Download Chapter 5 Integration

Chapter 5
Integration
5.1
Definition
Given a function f (x) (called the integrand) defined on a closed interval [a, b], the (defRb
inite) integral a f (x) dx of f over [a, b] is the area of the (x, y) plane bounded by the
graph of f , the x axis and the vertical lines x = a, x = b, where areas above and below
the x axis count as being positive and negative respectively.
Rb
NB a f (x) dx is just a number, not a function of x, since x is just a so-called “dummy”
variable which runs through all real values between
a and b,R and could be represented by
Rb
b
any convenient letter of the alphabet. Thus a f (x) dx = a f (t) dt, since both depend
only on the function f and the real numbers a and b.
5.2
Basic Properties of the Integral
Let a, b, c1 and c2 be any real constants with a < b, then
Z b
Z b
Z b
[c1 f (x) + c2 g(x)] dx = c1
f (x) dx + c2
g(x) dx
a
a
for any “reasonable” functions f and g defined on [a, b].
55
a
(5.1)
Let a, b and c be real numbers satisfying a < b < c, then
Z c
Z c
Z b
f (x) dx +
f (x) dx =
f (x) dx
a
b
(5.2)
a
for any “reasonable” function f on [a, c]. This holds for all values of the real numbers
a, b and c without the above restrictions, provided we define
Z a
Z b
f (x) dx = ¡
f (x) dx when a ¸ b.
(5.3)
a
5.3
5.3.1
b
The Fundamental Theorem of Calculus
Definition
Given a function f (x) (defined on some interval of the real line), a primitive of f (x) is a
function F (x) (on that interval) such that F 0 (x) = f (x) (on that interval).
5.3.2
Lemma
If F1 and F2 are both primitives of f (on some interval), then F2 ¡ F1 = constant (on
that interval).
Proof: Immediate from D I, since
d
(F2 ¡ F1 ) = F20 (x) ¡ F10 (x) = f (x) ¡ f (x) = 0.
dx
5.3.3
Fundamental Theorem
If f (x) is continuous on [a, b], then f (x) possesses at least one primitive (on [a, b]), and
Z b
f (x) dx = F (b) ¡ F (a) ´ [F (x)]ba for any primitive F of f .
(5.4)
a
56
“Proof”
Rt
For a variable t in the interval [a, b], consider the definite integral I(t) = a f (x) dx which
is a function of t. For δt small, it is clear from (5.2) and the picture that
Z t+δt
I(t + δt) ¡ I(t) =
f (x) dx ¼ f (t)δt
t
to a “good approximation”, so that in the limit we have
I(t + δt) ¡ I(t)
= f (t),
δt→0
δt
I 0 (t) = lim
i.e. I is a primitive of f .
Now let F be any primitive of f in [a, b]. Then I ¡ F = constant, by section 5.3.2,
Ra
Rb
I(b) ¡ F (b) = I(a) ¡ F (a). But I(a) = a f (x) dx = 0. Hence a f (x) dx = I(b) =
F (b) ¡ F (a).
QED
This is not a true proof since the “good approximation” is imprecise. A rigorous proof
using precise definitions of continuity and of the Riemann integral will be given in MATH
2011 Real Analysis 2.
The fundamental theorem provides a precise statement of the sense in which integration is the reverse of differentiation.
5.3.4
Definition
R
A primitive of f will henceforth be called an indefinite integral of f and written f (x) dx
(without limits). This is a function of x, but it is determined only to within an arbitrary
additive constant.
Once we have an indefinite integral of f on some interval, we can immediately evaluate
the definite integral of f over any subinterval using (5.4).
5.3.5
Examples
The list in Appendix B of derivatives of elementary functions may be “put into reverse”
to produce a corresponding list of indefinite integrals.
The first formula of Appendix B tells us that, for x > 0
d β
x = βxβ−1 .
dx
Replacing β by α + 1, we see that this is equivalent to the statement that xα+1 is an
indefinite integral of (α + 1)xα . Assuming α 6= ¡1, we may divide through by α + 1, to
deduce that
Z
xα+1
(i)
xα dx =
+ constant, for x > 0, α 6= ¡1
α+1
[valid also for x < 0 if α happens to be an integer (other than -1)].
