以 cosx) dx f(sin x)+f(cosx) 4 Jo f(sin x)+f(cosx 所以 dx (cosx) 1+(tan x) (cos x)+(sin x) 4 COSX 同理1 1+(cot x 三,已知x)在,1上连续,对任意x,y都有f(x)-fy)0,证明:对于满足0aff(x)dx 证明:令F(x)=x.f(1)dt-af(n)dt(x≥a),F(a)=af(1)dt>0 F(x)=f()dh-af(x)=[Df(t)-f(x)>0,(这是因为t≤ax≥a,且f(x)单减)
= (sin ) (cos ) 2 (sin ) (cos ) 2 0 2 0 p p p = = + + Ú Ú dx dx f x f x f x f x 所以 (sin ) (cos ) 4 (sin ) 2 0 p p = + Ú dx f x f x f x = Ú + 2 0 (sin ) (cos ) (cos ) p dx f x f x f x . 所以 Ú + = 2 0 1 (tan ) 1 p l dx x I (cos ) (sin ) 4 (cos ) cos sin 1 1 2 0 2 0 p p l l p l l = + = ˜ ¯ ˆ Á Ë Ê + = Ú Ú x x x dx x x 同理 1 (cot ) 4 1 2 0 p p l = + = Ú dx x I . 三．已知 f(x)在[0，1]上连续, 对任意 x, y 都有|f(x)－f(y)| 0, (0 0, 证明: 对于满足 0 b a a b f (x )dx a f ( x )dx 0 证明: 令 Ú Ú = - x F x x f t dt f t dt a a ( ) ( ) a ( ) 0 (x ³ a), ( ) ( ) 0 0 = > Ú a F a a f t dt . = - = Ú ' ( ) ( ) ( ) 0 F x f t dt a f x a Ú - > a 0 [ f (t) f (x )]dt 0 , (这是因为 t £ a, x ³ a, 且 f(x)单减)
所以F(B)>F()>0,立即得到Bf(x)dr>af(x)x 六设邱x)在ab上二阶可导,且(x)<0,证明 f(x)ax≤(b-a)/ 证明wxtelnbf(x)=f()+f()x-1)+'(2(x-0)sf()+f((x-) 令t=a+b a+b a+b 所以f(x)≤ 三边积分[/3(2+(2-2p r-q+b +b 七.设邱x)在[O,1上连续,且单调不增,证明:任给α∈(0,1),有 f(r)dx alf(x)dx 证明:方法一:令F(x)=af(an)d-a[f(t)t (或令F(x)=xf()dt-af()d) F"(x)=af(aa)-af(x)≥0,所以F(x)单增 又因为F(0)=0,所以F(1)≥F(0)=0.即 f(an)d-alf()d≥0,即 C//(x)dx a/(x)dx 方法二:由积分中值定理,存在∈a使∫。f(x)dx=af(5) 由积分中值定理,存在n∈,1使「f(x)x=f(m)1-a) 因为n≥5,所以f()≤f(5) 所以 5,xk=a∫。f(x+∫/(x)=a/(5)+a(m)1-a) saf(5)+af(5(1-a)=af(5)=f(x)dr 八设fx)在[a,b]上连续,∫(x)在a,b]内存在而且可积,fa)=fb)=0,试证 If(x)ksilIf'()ldx,(a<x<b) 证明:-|(x)(x)f(x)|,所以 f(o)ld≤f(x)-f(a)≤[|f(t)ldt
所以 F(b ) > F(a) > 0 , 立即得到 Ú Ú > b a a b f (x )dx a f ( x )dx 0 六. 设 f(x)在[a, b]上二阶可导, 且 f ' ' (x ) < 0, 证明: ˜ ¯ ˆ Á Ë Ê + £ - Ú 2 ( ) ( ) a b f x dx b a f b a 证明: "x, tŒ[a, b], 2 ( ) 2! ' ' ( ) ( ) ( ) ' ( )( ) x t f f x = f t + f t x - t + - x £ f (t ) + f ' (t )(x - t ) 令 2 a b t + = , 所以 ˜ ¯ ˆ Á Ë Ê + ˜ - ¯ ˆ Á Ë Ê + ˜ + ¯ ˆ Á Ë Ê + £ 2 2 ' 2 ( ) a b x a b f a b f x f 二边积分 Ú Ú Ú ˜ ¯ ˆ Á Ë Ê + ˜ - ¯ ˆ Á Ë Ê + ˜ + ¯ ˆ Á Ë Ê + £ b a b a b a dx a b x a b dx f a b f x dx f 2 2 ' 2 ( ) = ˜ ¯ ˆ Á Ë Ê + - 2 ( ) a b b a f . 七. 设 f(x)在[0, 1]上连续, 且单调不增, 证明: 任给a Œ (0, 1), 有 Ú Ú ³ 1 0 0 f (x )dx a f (x )dx a 证明: 方法一: 令 Ú Ú = - x x F x f t dt f t dt 0 0 ( ) a (a ) a ( ) (或令 Ú Ú = - x F x x f t dt f t dt 0 0 ( ) ( ) a ( ) a ) F'(x ) = af (ax ) -af (x ) ³ 0 , 所以 F(x)单增; 又因为 F(0) = 0, 所以 F(1) ³ F(0) = 0. 