(整理)定积分及其应用61289

第五章 定积分及其应用 第一部分 重点、难点及分析
一、 定积分的定义及几何意义：
由定积分的定义，⎰b
a dx x f )(=i n
i i x f ∆∑=→1
0)(lim ξλ
一个函数的定积分是由“和式的极限”来定义的，许多现实问题都归结于求和式极限问题，如：不规则图形的面积、变力作功、变速直线运动的距离，都可通过定积分的计算找到答案。

因此定积分的概念来源于实际，是解决实际问题的有力“工具”。

定积分的概念由四个步骤构成：分割、近似代替、求和、取极限。

求和i n
i i x f ∆∑=→1
0)(lim ξλ得到了近似的结果，而取极限改善了近似程度，
极限值给出了所要测度的量的精确定义。

应该说，定积分是一种特殊的“和式极限”，它比数列、函数的极限要复杂一些，在它的定义中，没有简单地写成以n 或λ为自变量。

定积分⎰b
a dx x f )(的值与积分区间[a ，
b ]和被积函数f 有关，而与积分
变量用哪一个字母来表示无关，也就是说在⎰b
a
dx x f )(中将x 改写成u
或t 得到的结果是一样的。

在f （x ）> 0时，⎰b
a dx x f )(的数值，在几何上等于曲线f （x ），x =a ，
x =b 和x 轴所围成的曲边梯形的面积。

在一般情况下，表示曲线f （x ），直线x =a ，x =b 和x 轴之间各部分面积的代数和。

其中，在x 轴上方的面积取正号，在x 轴下方的面
积取负号。

定积分的存在性，要满足两个任意性：一是对区间[a ，b ]的任意划分，可以是等距离的，也可以是非等距离的；另一个是在小区间],[1i i x x -上任意取一点i ξ，可以在区间端点上取，也可以在区间内取。

在“微分学”中知道函数的连续性不能保证可导性，连续性是可导的必要条件，但不是充分条件。

而连续函数的定积分一定存在，这是定积分存在的充分条件，而不是必要条件，也就是说对不连续的函数，定积分也可能存在。

事实上，在区间[a ，b ]上若有有限个间断点，而有界的函数f （x ）在这个区间上的定积分是存在的。

二、 定积分的性质：
1） 运算性质：⎰±⎰=⎰±b
a
b
a
b
a
dx x g dx x f dx x g x f )()()]()([βαβα
2） 区间性质：
♥⎰+⎰=⎰b
c
c
a
b
a
dx x f dx x f dx x f )()()(
♥ 若f （x ）在[a ，b ]上可积，则在[a ，b ]上的部分区间],[],[11b a b a ⊂ 上也是可积的。

3） 不等式性质：
♥ 若f (x )，g （x ）在[a , b ]上可积，且f (x )≤g （x ），那么
⎰≤⎰b
a
b
a
dx x g dx x f )()(
♥ 若f (x )在[a , b ]上可积，M x f m ≤≤)(，则
)()()(a b M dx x f a b m b
a
-≤≤-⎰
4)积分中值性质：],[),)(()(b a a b f dx x f b a
∈-=⎰
ξξ
三、 定积分的计算：
重要的公式：）的原函数（是x f x F a F b F dx x f b
a )(),()()(-=⎰
计算定积分也可以象计算不定积分时使用换元法，在用第二换元法时，一是，要注意引进的新变量应是单值、单调、连续的函数，且要有连续的导数。

二是进行变量代换要换积分限。

如：
单调的函数。

上是不连续的，也不是，在，关键的是引进的，这是一个错误的结果从而，，如果令]11[101111111
111,11
112112112
21121
12-==++-=+-
=+=+⎰⎰⎰⎰⎰-----t x dx x dt t
dt t t dx x t x dx x 计算定积分时，常用的几个结论： 1） 当被积函数f （x ）是奇函数时，⎰-=a
a
dx x f 0)( 。

2） 当被积函数f （x ）是奇函数时，⎰⎰-=a
a
a
dx x f dx x f 0
)(2)(
3）
若f （x ）是以T 为周期的连续函数，
⎰⎰+=T
T a a
dx x f dx x f 0
)()(
四、 定积分的应用：在几何中的应用：求平面图形的面积、求旋转
体的体积；在物理方面的应用：变力作功。

