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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)<Mx-y,证明 (xd n/ 2 [2+真[y k ∑[(x)-/()s∑ k n 1 M 2 四.设Ln=[tan”xdr,n为大于1的正整数,证明 <Ⅰ< 2(n+1) 证明令mx,则1,=myx= 因为(1+)=a+7)2011所以+F<1+P=2 于是 I"dt dh 立即得到 <l< 2 2n2(n-1) 五.设f(x)在[O,1连续,且单调减少,fx)>0,证明:对于满足0<a<B<1的任何a,B,有 BS/(x)dx>aff(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)| < M|x－y|, 证明 n M n k f n f x dx n k 2 1 ( ) 1 1 0 ˜ £ ¯ ˆ Á Ë Ê Ú - Â= 证明: Ú ÂÚ = = - n k n k n k f x dx f x dx 1 1 1 0 ( ) ( ) , Â = = n k n k f n 1 ( ) 1 dx n k f n k n k n ÂÚ k = - 1 1 ( ) n M n M x dx n k M dx n k dx M x n k f x f dx n k f x f n k f n f x dx n k n k n k n k n k n k n k n k n k n k n k n k n k n k 2 1 2 ( ) ( ) ( ) ( ) | ( ) ( ) | 1 ( ) 1 2 1 1 1 1 1 1 1 1 1 1 0 ˜ = = ¯ ˆ Á Ë Ê = - £ - £ - ˙ ˚ ˘ Í Î È - = - ÂÚ Â ÂÚ ÂÚ Ú Â ÂÚ = = - = - = - = - = 四. 设 Ú = 4 0 tan p I xdx n n , n 为大于 1 的正整数, 证明: 2 ( 1 ) 1 2 ( 1 ) 1 - < < + n I n n . 证明: 令 t = tan x , 则 Ú Ú + = = 1 0 2 4 0 1 tan dt t t I xdx n n n p 因为 2 2 2 ' 2 (1 ) 1 1 t t t t + - ˜ = ¯ ˆ Á Ë Ê + > 0, (0 < t < 1). 所以 2 1 1 1 1 1 2 2 = + < + t t 于是 Ú Ú Ú - < + < 1 0 1 1 0 2 1 0 2 1 2 1 1 dt t dt t t t dt n n n 立即得到 2 ( 1 ) 1 2 1 2 ( 1 ) 1 - < < < + n n I n n . 五. 设 f(x)在[0, 1]连续, 且单调减少, f(x) > 0, 证明: 对于满足 0 < a < b < 1 的任何 a, b, 有 Ú Ú > 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)