考研高数习题集(上)(4)

f(x)dx maxf"(x)

30 x 2

211122

[ (f'(1)(x 1) f"( )(x 1))dx maxf" (x 1)dx maxf"]

00220 x 230 x 2

2

6. 设 f(x) 在 [ 1,1] 上有三阶连续导数, 且 f( 1) 0,f(1) 1,f'(0) 0, 证明: ( , )使f"'( ) 3 1,1

1111

f"(0)( 1)2 f"'( 1)( 1)3;f(1) f(0) f"(0) f"'( 2) 23!23!

111

1 0 f"'( 1) f"'( 2),f"'( ) [f"'( 1) f"'( 2)] 3]

3!3!2

[f( 1) f(0)

7. 设f(x)在[ 2,2]上有一阶连续导数, 在( 2,2)内有二阶导数, 且f(x) 1,f'(0) 1

证明: ( 2,2), 使得f"( ) 0.

1 2

f( 2) f(0) f'(0)( 2) f"( )( 2)1 f"( 1) 0 2

[ f"( ) 0]

f"( ) 01 2 f(2) f(0) f'(0)(2) f"( )(2)2

2

2

8. 设f(x)在[a,b]上二阶可导, 且f"(x) 0, 证明:

a b1b1

) f(x)dx [f(a) f(b)] a2b a2

a ba ba b1a b2

[(1)f(x) f() f'()(x ) f"( )(x )

2222!2

f(

(2)f(x) f(a) f'( )(x a) f(a) f'(x)(x a)]

第三讲: 一元积分学

单元一: 原函数与不定积分 1. 设函数f(x)连续, 且 [x

x

tf(x t)dt 1 cosx, 求:f(x).

x

x

x

f(u)du uf(u)du 1 cosx f(u)du sinx f(x) cosx]

2. 设xf(x)dx arcsinx c, 求:

1

f(x)

3

122

(1 x) c]

[f(x)

33. (1)

2

[ 3c] 95

3

1312

[ x (x 1)2 c]

(2)331ex 1dx

(3) x [ lnx c]

2e 1e e x

(4)

dx1 2x

[ ln(1 4e) c] e2x 48

(5)

c]

[

(6)

[ c]

x(1 x2)1124

dx (7) . [ arctanx ln(1 x) c] 4

1 x24dx1x6

ln6 c] (8) [ 6

x(x 4)24x 4

(9)

2

[ (x3lnx 2) c]