第二部分 书后习题
1)①
).
()(lim )
()(121][)(1
01
1a b k x x f a b k n a b nk n a b k x x f k x f k x f x n
a
b x n n i n
a b i x n b a a b k kdx i n
i i n i i n i i i i i i i b
a
-=∆-=-⋅=-=∆===-=
∆-=-=-=∑∑∑⎰=→==λξ求极限：和式）（，从而）（；而，取的长度；每个小区间，，，，）
（等分，分点为分成，将。

用定义证明：Λ2）②121)1(222)12(1
0101
1
10
-=+-=+=+=+⎰⎰⎰e e x e
dx dx e
dx e x x
x
④2
)4(4arctan 11
11
12
π
ππ
=--=
=+
--⎰x x dx
⑥4
1arctan 1111
111
0101
02
1
10
2
21
02
2π
-=-=+
-=+-+=+
⎰
⎰⎰⎰
x x dx x
dx dx x
x dx x
x
3）②。

对第一个积分，⎰⎰⎰⎰⎰⎰⎰⎰⎰=∴=-=--−−→−+=--=---a
a
a
a
a
a
t
x a
a
a
a
a
dx x f dx x f dx x f dt t f dt t f dx x f dx
x f dx x f dx x f 0
0)(2)()()()()()()()( 4）②
72
7)3635(101)1361(101)21(5151)511()511(151)
511(1
6126131231
23=-⋅-=--=-⋅==++=+-----⎰⎰⎰u u du x d x dx x ④
3ln 2)]11(2
1
3[ln 2])1(21)1[ln(2))1(11(21112122112
2
2
020
202022022
02,4
0=-+=-++=--+==+-+=+=+−−−−−→−+⎰⎰⎰⎰⎰⎰
==t t dt t dt t dt t t dt t t tdt t t
dx x x
tdt
dx t x ⑥2121)(ln ln ln
10210
ln 11==−−→−=⎰⎰⎰=t tdt x xd dx x x t
x e e
⑧3
1
31sin sin cos sin 1
3
10
2
2
0sin 2
2
02
==−−
−→−=⎰⎰⎰=t dt t x xd xdx x t
x π
π
⑩
72
)036(2121)(arcsin arcsin 1arcsin
2260
2
6
arcsin 2
1
021
02πππ
π
=-==−−−−→−=-⎰⎰⎰=t tdt x xd dx x x
t
x ○
12的结果。

