In this section we extend our notion of integral to include integrals for which one or both of these conditions break down.. Such an integral is called an improper integral.. What motiva
Trang 1EXAMPLE 29.19 Find x−2x−12(x2
+1)dx
SOLUTION The integrand can be broken into a sum as follows,
−2x − 1
x2(x2+ 1)=
A
x +xB2+Cx + Dx2
+ 1
A repeated factor contributes multiple fractions, as shown.3
We can solve for A, B, C, and D One method is to clear denominators and evaluate both sides at four different x-values This leaves us with four equations and four unknowns
We can show that
−2x − 1
x2(x2+ 1)=
−2
x +−1
x2 +2x + 1
x2+ 1.
−2x − 1
x2(x2+ 1)dx = −2
1
x dx −
1
x2dx +
2x + 1
x2+ 1 dx
= −2 ln |x| −x
−1
−1 +
2x
x2+ 1 dx +
1
x2+ 1 dx
= −2 ln |x| +1
x + ln(x2+ 1) + arctan x + C
General Procedure for Partial Fraction Decomposition of P (x)Q(x)
If Q(x)P (x) is improper, do long division to express it as the sum of a polynomial and a proper rational function
Factor Q(x) into the product of linear and irreducible quadratic factors
(x − a) is a factor of Q(x) if Q(a) = 0 Rewrite the proper rational function as a sum of simpler rational functions as follows For every nonrepeated linear factor (x − a) of Q there is a termx−aA
For a repeated linear factor (x − a)nof Q there are n partial fractions All numerators are constant; denominators consist of (x − a) raised to successively higher powers E.g., for (x − a)3a factor of Q there are fractions x−aA +(x−a)B 2 +(x−a)C 3
For a nonrepeated irreducible quadratic factor of Q, (x2+ bx + c), there’s a term Ax+B
x 2 +bx+c
If the quadratic factor of Q is repeated, there are multiple partial fractions, each with a linear expression in the numerator E.g., (x2+ 4)2as a factor of Q would result in the partial fractionsAx+Bx2 +4 +(xCx+D2 +4) 2
Finding numerical values for the constants is an algebra problem Integrating the partial fractions coming from linear factors of Q is straightforward Some quadratics and repeated quadratics are straightforward while others require both completing the square and trigono-metric substitution
3 The denominators are x and x 2 , not x and x.
Trang 2P R O B L E M S F O R S E C T I O N 2 9 3
In Problems 1 through 6, write out the partial fraction decomposition of each rational function You need not determine the coefficients; just set them up.
1 (a) x(x−1)(x+5)x2+3 (b) x3x+x
2 (a) x3 −4x3 (b) x3 +2x4
3 (a) x3x3
+2x 2 +1
4 (a) x+5
x 2 −4
5 3
x 3 +4x
6 (x2 +x+1)(x−1)x2+1
In Problems 7 through 17, evaluate the integrals.
7. x23x+9
−6x+5 dx
8. 2 x(x−1) 2 dx
9.012x+3x2 dx
10. 2
x 4 −1 dx
11. (x23x−1)(x−2)2+3 dx
12. x(x−3)1 dx
13. e2x (e x +2)(e x −1) 2 dx
Hint: Use substitution Do not attempt partial fractions until the integral is in the form
P (x) Q(x) where P (x) and Q(x) are polynomials.
