2009高等数学竞赛题参考解答(专业组)

¦ÐÇÚݲ
غ²
3 −1
′ ′ ′ f (x) f (x) 0 2 f (x) 3 f (x) dx = dx + dx + 2 2 −1 1+f (x) 0 1+f (x) 2 1+f 2 (x) dx 1 + f 2 (x) 32 2 3 = arctgf (x)|0 −1 + arctgf (x)|0 + arctgf (x)|2 = arctg 27 − 2π
ÐÇ
«Á ¹¦Ä
¦
I2 =
x · cosx dx = π 1 + sin2 x 2 π 2 cost =π dt − 1 + sin2 t 0
π 2
π
2 (π − t)cost (π − t)cos(π − t) dt = dt 2 π 1 + sin (π − t) 1 + sin2 t 0 2 π π 2 2 tcost cost dt = π dt − I1 2t 1 + sin 1 + sin2 t 0 0 π 2
£ Þ ¾
1. lim (x3 − x2 +
x→∞
¥
§
Taylor
§
x 1 )e x − 2
Hale Waihona Puke Baidu
ß
x6 + 1.
2.
º² § ¡
x 1 )e x − x6 + 1 x→∞ 2 x 1 1 1 1 1 1 = lim (x3 − x2 + )(1 + + ( )2 + ( )3 + o( )3 ) x→∞ 2 x 2 x 6 x x 1 1 1 1 −x3 (1 + 1 2 x6 − 8 x1 2 + 0( x1 2 )) 1 1 1 = lim ( + 0( )) = x→∞ 6 x 6 π x · cosx dx. 2 0 1 + sin x π π x · cosx π x · cosx 2 x · cosx dx = dx + dx = I1 + I2 2 2 π 1 + sin2 x 0 1 + sin x 0 1 + sin x 2 I2 , x=π−t lim (x3 − x2 +
′
f (η) η .
′
¥ [a, b] Ê
1 f (x)dx)2 ≤ f (x)
b a
Ï f (a) = f (b), ªÉ ¥ ξ, η
′
1 dx 2 f (x)
b a
f 2 (x)dx ∈ (a, b)
¦
Æ­
[a, b]
′
« f (b) − f (a) = f (ξ)(b − a), Ó
f (b) − f (a) f (η = 2 2 b −a 2η
′
b
′
b
′
′
′
′
′
′
b a
|f (x)f (x)|dx =
a
′
b b a
|f (x)||f (x)|dx ≤ |f (x)|dx)2 ≤
′
′
b
g(x)g (x)dx = 1 2
a b a
′
b
g(x)dg(x)
b a
= =
1 2 1 1 2 g (x)|b a = g (b) = ( 2 2 2 b ′ 1 2 |f (x)| dx (b − a) 2 a f (x)
|f (x)|2 dx
′
12 dx
a
Å · Õ¦ÍÎ
(1)
¥ (−∞, +∞) Ê Þ Û ¦ L (a, b), ¯ (c, d), ¿ Õº²
I=
L
©ÌÈ (y > 0) Ê ×² µ¸
1 x [1 + y 2 f (xy )]dx + 2 [y 2 f (xy ) − 1]dy y y
ªÉ Õº² I ¢Æ L Ñ´¨ (2) ab = cd ¦ I ­¡ 1 x (1) ªÉ§ P = (1 + y 2 f (xy )), Q = 2 (y 2 f (xy ) − 1) y y ¥ ©ÌÈ ¯ ¦ ߥ ©ÌÈÊÕº² I ¢Æ L Ñ´¡ ¢ 2 ¤ §Ü º²Æ L (a, b) (c, d) ¨Õ¦¦ÐÇ
¼
f (η) 2 2 2η (b − a ) ′ ′ f (ξ ) f (η) ξ = η .