例书注：第一个等号是根据890.)002
(2)
cos sin (2cos 4420
2
2
sin 220
2p t t t dt t dx x t
x ππ
π
π
=-+=+=−−−→−-⎰⎰=
5)②
1
2ln 21)12(ln )1(2ln 2ln 2ln 2ln 2ln 2ln 0
2
ln 2
ln 0
2ln 0
2
ln 0
2ln 0
-=+-==
--=-=-=
=⎰⎰⎰e e e e
e
dx e xe xde dx xe x x x x
x
④)1(2
1
)sin (cos 21
cos cos 22
2
2
0-=+==⎰⎰π
π
π
π
e e x x xde xdx e x
x x 6）①
.
21)2121
(lim )2
1(lim lim 1
3
21
2
131
3
收敛⎰
⎰⎰
∞++∞
→-+∞→+∞→∞+∴=+-
=-==x
dx
a
x x dx
x dx a a
a a a
②
.
2
1
)11(lim 21)(lim 2121lim lim 0
020
2
2222
2
收敛dx xe
e e dx e dx xe
dx xe
x a a a
x a a x a a
x a x ⎰⎰⎰⎰∞+-+∞→-+∞→-+∞→-+∞→∞+-∴=+-=
=
-===
7）①所围面积：
3
13132)313
2()(10323
1
02
=-=-=-
=⎰x x
dx x x S ②所围图形有两个交点：8,22)4(4222=⇒=-⇒⎪⎩⎪⎨⎧-==x x x x y x
y ；
两个交点：（2，－2）、（8，4）
18
24)464(2
1
)22216(32222324)28(42
13
223
222)]4(2[])2(2[18
24126)24(4)6121()24(82
282
2320
23
8
2
20
4
2324
2
2=+---+⋅=-+-⋅
+⋅
=--+--==+-=++-=-+=⎰⎰⎰--x x
x
dx x x dx x x S y y dy y y S 围面积：
可以用两种方法计算所8）①球的方程是：
3333
32
2
2
222
22222223
4)
(3
1
231)]([)(R R R R x
R R R dx x R V x R x f x A x R y x R z y x R
R
R
R
πππππππππ=+-=---=-=-==⇒=+=++--⎰）（）
（）（轴旋转得到的
绕，它可以看成是由圆②22
]2sin 2
1)([2)2cos 1(2sin πππππ
ππ
ππ
ππ
π
=---=-⋅=
=---⎰⎰x dx x xdx V 第三部分 附加习题
1）求定积分： ①
2
)01()10(212212222221
020121
01100111=-+--=⋅+⎰⋅-=⎰+-=⎰+⎰-=⎰----x x xdx xdx xdx xdx dx x ②
)1(21)1ln (ln )1(21)(ln 21221)(ln ln 221ln 2ln 2
2121
121121122+=-+-=⋅⎰+=⎰+=⎰+=⎰
+e e e x x dx x xd x dx x x xdx dx x x x e e e e e e e
③ ⎰2
5
sin cos π
xdx x ,令sixdx dt x t -==,cos 当x= 0时,t = 1, 当2
π
=
x 时,t = 0
=61
611061
5
015
==⎰=⎰-t
dt t dt t
；
时，，当时，，当，，令341021
12,1
2224
0=====-==+++⎰
t x t x tdt dx t x t x dx x x 原积分化为：
322)]331()9327[(21)33(21)3(21221
31331231
2=+-+=+=⎰+=⎰
+-t t dt t tdt t t
2)设f （x ）在[0，1]上连续并单减，试证：对任何ξ∈（0，1）有
.)()(1
⎰≥⎰dx x f dx x f ξξ
⎰⎰⎰⎰⎰⎰⎰--==
--=-1
1
1
)()()1()()()()()(ξ
ξ
ξ
ξ
ξξ
ξξξξξdx
x f dx x f dx x f dx x f dx x f dx x f dx x f
因为f （x ）在[0，1]上单减，所以 ⎰⋅=≥⎰ξ
ξξξξ0
)()()(f dx f dx x f
)()1()()(1
1
ξξξξ
ξ
f dx f dx x f -=⎰≤⎰
故 0)()1()()1()()(1
=---≥⎰-⎰ξξξξξξξξ
f f dx x f dx x f
即.)()(1
⎰≥⎰dx x f dx x f ξξ
3）设f （x ）∈[0，1]，且,3)(1
=⎰dx x f 求 ⎰2
22sin )(cos πxdx x f 。

⎰
2
2
2sin )(cos πxdx x f =
3)()()(cos )(cos 1
1
2
2
2=⎰=⎰-=⎰-dt t f dt t f x d x f π 。

4）求⎰-1
22dx x x 。

该积分的被积函数非负，积分上限大于下限，该积分值为积分区间
上曲边梯形的面积，由于22)1(12--=-x x x ，积分值应为圆
1)1(2
2
=+-y x 的面积的四分之一，于是积分值是4
π。

5）设非负二阶可导函数f （x ）在[0，+∞]
内,).2()(:,00)(0
a
af dx x f a x f a
≥⎰>>''证明，对常数
设 ],0[),2
()()(0a x x
xf dt t f x F x
∈-⎰=
)]2
()([2)2(22)()2(2)2()()(x
f f x x
f x x f x f x x f x f x F '-'==
'-⋅'−−−−−−−→−'--='ξξ由拉格郎日中值定理
其中),2
(x x ∈ξ
由0
)0()(,)(,0)(),2
()(,)(0)(=≥∴≥''≥''>''F a F x F x F x
f f x f x f 单增即于是即单增，知ξ 因此).2()(0
a
af dx x f a
≥⎰
6)计算：)0(21lim
1
>++++∞
→p n
n p p
p p n Λ
11
)(1lim 21lim
110
1
+=
∑⎰==+++=∞→+∞
→p dx x n k n n n n k p p n p p
p p n Λ 7）计算：⎰-→x
a
a x dt t f a x x )(lim ，f （t ）是连续函数
令⎰=x
a
dt t f x x F )()(，⎰-→x a a x dt t f a x x )(lim =)()
()(lim
a F a x a F x F a x '=--→ 而⎰+='x
a dt t f x xf x F )()()(，)()(a af a F ='
所以，)()(lim a af dt t f a x x x
a
a x =⎰-→ 。