14. x−2arctan x dx
15. x2arctan x dx
16. √x3 1−x 2 dx
17. ln(x2
− 1) dx
18. 2x3−2x2+4x+8 (x−2) 2 (x 2 +3) dx
19. x2 +x−6x3 dx
Trang 329.4 IMPROPER INTEGRALS
Mathematicians make a habit of testing and pushing the boundaries of their theorems and definitions In this section we’ll push the boundaries of the definition of the definite integral The definition of the definite integralabf (x) dxrequires that
the interval of integration, [a, b], is finite and the integrand, f (x), is defined and bounded everywhere on [a, b]
In this section we extend our notion of integral to include integrals for which one or both of
these conditions break down Such an integral is called an improper integral The integral
∞ 1 1
x 2 dx is improper because the interval of integration is unbounded; 01x−1/2dx is improper because the integrand is unbounded at x = 0
What motivates us to try to make sense of improper integrals? A perfectly good answer
is “to see if we can.” But in fact, physicists and statisticians work with such integrals on
a regular basis For example, probabilities are computed by calculating the area under a distribution p(x), called a probability density function The probability of a randomly chosen data point lying to the right of the marker “a” on the distribution is given by
∞
a p(x) dx, where p(x) is a nonnegative function
p(x)
x a
Figure 29.3
The probability of lying somewhere must be 1, (100 %), so if p(x) is a probability density
function, then−∞∞ p(x) dx = 1
Physicists use improper integrals in their models with great regularity, moving particles
in “from infinity” in particle physics, and computing integrals such as01
1+x 1−xdxwhen studying aerodynamics The integral 01
1+x 1−xdx is improper because the integrand is unbounded, “blowing up” at x = 1 We’ll make sense of improper integrals by taking up the types of improprieties, infinite interval and unbounded integrand, one at a time Our method
of dealing with an impropriety at an endpoint of integration is to excise it, constructing
an ordinary integral with an endpoint that we allow to approach the trouble spot If the appropriate limit exists and is finite, we define the improper integral to be that limit If the impropriety does not occur at an endpoint, then we split up the integral so that it does We’ll begin with a few examples
Infinite Interval of Integration
EXAMPLE 29.20 Compute1∞ 1
x 2 dx, if possible
SOLUTION 1∞x12 dxcorresponds graphically to the area underx12and is bounded on the left by x = 1
We approach this problem by looking at the proper integral1b 1
x 2 dx, and letting b increase without bound As b increases the accumulated area increases; the question is whether or
Trang 4not the accumulated area increases without bound.
b
1
1
x2dx =−1x
b 1
= −1b + 1
lim b→∞
b
1
1
x2dx = lim
b→∞
−1b + 1
= 1
So the area of the shaded region approaches 1 as b increases without bound We say
∞ 1 1
x 2 dxis convergent, and converges to 1.
x y
b
1
y = 1
x2
Figure 29.4
EXAMPLE 29.21 Compute1∞ 1xdx, if possible
SOLUTION Again, we integrate from 1 to b (we compute a proper integral) and take the limit as b → ∞
b
1
1
x dx = ln x
b 1
= ln b − ln 1 = ln b
lim b→∞
b
1
1
x dx = lim b→∞ln b = ∞
We say1∞ 1x dxdiverges.
x y
b
1
y = 1x
Figure 29.5
REMARKIt is not at all clear from looking at the graphs of x12 and1xthat1∞ x12 dxshould converge to 1 and1∞ 1xdxshould be unbounded In fact, it might surprise you After all, as
x → ∞ bothx1and 1
x 2approach 0 The crucial difference is that 1
x 2approaches 0 much faster What should not surprise you is that 0 <1∞ 1
x 3 dx ≤ 1, because 0 < x13 ≤ x12 for x ≥ 1 More generally, if f is continuous and f (x) ≥ 0 for all x ≥ a, then for b > aabf (x) dx increases with b Therefore, limb→∞abf (x) dxis either finite or increases without bound
In other words, if limb→∞abf (x) dxis bounded, thena∞f (x) dxconverges, provided
f is nonnegative on [a, ∞)
Trang 5Similarly, we can argue that because
0 <1
x ≤√1
x for x ≥ 1,
0 <
b
1
1
x dx ≤
b
1
1
√
x dxfor b ≥ 1
As b increases without bound1b1xdxincreases without bound, so1b√1
xdxincreases without bound as well
This type of comparison was not useful when dealing with 1x and x12 0 <x12 ≤1x for
x ≥ 1, but being larger than a convergent integral is an inconclusive characteristic, as is being smaller than a divergent integral
1
x2
1
x
1
√x
1
√x
1
x2
1
x3
1
x3
x y
y
1
0 < ≤ for x ≥ 1 1
x
0 < ≤ for x ≥ 1
Figure 29.6
D e f i n i t i o n
Considera∞f (x) dx, whereabf (x) dx exists for all b ≥ a.a∞f (x) dx = limb→∞abf (x) dxprovided this limit exists and is finite If the limit is finite, thena∞f (x) dxconverges Otherwise the integral diverges.