For α = ¡1, note that
Z
d
−1
for x > 0,
[ln (x)] = x =) x−1 dx = ln x + c,
dx
Z
d
−1
[ln (¡x)] = x =) x−1 dx = ln (¡x) + c,
for x < 0,
dx
57
giving the general result
(ii)
Z
x−1 dx = ln jxj + constant, for x 6= 0.
Similarly,
R x
e dx = ex + constant,
(ex )0 = ex ) (iii)
R
sin x dx = ¡ cos x + constant,
(cos x)0 = ¡ sin x ) (iv)
R
0
cos x dx = sin x + constant.
(sin x) = cos x ) (v)
See Appendix I, for all of these and more!
5.4
Methods of Integration
Functions which appear in Appendix I may be integrated straight off, and linear combinations of such functions may be integrated using the basic property of equation (5.1)
(which has an obvious counterpart for indefinite integrals). In this section we deal with
four methods enabling other functions to be integrated. Three of these methods convert
the original integral into a new (and hopefully easier!) integral, while the fourth provides
a systematic method for integrating rational functions.
5.4.1
Substitution = Change of Variable (cf Chain Rule, D V)
This method exists in two versions, which we shall deal with in turn.
Version 1: spotting directly
Let f be a continuous function with a known indefinite integral F , g a function with
continuous derivative. Recall the chain rule
DV
d
[f fg(x)g]
dx
= f 0 fg(x)g g0 (x) for any (reasonable) functions f and g
Then replacing f by F gives
d
[F fg(x)g] = F 0 fg(x)gg 0 (x) = f fg(x)gg 0 (x),
dx
i.e. F fg(x)g is an indefinite integral of f fg(x)gg 0 (x), i.e.
Z
f fg(x)gg0 (x) dx = F fg(x)g + constant
(5.5)
This gives an indefinite integral very simply, provided you can spot that the integrand is of the form f fg(x)gg 0 (x) where f has a known indefinite integral.
A useful special case
Suppose g(x) = px + q, where p and q are constants and p =
6 0. Then g 0 (x) = p, so
(5.5) reduces to
Z
Z
F (px + q)
+ const (5.6)
f (px + q) p dx = F (px + q) + const, i.e. f (px + q) dx =
p
58
where F is an indefinite integral of f . For example,
Z
Z
dx
epx
= ln jx + qj + const,
epx dx =
+ const,
x+q
p
Z
cos px dx =
sin px
+ const
p
Warning!
Unless g 0 (x) = constant, it cannot be taken outside the integral sign. It is NOT (repeat
NOT!) true that
Z
F fg(x)g
+ const.
f fg(x)g dx =
g 0 (x)
Examples
Z
Z
(i)
2x + 2
1
I =
dx =
g0 (x) dx where g(x) = x2 + 2x + 2
2
x + 2x + 2
g(x)
Z
=
f fg(x)gg0 (x) dx where f (g) = 1/g, F (g) = ln jgj.
So I = ln(x2 + 2x + 2)+const.
Similarly
Z
Z
Z
1 d
sin x
dx = ¡
(cos x) dx = ¡ ln j cos xj + const
tan x dx =
cos x
cos x dx
Z
Z
x dx
1
p
I =
=¡
[g(x)]−1/2 g 0 (x) dx where g(x) = 1 ¡ x2
2
2
Z1 ¡ x
1
f fg(x)gg 0 (x) dx where f (g) = g −1/2 , F (g) = g 1/2 /(1/2) = 2g1/2 ,
= ¡
2
p
p
so I = ¡ 12 2 1 ¡ x2 + const = ¡ 1 ¡ x2 + const.
A brief word about improper integrals
The above result can be used to evaluate a definite integral with particular limits,
e.g.
Z 1
i1
h p
x dx
2
p
=
¡
1
¡
x
= 0 ¡ (¡1) = 1.
0
1 ¡ x2
0
(ii)
59
The integrand tends to +1 as x tends to the upper limit 1 of integration. This is an
example of an improper integral, which has a finite value although it is the area of a
region extending infinitely upwards in the y direction. The integral is strictly speaking
defined as the limit
Z 1−²
x dx
p
lim
.