8）设f （t ）连续，。

求)0(,)()(0g dt t f x x g x
''⎰=
)
0(2)0()0()0()(lim )0()(lim
)]()([lim 0
)
0()()(lim 0
)0()(lim )0(0
)0(),()()(0
000
f f f f x f f x
dt t f x f x
dt
t f x g x xf dt t f x g x g g g x xf dt t f x g x x
x x
x x
x x x
=+==
+=+⎰=+⎰==
-'-+⎰=-'-'=''='+⎰='→→→→→
9）设,6
1
2
ln π
=
⎰
-x
t
e dt 求x 。

先计算：
c e u du u udu u e dt
t
u t e u t t +-=⎰+=+⋅⎰⎰−−−−−−−→−-+=-=1arctan 2121211
22)
1ln(,12 所以，
313
)62(211arctan ,
6
2
1arctan 24
21arctan 21
2
ln =-⇒=+=
-∴=
-
-=⨯
--=⎰
-x x x x x t e e e e e dt π
πππ
π
π
由此可得，x =ln4 。

10）求极限： ①
6
1
cos 1
sin lim 6161cos sin lim 3cos ln lim
cos ln lim
00
,cos ln lim
00
2
03
3
0-
=⋅
-=⋅-==⎰⎰+++
+
+
→→→→→x x x x x x x x x
xdx
x xdx
x x x x
x x
x 型不定式
这是② ③
11）若当()x F t
dt
t
dt
x F x x
x
求,11)(,0102
02
⎰
++⎰
+=> 。

0111)1(11)(2
2
2
=+
⋅
-
++=
'x
x
x
x F ，请注意：)]([)(])([)(x a f x a dt t f x a b
'='⎰
所以，F (x )是常数。

2
arctan 212)1()(1
01
02
π
=
=⎰
+==t t
dt
F x F 。

12）设⎪⎪⎩⎪⎪⎨
⎧=≠⎰=0
0)()(20
x c
x x dt t tf x F x
，其中f （x ）是一个在x =0处连续的已知函数，如果)(x F 在x =0处连续，求c 。

由c F f x x xf x dt t tf x F x x
x x ====⎰=→→→)0()0(2
1
2)(lim )(lim
)(lim 02
即c =
)0(2
1
f 。