a
−∞f (x) dx = limb→−∞
a
b f (x) dxprovided this limit exists and is finite, and
a
b f (x) dxexists for all b ≤ a
Ifa∞f (x) dxand−∞a f (x) dxboth converge, then
∞
−∞f (x) dx =
a
−∞f (x) dx +
∞ a
f (x) dx
If one or both ofa∞f (x) dx and−∞a f (x) dx diverges, then −∞∞ f (x) dx diverges
EXAMPLE 29.22 Evaluate−∞∞ 1
1+x 2 dxif this integral is convergent
SOLUTION Split this into two improper integrals x = 0 makes a convenient breaking point
Trang 6x
y = 1
1 + x2
Figure 29.7
∞
−∞
1 1+x 2 dx =−∞0 1+x12 dx +0∞ 1+x12 dx, provided both these integrals converge
∞ 0
1
1 + x2dx = lim
b→∞
b
0
1
1 + x2 dx
= lim b→∞arctan x
b 0
= lim b→∞arctan b − arctan 0
= lim b→∞arctan b
=π 2
y
x
y = tan x
x = – π
2 x =
π 2
Figure 29.8
1 1+x 2 is an even function, so−∞0 1
1+x 2 dx =π2, and−∞∞ 1
1+x 2 dx =π2+π2 = π
So far the divergent improper integrals we’ve looked at diverge by growing without bound The next example illustrates another way of diverging
EXAMPLE 29.23 Determine whether0∞sin x dx converges or diverges
SOLUTION
∞
0 sin x dx = lim
b→∞
b
0 sin x dx
= lim b→∞− cos x
b 0
= lim b→∞− cos b + 1
Trang 7This limit doesn’t exist The value of − cos b + 1 oscillates between 0 and 1 This improper integral diverges by oscillation
EXERCISE 29.10 Show that1∞e−xdx =1e
The solution is embedded in the solution to Example 29.28.
Observation
If limx→∞ 0∞f (x) dxdiverges
If limx→∞f (x) = 0, then work must be done in order to determine whether or not
∞
a f (x) dxdiverges or converges
Unbounded and Discontinuous Integrands
We begin with an example
EXAMPLE 29.24 Find01√1
x
b 1
y = 1
√x
Figure 29.9
SOLUTION This integral is improper because √1
x blows up at x = 0 To get around this problem we compute
1
b
1
√
x dx for b > 0 and take the limit as b → 0+
lim b→0 +
1
b
x−1 dx = lim
b→0 +2x1
1 b
= lim b→0 +2 · 1 − 2√b
EXAMPLE 29.25 Show that01x1dxdiverges
SOLUTION
1
0
1
x dx ⇒ lim
b→0 +
1
b
1
x dx = lim b→0 +ln |x|
1 b
= lim b→0 +0 − ln |b| = −(−∞)
y
x
B A
1 1
y = 1x
Figure 29.10
Trang 8One could also argue that the integral diverges by the symmetry of the shaded regions in Figure 29.10 Area A is infinite, so area B is infinite But01x1dx = 1+ (the area of B) Consequently01x1dxis divergent
D e f i n i t i o n
If f is continuous on (c, d] and discontinuous at c, then
d
c f (x) dx = lim
b→c +
d
b
f (x) dx provided this limit exists and is finite
If f is continuous on [c, d) and discontinuous at d, then
d
c f (x) dx = lim
b→d −
b
c
f (x) dx provided this limit exists and is finite
If f has a discontinuity at p, c < p < d, then
d
c f (x) dx =
p
c f (x) dx +
d
p
f (x) dx provided both integrals are convergent
It can be shown that
i.1∞ x1p dxconverges for p > 1 and diverges for p ≤ 1,
ii.01x1pdxconverges for p < 1 and diverges for p ≥ 1
This is left as an exercise
EXERCISE 29.11 Prove statement (i) above
EXERCISE 29.12 Prove statement (ii) above.4
How to Approach an Improper Integral: An Informal Summary
1 Determine all improprieties: Infinite interval, discontinuous integrand, unbounded
integrand
2 If necessary, split up the integral into the sum of integrals so that
(a) each integral has only one impropriety, (b) the impropriety is exposed; it occurs at the endpoint of an integral
