第6.7节 反常积分
标签:
it |
6.7
What is the area of the region under the curve y=1/______from x= 0 to x=1 (Figure 6.7.1(a)) ? The function 1/_____is not continuous at x= 0, and in fact 1/_____is infinite for infinitesimal ε>0. Thus our notion of a definite integral does not apply. Nevertheless we shall be able to assign an area to the region using improper integrals. We see from the figure that the region extends infinitely far up in the vertical direction. However, it becomes so thin that the area of the region turns out to be finite.
The region of Figure 6.7.1(b) under the curve y=x -3from x=1 to x=∞


Figure 6.7.1
extends infinitely far in the horizontal direction. We shall see that this region, too, has a finite area which is given by an improper integral.
Improper integrals are defined as follows.
DEFINITION
|
|
If the limit exists the improper integral is said to converge. Otherwise the improper integral is said to diverge.
The improper integral can also be described in terms of definite integrals with hyperreal endpoints. We first recall that the definite integral
|
|
is a real function of two variables uand v. If u and v vary over the hyperreal numbers instead of the real numbers, the definite integral ____f(x)dx stands for the natural extension of Devaluated at (u,v),
|
|
Here is description of the improper integral using definite integrals with hyperreal endpoints.
Let f be continuous on (a, b].
(1) ___
(2) ______ f(x) dx = ∝ (or -∝ ) if and only if ___ f(x) d(x) is positive infinite (or negative infinite) for all
EXAMPLE 1
|
|
|
|
EXAMPLE 2
|
|
This time
|
|
The improper integral diverges. Since the limit goes to infinity we may write
The region under the curve in Figure 6.7.2 is said to have infinite area.
Warning: we
remind the reader once again that the symbols
∞and -∞are not
real or even hyperreal numbers. We use them only to indicate the
behavior of a limit, or to indicate an interval without an upper or
lower endpoint.
6.7.2
EXAMPLE 3
is undefined at x=0. Thus the length formula gives an improper integral,
|
|

Figure 6.7.3
Let u=9x2/3+ 4, du = 6x-1/3dx. The indefinite integral is
|
|
Therefore
|
|
Notice that we use the same symbol for both the definite and the improper integral. The theorem below justifies this practice.
THEOREM 1
If f is continuous on the closed interval [a,b] then the improper integral of f from a to b converges an equals the definite integral of f from a to b.
PROOF We have shown in Section 4.2 on the Fundamental Theorem that the function
|
|
We now define a second kind of improper integral where the interval is infinite.
DEFINITION
|
|
Here is description of this kind of improper integral using definite integrals with hyperreal endpoints.
EXAMPLE 4
|
|
For
Thus
|
|
So the improper integral converges and the region has area___. The region is shown in Figure 6.7.1(b) and extends infinitely far to the right.
EXAMPLE 5
|
|
The region is shown in Figure 6.7.4 and has infinite area.

Figure 6.7.4
EXAMPLE 6
|
|

Figure 6.7.5
The last two examples give an unexpected result. A region with infinite area is rotated about the x-axis and generates a solid with finite volume! In terms of hyperreal numbers, the area of the region under the curve y= x-2/3from 1 to an infinite hyperreal number His equal to 3( H1/3-1), which is positive infinite. But the volume of the solid of revolution from 1 to His equal to
Which is finite and has standard part 3π.
We can give a simpler example of this phenomenon. Let Hbe a positive infinite hyperinteger, and form a cylinder of radius 1/Hand length H²(Figure 6.7.6). Then the cylinder is formed by rotating a rectangle of length H², width 1/H, and infinite area H²/H=H. But the volume of the cylinder is equal to π,

Imagine a cylinder made out of modelling clay, with initial length and radius one. The volume is π. The clay is carefully stretched so that the cylinder gets longer and thinner. The volume stays the same, but the area of the cross section keeps getting bigger. When the length becomes infinite, the cylinder of clay still has finite volume V=π, but the area of the cross section has become infinite.
There are other types of improper integrals. If f is continuous on the half - open interval [a,b] then we define
If f is continuous on (-∝,b] we define
We have introduced four types of improper integrals corresponding to the four types of half - open intervals
[a,b),
By piecing together improper integrals of these four types we can assign an improper integral to most functions which arise in calculus.
DEFINITION
We can introduce the improper integral ____ f(x)dx whenever f is piecewise continuous on I and a,b are either the endpoints of I or the appropriate infinity symbol. A few examples will show how this can be done.
Let f be continuous at every point of the closed interval[a,b] except at one point c where a We define
EXAMPLE 7
Then
|
|
Similarly,
|
|
So
|
|
and the improper integral converges. Thus, the region shown in Figure6.7.7 has finite area.