13)已知)()(10
x nf dt xt f =⎰ ,求f (x ) . 令)()()(,0
1
x nf x
du
u f dt tx f tx u x
=⎰⋅
=⎰= 即)()(0
x nxf du u f x
=⎰,
对等式两边关于x 求导,)()()(x f nx x nf x f '+=
另写成方程:
x
n n x f x f 1
1)()(⋅-=' 两边求积分:⎰-=⎰⇒⎰⎰-='dx x
n n x f x f d dx x n n dx x f x f 1
1)()]([11)()( n n x c x f c x n
n
x f -=⇒+-=1)(ln ln 1)](ln[ .
14)设).1
()(,1ln )(1x f x f dt t
t x f x
+⎰
+=求
⎰⎰+=+=-⎰+−−→
−⎰+==x x
x t u x
dt t
t t du u u u du u u u dt t
t
x f 1122211
11ln ln )1(111
ln
1ln )1(
21
2
1
11121)(ln 2
1
)(ln 21
)(ln ln ln )1(ln )1(ln 1ln )1
()(x t t td dt t
t dt t t t t dt t t t dt t t x f x f x
x
x x x x ==⎰==⎰=⎰++=⎰++⎰+=+
第四部分 英语参考资料
Fundamental theorem of calculus A sound approach to integration defined the integral ⎰b
a dx x f )(
as the limit ,in a certain sense of a sum .That this can be evaluated ,when f is continuous, by finding an antiderivative of f , is the result embodied in the so-called Fundamental Theorem of Calculus. It establishes that integration is the reverse process to differentiation:
Theorem: If f is continuous on [a ,b ] and φis a function such that
)()(x f x ='φ for all x in [a ,b ],then )
()()(a b dx x f b
a
φφ-=⎰
Area under a curve Suppose that the curve y=f (x ) lies above the x -axis, so that 0)(≥x f for all x in [a ,b ]. The area under a curve, that is ,the area of the region bounded by the curve , the x -axis and the lines x=a and x=b ,equals ⎰b
a dx x f )(.
The definition of integral is made precisely in order to achive this result.If
0)(≤x f for all x in [a ,b ],the intagral above is negative. However, it is still the case that its absolute value is equal to the area of the region bounded by the curve, the x -axis and the lines x=a and x=b .
Integral Let f be a function defined on the closed interval [a ,b ]. take points n x x x x Λ,,,210 such that n n x x x x x a <<<<<=-1210Λ=b ,and in each subinterval ],[1+i i x x take a point c i . From the sum
∑--=+10
1))((n i i i i x x c f ; that is ,
f (c 0)))(())(()(1112101---+-+-n n n x x c f x x c f x x Λ. Such a sum is called a Rieman sum for f over . Geometrically, it gives the sum of the area of n rectangles and is an approximation to the area under the curve y=f (x ) between x=a and x=b.
The integral of f over [a ,b ] is defined to be the limit I (in a sense
that needs more clarification than can be giveb here) of such a Rieman sumas n , the number of points, increases and the size of the subintervals gets smaller. The value of I is denoted by ⎰b a
dx x f )(or ⎰b
a
dt
t f )(
Where it is immaterial what letter ,such as x or t ,is used in the integral .The intention is that the value of the integral is equal to what is intuitively understood to be the area under the curve y=f (x ) .Such a limit does not always exist ,but it can be proved that is does if for example ,f is a continuous function on [a ,b ].
If f is continuous on [a ,b ] and F is defined by ⎰=x
a
dt
t f x F )()( ,then
)
()(x f x F ='
for all x in [a ,b ] ,so that F is an antiderivative of f . Moreover, if an antiderivative φ of f is known the integral can be easily evaluated: the Theorem of Caculus gives its value as φ(b )－φ(a ). Of the two integrals
⎰b
a
dx x f )( and ⎰dx x f )( ,the first, with limits, is called a definite
integral and the second ,which denotes an antiderivative of f ,is an indefinite integral.
Intermediate value theorem The following theorem stating an important property of continuous functions: If the real function f is continuous on the closed interval [a ,b ] and η is a real number between f (a ) and f (b ), then for some c in (a ,b ), f (c )= η.
The theorwm is useful for locating roots of equation. For example , suppose that f (x )=x-cosx . Then f is continuous on [0,1],and f (0)<0 and f (1)>0 ,so it follows from the Intermediate value theorem that the equation f (x )=0 has a root in the interval (0,1).
Density The average density of a body is the ratio of its mass to its volume. In general ,the density of a body may not be constant throughout the body. The density at a point P , denoted by )(P ρ, is equal to the limit as
0→∆V of
V
m
∆∆, where V ∆ is the volume of a small region containing P and m ∆ is the mass of the part of the body occupying that small region.
Consider a rod of length l , with density )(x ρ at the point a distance x from one end of the rod. Then the mass m of the rod is given by
⎰=l
dx x m 0
)(ρ .In the same way , the mass of a lamina or a 3-dimensional
rigid body, with density )(r ρ at the point with position venctor r ,is given by ⎰V
rdV )(ρ, with, as appropriate ,a double or triple integral over the
region V occupied by the body .
e The number that is the base o
f natural logarithms .There are several ways of definin
g it .Probably the most satisfactory is this . First ,define ln as in approac
h 2 to the logarithmic function . Then define exp as the inverse function of ln . Then define e as equal to exp 1 .This amounts to saying that
e is the number that makes 11
1=⎰e dt t
.It is necessary to go on to show that
x e and exp x are equal and so are idetial as function s ,and also that ln and
log e are identical functions.
The number e has important properties derived from some of the properties of ln and exp. For example ,n
n h h n
h e )11(lim )1(lim 1
+=+=
∞→→ .Also ,e is the sum of the series ΛΛ+++++!
1
!21!111n .
Another approach ,but not a recommended one ,is to make one of these properties the difinition of e . Then exp x would be defined as x e , ln x would be defined as its inverse function , and the properties of these functions would have to be proved .
The value of e is 2.71828183. The proof that e is irrational is
comparatively easy . In 1873 Hermite proved that e is transcendental , and his proof was subsequently simplified by Hilbert .。