英文版-微积分试卷答案-(1)
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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.
1
1d 1x x x C α
αα+=
++⎰ D. 5511d d 11b
b a a x y x y =++⎰⎰ (9) For continuous function ()f x , the 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4.4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4.4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5.3节)
5、 Find 2
2
arctan 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
311=-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 -⎰() -122
11
=-arctan +d d arctan 1+2
x x x x x x x x --⎰⎰() -12211
=-arctan +In In 1+arctan 22
x x x x x --
-121
=-arctan arctan +C 2
x x x - (凑微分求不定积分,积分是微积分的重点及难点,大家一定要掌握透彻。)
6、 Given that 2
2()1
x f x x =+.