²→0+ 0
1 ¡ x2
(iii)
Z
−x2
1
dx = ¡
2
Z
xe
eg(x) g 0 (x) dx
Z
1
[exp g(x)] g 0 (x) dx,
= ¡
2
I =
where g(x) = ¡x2
2
so that I = ¡ 12 e−x +const. We can use this to evaluate another type of improper
integral:
¸∞
·
µ ¶
Z ∞
1
1 −x2
1
−x2
= ,
(5.7)
xe dx = ¡ e
=0¡ ¡
2
2
2
0
0
giving the finite value of the area of a region extending infinitely to the right in the x
direction. The integral should properly be evaluated as the limit
÷
µZ X
¸X !
¶
1
2
2
¡ e−x
xe−x dx
= lim
lim
X→+∞
X→+∞
2
0
0
µ
¶
1 −X 2 1
¡ e
+
= lim
X→+∞
2
2
1
.
=
2
2
The 0 in (5.7) is a shorthand for the limit of ¡ 12 e−x as x tends to +1.
Version 2: explicit substitution
In Version 1 we effectively changed from the variable x to a new variable g(x). We now
make this change of variable explicit, but use a different notation. Express x as a function
x(u) of a new variable u, where
either
or
dx
du
dx
du
> 0 (x is an increasing function of u, as in the graph)
< 0 (x is a decreasing function of u)
60
and let x(α) = a, x(β) = b, so that α and β are the values of u corresponding to the
values a and b of x respectively. Because x is either an increasing or decreasing function
1
of u, it can be inverted to express u as a function u(x) of x, with du/dx = (dx/du)
.
Now suppose that F (x) is an indefinite integral of a continuous function f (x). Then,
by D V,
d
dx
F fx(u)g = f fx(u)g
.
du
du
Integrating both sides of this equation with respect to u between u = α and u = β we
obtain
Z β
Z β
d
dx
du =
F fx(u)g du
f fx(u)g
du
α
α du
= [F fx(u)g]βα
= F fx(β)g ¡ F fx(α)g
(5.8)
= F (b) ¡ F (a)
(5.9)
Z b
f (x) dx
(5.10)
=
a
Z b
Z β
dx
du =
f fx(u)g
f (x) dx
(5.11)
du
α
a
The new integral (with respect to u) on the LHS might be easier than the old one (with
respect to x) on the RHS.
Note that the new integrand is obtained from the old one by expressing in terms of
the new variable u and multiplying by the factor dx/du. The latter process may be
thought of as resulting from “transformation of the differential” according to the rule
dx ¡!
dx
du.
du
(5.12)
For an indefinite integral, we simply leave b free to vary, making sure we express both
sides of the equation in terms of the same variable:
Z
Z
dx
du
expressed as a function of x.
(5.13)
f (x) dx = f fx(u)g
du
61
The rules (5.11) and (5.13) for change of variable in definite and indefinite integrals may
be summarized as follows.
² In both cases express the original integrand in terms of the new variable and transform the differential via (5.12).
² In the indefinite case (5.13), reexpress the new indefinite integral (with respect to
u) in terms of x.
² In the definite case (5.11), transform the limits of integration to the corresponding
ones for the new variable.
[In cases where x is a decreasing function of u, the transformed upper limit will be less
than the transformed lower limit, but the convention (5.3) ensures that the answer will
come out right in the end!]
Examples
Z 5 2
p
x ¡ 2x
(i)
p
I=
dx. Take u = x ¡ 1, i.e. substitute x = u2 + 1.
x¡1
1
p
We have dx = 2u du, while for the limits, x = 1 , u = 0, x = 5 , u = 5 ¡ 1 = 2. It
follows by (5.11) that
Z 2 2
Z 2
Z 2
(u + 1)2 ¡ 2(u2 + 1)
4
2
2
I =
(u + 2u + 1 ¡ 2u ¡ 2) du = 2
(u4 ¡ 1) du
2u du = 2
u
0
0
0
·
¸2
¸
· 5
32
44
u
¡u =2
¡2 =
= 8.8.
= 2
5
5
5
0
Note that the change of variable here has converted an improper integral with respect to
x (with integrand infinite at the lower limit) into a proper integral with respect to u.