′
= f (ξ )(b − a)
′
2
a Ò f (x) ∈ C [a, b], f (a) = 0, ªÉ§ a |f (x)f (x)|dx ≤ b − (f (x))2 dx. 2 a x ªÉ§Ä g(x) = a |f (t)|dt, ¦ g (x) = |f (x)|, x x ÃÞ±È |f (x)| = |f (x) − f (a)| = | a f (t)dt| ≤ a |f (t)|dt = g(x),
¦ ß ©Ö½ ¾ ˬ°«¤½ ¡ §ÐÇ¥ »Ô ¶ Æ Â Stolz ¡ b b dx ℄ f (x) ∈ C [a, b] Ï f (x) = 0, ∀x ∈ [a, b], ªÉ§ a f 21 f 2 (x)dx ≥ (b − a)2 . (x) a b b 1 2 ªÉ§±°Þ§Ä I = a f 2(x) dx a f (x)dx b b b b 1 ¦ I = a f 21 dx f 2 (y )dy = dy f 2 (x)dx 2 (x) a a f (y ) a
0
π
cost π ¤ = π 0 1 + dt = πargtnsint|0 = 2 sin t 4 3. ¾ º² I = D sinxsiny max{x, y}dxdy ¦ÍƧ D = [0, π] × [0, π] ¡ § ß º² £ D ²¯ÀƲ D1,D2 ¦ÍÆ
2
D1 : 0 ≤ y ≤ π, 0 ≤ x ≤ y, D2 : 0 ≤ x ≤ π, 0 ≤ y ≤ x,
I = =
0
D1 π
sinxsinyydxdy +
y
D2 sinxsinyxdxdy π x
sinyydy
0 3 −1
sinxdx +
0
′
sinxxdx
0
5 sinydy = π 2
4.
§
f (x) =
¥ x = 0, x = 2 ªº·
′
(x+1)2 (x−1) , x3 (x−2)
f (x) dx. 1 + f 2 (x)
1
1 ˬ½ lnn ! n=2 1 §Ä un = lnn! , ¦
∞
Ù¡
1 nlnn! (n + 1)ln(n + 1) − nlnn = lim = lim n→∞ n→∞ nlnn n→∞ lnn! ln(n + 1) 1 n ln(1 + n ) + ln(n + 1) nln(n + 1) − nlnn + ln(n + 1) = lim = lim n→∞ n→∞ ln(n + 1) ln(n + 1) 1 n ln(1 + n ) ln(n + 1) = lim + lim =0+1 =1 n→∞ ln(n + 1) n→∞ ln(n + 1) lim un /
I =
d c 1 [1 + b2 f (bx)]dx + [y 2 f (y ) − 1]dy 2 a b b y c d c−a c c = + bf (bx)dx + cf (y )dy + − b d b a b bc dc c a = − + f (t)dt + f (t)dt d b ab bc dc c a c a = − + f (t)dt = − + 0 (ab = cd) d b d b ab c ∂P ∂y
1 nlnn n=2
∞
½
I =
b b f 2 (x) 1 b b f 2 (y ) 1 b b f 2 (y ) f 2 (x) ( dxdy + dxdy ) = ( + )dxdy 2 2 a a f 2 (x) 2 a a f 2 (x) f 2 (y ) a a f (y ) 1 b b f 4 (y ) + f 4 (x) 1 b b 2f 2 (x)f 2 (y ) = dxdy ≥ dxdy 2 a a f 2 (x)f 2 (y ) 2 a a f 2 (x)f 2 (y ) b b 2 = a a 1dxdy = (b − a)
= f (xy ) −
1 y2
+ xyf xy =
′
∂Q ∂x
ad ¼ I = bc − . db
3
±° §ÐǬ
(b − a)2 = (
Schwarz
b a
­ ¦
b a
1dx)2 = (
f (ξ ) ξ
′
=
f (x) ∈ C [a, b],f (x)
ªÉ§ · f (x) ¥ [a, b] ³ ÃޱȦ · f (x)g(x) = x2 ¥ ¼ f (b) − f (a) = f 2(ηη) (b2 − a2)