Figure 6.7.7
If f is continuous on the open interval (a, b), the improper integral is defined as the sum
|
|
where c is any point in the interval (a,b). The endpoints a and b may be finite or infinite. It does not matter which point c is chosen, because if e is any other point in (a,b), then
|
|
EXAMPLE
|
|
The function 2/______ +1/____ is continuous on the open interval (0, 2)but discontinuous at both endpoints (Figure 6.7.8). Thus
|
|

Figure 6.7.8
First we find the indefinite integral.
|
|
Then
|
|
Also
|
|
Therefore
|
|
EXAMPLE 9
|
|
The function 1/x² + 1/(x-1)² is continuous on the open interval (0,1) but discontinuous at both endpoints. The indefinite integral is
|
|
We have
|
|
Similarly we find that
|
|
In this situation we may write
|
|
and we say thttp://www.google.com/hat the region under the curve in Figure 6.7.9 has infinite area.

Figure 6.7.9
Remark In Example 9
We are faced with a sum of two infinite limits. Using the rules for adding infinite hyperreal
numbers as a guide we can give rules for sums of infinite limits.
H-K can be either finite, positive infinite, or negative infinite.
By analogy, we
use the following rules for sums of two infinite limits or of a
finite and an infinite limit. These rules tell us when such a sum
can be considered to be positive or negative infinite. We use the
infinity symbols as a convenient shorthand, keeping in mind that
they are not even hyperreal numbers.
EXAMPLE 10 Find _____
xdx. We see that
|
|

Figure 6.7.10
It is tempting to argue that the positive area to the right of the origin and the negative area to the left exactly cancel each other out so that the improper integral is zero. But this leads to a paradox.
Wrong: _____ xdx =0. Let v= x+2, dv= dx. Then
|
|
|
|
Subtracting
But
So we do not give the integral_____ xdx the value 0, and instead leave it undefined.
PROBLEMS
In Problems 1-36, test the improper integral for convergence and evaluate when possible.
1_________x
-2dx
3________x
-1/2dx
5______(2x-
1)-3dx
7______x2+
2x-1dx
9__________
x(1+x2)-2 dx
11________x -1/2+ x-2dx ??
13 _______(x-1) -2/3
dx
15 _____x -2/3
dx
17____2x(x2-1)
-1/3
dx
19_____(2x-1) -2/3
dx
21______x²dx
23_________dx
25
________ x3
dx
27________
dx
29_________dx
31____
(x-1)-1/2 + (3-x)
-1/2
dx
33__________
35_________f(x)dx
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
EXTRA
1
2
3
4
In Problems 5-8, the region bounded by the given curves is rotated about (a) the x-axis, (b) the y-axis. Find the volumes of the two solids of revolution.
5
6
7
8
9
10 The region under the curve y=(x²+4 )1/3, 0 ≤ x ≤ 2, is rotated about the y-axis. Find the volume of the solid of
11 Find the length of the
curve
12 Find the length of the
curve
13 Find the length of the
curve
14 Find the length of the
curve
15Find the length of the
line
16 Find the area of the surface generated by rotating the curve y=3x²-2, 0 ≤ x ≤1, about the y-axis.
17 Find the area of the surface generated by rotating the curve x=At²+Bt, y=2At +B,0 ≤ t ≤ 1, about the x-axis.
18 Find the average value of f(x)= x/ _______ , 0≤ x ≤4.
19Find the average value of
f(x)= xp,
20 Find the average distance from the origin
of a point on the parabola y=x²,
21 Given that f(x) = xp, 0≤ x ≤ 1, p a positive constant, find a point cbetween 0 and 1 such that f(c)equals the average value of f(x)
22 Find the center of mass of a wire on the x-axis, 0≤ x ≤ 2, whose density at a point xis equal to the square of the distance from (x,0) to (0,1).
23 Find the center of mass of a length of wire with constant density bent into three line segments covering the top, left, and right edges of the square with vertices (0,0), (0,1), (1,1), (1,0).
24 Find the center of mass of a plane object bounded by the lines y=0,y=x,x=1, with density p(x)=1/x.
25 Find the center of mass of a plane object bounded by the curves x=y², x=1, with density p(x)=y².
26 Find the centroid of the triangle bounded by the x-and y-axes and the line ax+by = c, where a, b, and c are
27 A spring exerts a force of 10x1bs when stretched a distance x beyond its natural length of 2ft. Find the work required to stretch the spring from a length of 3 ft to 4ft.
In Problems 28-36, test the improper integral for convergence and evaluate if it converges.
28
30
32
34
36
37
|
|
38
39
shape of
the semicircle y=________,
□40 An object fills the solid generated by rotating the region under the curve y=f(x), a ≤x ≤b, about the x-axis.
Its density per unit volume is p(x). Justify the following formula for the mass of the object.
□41 A container filled with water has the shape of a solid of revolution formed by rotating the curve x=g(y),
|
|
http://www.google.com/

加载中…