Z
x2 dx
p
(indefinite)
4 ¡ 9x2
¶
µ
2
2
.
¡ <x<
3
3
I=
p
p
Since 4 ¡ 9x2 = 2 1 ¡ (3x/2)2 , we make a trigonometric substitution using the identity cos2 θ + sin2 θ = 1. Put
(ii)
3x/2 = sin θ with (¡π/2 < θ < π/2), i.e. x =
dx =
2
cos θ dθ,
3
4 ¡ 9x2 = 4 ¡ 4 sin2 θ = 4 cos2 θ
62
2
sin θ
3
p
so that 4 ¡ 9x2 = 2 cos θ (positive square root, since ¡π/2 < θ < π/2).
Thus we get [using the double angle formulae from Appendix C]
µ
¶
Z
Z
Z 4 2 2
sin θ 3 cos θ dθ
4
2
1
2
2
9
=
sin θ dθ =
(1 ¡ cos 2θ) dθ =
θ ¡ sin 2θ + const
I =
2 cos θ
27
27
27
2
2
=
(θ ¡ sin θ cos θ) + const
27
p
¡ ¢
,
cos
θ
=
1 ¡ (3x/2)2 and θ = sin−1 3x
so
and sin θ = 3x
2
2
"
#
r
µ ¶
2
3x
3x
9x
2
sin−1
¡
+ const
1¡
I =
27
2
2
4
µ ¶
x p
3x
2
−1
sin
¡
4 ¡ 9x2 + const
=
27
2
18
Z
x dx
p
I=
(indefinite)
(iii)
4x2 + 12x + 13
First, “complete the square” (cf Appendix H) in the quadratic under the root sign,
giving 4x2 + 12x + 13 = (2x + 3)2 + 4. In terms of the variable 2x + 3, this is similar
to (ii), but with the sign of the squared term reversed. We use the fundamental identity
cosh2 u ¡ sinh2 u = 1 by making the hyperbolic substitution
(2x + 3)2 + 4 = 4(1 + sinh2 u) = 4 cosh2 u
1p
1p 2
=) cosh u =
(2x + 3)2 + 4 =
4x + 12x + 13
2
2
dx = cosh u du,
x = sinh u ¡ 3/2, so
2x + 3 = 2 sinh u
¢
¶
Z ¡
Z µ
sinh u ¡ 32 cosh u du
1
3
1
3
I =
=
sinh u ¡
du = cosh u ¡ u + const
2 cosh u
2
4
2
4
µ
¶
1p 2
3
3
=
4x + 12x + 13 ¡ sinh−1 x +
+ const.
4
4
2
Remark
Integrals involving square roots of quadratics are usually tackled by first completing
the square and then making a suitable trigonometric or hyperbolic substitution. After
completing the square we obtain one of the following three forms:
q
if of form γ 2 ¡ (αx + β)2 then substitute αx + β = γ sin u (c.f. example (ii) above),
q
if of form (αx + β)2 + γ 2 then substitute αx + β = γ sinh u (c.f. example (iii) above),
q
if of form (αx + β)2 ¡ γ 2 then substitute αx + β = γ cosh u.
5.4.2
Integration by Parts (cf Product Rule, D III)
Let u(x) and v(x) be functions with continuous derivatives. Then D III tells us that
d
dv
du
(uv) = u
+v
dx
dx
dx
63
i.e. uv is an indefinite integral of u(dv/dx) + v(du/dx),
¶
Z µ
dv
du
u
+v
dx,
uv =
dx
dx
which rearranges via (5.1) to
Z
dv
u
dx = uv ¡
dx
Z
du
v dx.
dx
(5.14)
Subtracting (5.14) at x = a from (5.14) at x = b gives a corresponding result
Z
b
a
dv
u
dx = [uv]ba ¡
dx
Z
b
a
du
v dx, where [uv]ba = u(b)v(b) ¡ u(a)v(a),
dx
(5.15)
for definite integrals.
² These formulae enable us to tackle integrals of a product of two functions u and
dv/dx.
² The factor dv/dx must be a function of which we know an indefinite integral v.
² The first term on the RHS of (5.14) is obtained from the original integrand by
integrating its second factor.
² The integrand of the new integral on the RHS (carrying a minus sign) is obtained
by integrating the second factor and differentiating the first factor.
