英文版微积分试卷答案(1)

1、 (1) sin 2lim

x x

x

→∞= 0 .

(2) d(arctan )x = 2

1

d 1+x x

(3)

21

d sin x x =⎰ -cot +C x x

(4).2()

()x n e = 22n x e .

(5)

x =⎰

26/3

2、

(6) The right proposition in the following propositions is ___A_____.

A. If lim ()x a

f x →exists and lim ()x a

g x →does not exist then lim(()())x a

f x

g x →+does

not exist.

B. If lim ()x a

f x →,lim ()x a

g x →do bot

h not exist then lim(()())x a

f x

g x →+does not

exist.

C. If lim ()x a

f x →exists and lim ()x a

g x →does not exist then lim ()()x a

f x

g x →does not

exist.

D. If lim ()x a

f x →exists and lim ()x a

g x →does not exist then ()

lim

()

x a

f x

g x →does not exist.

(7) The right proposition in the following propositions is __B______.

A. If lim ()()x a

f x f a →=then ()f a 'exists.

B. If lim ()()x a

f x f a →≠ then ()f a 'does not exist.

C. If ()f a 'does not exist then lim ()()x a

f x f a →≠.

D. If ()f a 'does not exist then the cure ()y f x =does not have tangent at (,())a f a .

(8) The right statement in the following statements is ___D_____.

A. sin lim 1x x

x

→∞= B. 1

lim(1)x x x e →∞+=

C. 11d 1x x x C ααα

+=

++⎰ D. 5511

d d 11b b a a x y x y =++⎰⎰ (9) For continuous function ()f x , th

e erroneous expression in the

following expressions is ____D__.

A.d (()d )()d b a f x x f b b =⎰

B. d (()d )()d b

a f x x f a a =-⎰ C. d (()d )0d

b a f x x x =⎰ D. d (()d )()()d b

a

f x x f b f a x =-⎰

(10) The right proposition in the following propositions is __B______.

A. If ()f x is discontinuous on [,]a b then ()f x is unbounded on [,]a b .

B. If ()f x is unbounded on [,]a b then ()f x is discontinuous on [,]a b .

C. If ()f x is bounded on [,]a b then ()f x is continuous on [,]a b .

D. If ()f x has absolute extreme values on [,]a b then ()f x is continuous on [,]a b .

3、Evaluate 2011lim()x x e x x →-- 201=lim(

)x x e x x →--01=lim()2x x e x →-01

=lim =22x x e →

（考点课本节洛比达法则，每年都会有一道求极限的解答题，大多数都是用

洛比达法则去求解，所以大家要注意节的内容。注意洛比达法则的适用范围。）

4．Find 0d |x y =and (0)y ''if 2

0x x x y y t e +=+⎰.

2

'()'

x x x y y t e +=+⎰（）

1'2()'2()1

x x y x y x e y x y x e +=⋅+⇒=⋅+-

0(20(0)1)0x dy y e dx dx

==⋅⋅+-=

''(2()1)'2()2'()x

x

y x y x e y x xy x e =⋅+-=++

2

00

-(0)0-01

x x y y t e x y e =+⇒=+=⎰

0''02(0)20'(0)=3

y y y e =+⋅+（） （考察微积分基本定理与微分，书上节）

5、 Find 2

2arctan d (1)x

x x x +⎰=22221)arctan d (1)x x x x x x +-+⎰（

22arctan arctan =d d (1)

x x x x x x -+⎰

⎰

-12

3

11=-arctan +d arctan +2

x x x x x x -⎰

22-1

2

2

1++1=-arctan +d arctan 1+2

x x x x x x x x -⎰（）