3 Compute each improper integral as the limit of a proper integral Replace the endpoint
at which the impropriety occurs and compute the appropriate one-sided limit
4 The original integral converges only if each of the summands converges independently.
4 Both exercises are also included as problems at the end of the chapter.
Trang 9EXAMPLE 29.26 Compute−∞∞ x13 dx, if possible.
SOLUTION This integral is improper at x = 0 and at the endpoints of integration Split the interval up:
(−∞, −1], [−1, 0], [0, 1], [1, ∞)
–1
X X X
1 0
∞
−∞
1
x3 dx =
−1
−∞
1
x3dx +
0
−1
1
x3dx +
1
0
1
x3dx +
∞ 1
1
x3dx
We need to compute four different limits We know1∞ x13 dxwill converge, either having completed Exercise 29.11 or because
0 < 1
x3≤ 1
x2 for x ≥ 1 and
∞ 1
1
x2dxconverges So
∞ 1
1
x3 is increasing and bounded
g
x
y = 1
x3
Figure 29.11
(By symmetry, we know−∞−1 x13 dxis convergent, converging to −1∞
1
x 3 dx.) If you’ve completed Exercise 29.12, you know01 1
x 3 dxdiverges We show this below
1
0
1
x3dx = lim
b→0 +
1
b
1
x3dx
= lim b→0 +
x−2
−2
1 b
= lim b→0 +−1
2 + 2 2b2
= ∞ ⇒ The integral
1
0
1
x3dxdiverges
If any one of the summands diverges the original integral is divergent, so we’re done.
Trang 10
EXAMPLE 29.27 Compute03x−21 dx, if possible.
SOLUTION The integrand blows up at x = 2, and 2 ∈ [0, 3] Therefore, we must split the interval into
[0, 2] and [2, 3]
3
0
1
x − 2dx =
2
0
1
x − 2 dx +
3
2
1
x − 2dx You can compute either of these summands independently Both diverge, so the original integral diverges as well
Alternatively, you can notice that
3
2
1
x − 2dx =
1
0
1
udu and this latter integral diverges (This was shown in Example 29.25.)
y
x = 2
y = 1u
y
x u
y = 1
x –2
3
Figure 29.12
CAUTIONIf you aren’t careful you might not notice that03x−21 dxis an improper integral Failure to notice this makes your work fatally flawed, as you would erroneously write
3
0
1
x − 2 dx = ln |x − 2|
3 0
= ln |1| − ln | − 2| = − ln 2
EXAMPLE 29.28 Determine whether0∞e−x2dxconverges or diverges
SOLUTION We have an interesting dilemma here because we can’t find an antiderivative for e−x2
Therefore, we’ll argue by comparison e−x2 approaches zero very quickly, so we expect the integral to converge Therefore, we’ll show that it is less than some convergent integral For x ≥ 1,
0 < ex≤ ex2 so 1
ex ≥ 1
ex 2
∞ 0
e−x2dx =
1
0
e−x2dx +
∞ 1
e−x2dx