即 ( ) ( ) 0 1 0 1 0 - ³ Ú Ú a f a t dt a f t dt , 即 Ú Ú ³ 1 0 0 f (x )dx a f (x )dx a 方法二: 由积分中值定理, 存在xŒ[0, a], 使 ( ) ( ) 0 a x a f x dx = f Ú ; 由积分中值定理, 存在hŒ[a, 1], 使 ( ) ( )(1 ) 1 h a a = - Ú f x dx f 因为 h ³ x , 所以f (h ) £ f (x ) . 所以 ( ) ( ) ( ) ( ) ( )(1 ) 2 1 0 1 0 a a a a x a h a a a = + = + - Ú Ú Ú f x dx f x dx f x dx f f Ú £ + - = = a a x a x a a x 0 2 f ( ) f ( )(1 ) f ( ) f (x )dx 八.设 f(x)在[a, b]上连续, f '( x) 在[a, b]内存在而且可积, f(a) = f(b) = 0, 试证: Ú £ b a f x | f ' (x ) | dx 2 1 | ( ) | , (a < x < b) 证明: - | f ' ( x) | £ f ' ( x) £ | f ' ( x) | , 所以 Ú Ú - £ - £ x a x a | f ' (t) | dt f (x ) f (a) | f ' (t) | dt
r(o)dsf(x)≤∫r(o)ld f()ld≤f(b)-f(x)≤[f(o)ldt 即-[f(o)|d≤f(x)≤f(o)ldt 所以-f()|d≤2f(x)sr(o)ldr B If(x)k If(x)Idx,(a4 f∫(x) 证明:因为(0,1)上f(x)≠0,可设x)>0 因为f(0)=f()=0 3xo∈(0,1)使fx0)=max(fx) 所以 f"'(x) dx> f(x) f(xo) "(x)x 在(0,x0)上用拉格朗日定理 (0,x0) 在(x.1)上用拉格朗日定理 r(B)=-/(x)B∈(xn,1) 所以 I'v(x ldx 25 "(x )dk =(rr "()dr =1/(B)-f'(a) f(x0) ≥4f(x0) a (因为( b) 所以 f"(x) dx 4 十.设x)在[a,b]上连续,且f(x)>0.,则 f(x)dx In f(x)dx 证明:将lnx在x0用台劳公式展开 Inx=In x+ Co x0)-(5-x)2≤hx0+-( (1)
即 Ú Ú - £ £ x a x a | f ' (t) | dt f (x ) | f ' (t) | dt ; Ú Ú - £ - £ b x b x | f ' (t) | dt f (b) f (x ) | f ' (t) | dt 即 Ú Ú - £ £ b x b x | f ' (t) | dt f (x ) | f ' (t) | dt 所以 Ú Ú - £ £ b a b a | f ' (t) | dt 2 f (x ) | f ' (t) | dt 即 Ú £ b a f x | f ' (x ) | dx 2 1 | ( ) | , (a Ú dx f x f x 证明: 因为（0，1）上 f(x) ¹ 0, 可设 f(x) > 0 因为 f(0) = f(1) = 0 $x0 Œ (0,1)使 f(x0) = 0 1 max £x£ (f(x)) 所以 dx f x f x Ú 1 0 ( ) ' ' ( ) > f x dx f x Ú 1 0 0 ' ' ( ) ( ) 1 （1） 在（0，x0）上用拉格朗日定理 0 0 ( '( ) x f x f ） a = (0 , ) 0 a Œ x 在（x0, 1）上用拉格朗日定理 0 0 1 ( ) '( ) x f x f - b = - ( , 1 ) 0 b Œ x 所以 4 ( ) (1 ) ( ) ' ' ( ) ' ' ( ) ' ' ( ) ' ( ) ' ( ) 0 0 0 0 1 0 f x x x f x f x dx f x dx f x dx f f ³ - = ³ £ = - Ú Ú Ú b a b a b a （因为 ab a b ³ + 2 ) 2 ( ） 所以 Ú ³ 1 0 0 ' ' ( ) 4 ( ) 1 f x dx f x 由（1）得 Ú > 1 0 4 ( ) ' ' ( ) dx f x f x 十．设 f(x)在[a, b]上连续, 且 f(x) > 0,则 Ú Ú - ³ ˙ ˚ ˘ Í Î È - b a b a f x dx b a f x dx b a ln ( ) 1 ( ) 1 ln 证明: 将 lnx 在 x0用台劳公式展开 ( ) 1 ( ) ln 1 ( ) 1 ln ln 0 0 0 2 2 0 0 0 0 0 x x x x x x x x x x = x + - - x - £ + - （1）