Examples
(i) A ‘trick’ example, where the factor dv/dx in the original integrand is taken to be just
the constant number 1.
Z
Z
Z
−1
−1
−1
I =
tan x dx = |tan{z x} |{z}
1 dx = |{z}
x |tan{z x} ¡ (1 + x2 )−1 |{z}
x dx
| {z }
0
u
= x tan−1 x ¡
v
v
u
v
u0
1
ln (1+x2 ) + const
2
where the new integral was evaluated by substitution, making use of the fact that x =
1 d
(1 + x2 ).
2 dx
(ii) An example requiring integration by parts twice.
Z
Z
Z
2
2
2
I = |{z}
x sin
x dx = x (¡ cos x) ¡ 2x(¡ cos x) dx = ¡x cos x + 2 |{z}
x cos
|{z}
| {zx} dx
diff
int
diff
int
½
¾
Z
= ¡x cos x + 2 x sin x ¡ sin x dx = ¡x2 cos x + 2 (x sin x + cos x) + const.
2
64
5.4.3
Reduction Formulae
In the above example, we had to use integration by parts twice. Suppose we had asked
for
Z π
x6 sin x dx.
0
We would have to carry out integration by parts six times to complete this integral!
Instead, look at the general case:
Z
I(n) = xn sin x dx
and integrate by parts twice:
Z
Z
Z
n
n
n−1
n
n−1
I(n) =
x sin
x dx = x (¡ cos x) ¡ nx (¡ cos x) dx = ¡x cos x + n x
|{z}
|{z} cos
|{z}
| {zx} dx
int
int
diff
diff
½
¾
Z
= ¡xn cos x + n xn−1 sin x ¡ (n ¡ 1)xn−2 sin x dx
= ¡xn cos x + nxn−1 sin x ¡ n(n ¡ 1)I(n ¡ 2).
R
This is a reduction formula, which we can use with the easy integral I(0) = sin x dx =
¡ cos x to deduce I(n) for any even n.
For our specific case, we use the limits 0 and π to calculate I(0) = [¡ cos x]π0 = 2 and
to simplify the reduction formula, then apply the formula repeatedly to derive the actual
value of I(6):
I(n) = ¡[xn cos x]πx=0 + n[xn−1 sin x]π0 ¡ n(n ¡ 1)I(n ¡ 2)
I(n) = π n ¡ n(n ¡ 1)I(n ¡ 2).
Z
π
0
I(2) = π 2 ¡ 2I(0) = π 2 ¡ 4
I(4) = π 4 ¡ 12I(2) = π 4 ¡ 12[π2 ¡ 4] = π 4 ¡ 12π 2 + 48
I(6) = π 6 ¡ 30I(4) = π 6 ¡ 30[π4 ¡ 12π 2 + 48]
x6 sin x dx = π 6 ¡ 30π4 + 360π 2 ¡ 1440.
Another Example A list of useful reduction formulae is given in Appendix J; here we
derive another example.
Z
Z
Z
Z
n
n−2
2
n−2
2
tan
x(1 + tan x ¡ 1)dx = tan
x(1 + tan x)dx ¡ tann−2 xdx
tan xdx =
Z
Z
Z
tann−1 x
n−2
2
n−2
=
tan
¡ tann−2 xdx
x sec xdx ¡ tan
xdx =
n¡1
where the integration was carried out using the substitution u = tan x with du =
sec2 x dx. Note that in this case we did not use integration by parts; reduction formulae
can come from any source!
65
5.4.4
Partial Fractions (for Rational Functions only)
This is a purely algebraic technique for expressing a rational function as a sum of simple
terms. Its relevance to calculus is that it provides a means of integrating rational functions. The most general case of partial fractions is dealt with in Appendix K. Here we
shall do three simple examples illustrating the general principles. You will not be asked
to solve any problems more complicated than these.
Example
I=
Z
1
0
x2
3x + 7
dx
+ 5x + 6
Solution
p
The roots of the denominator are (¡5 § 25 ¡ 24)/2 = (¡5 § 1)/2 = ¡2 and ¡3,
so the integrand equals (3x + 7)/(x + 2)(x + 3), which the general theory tells us can be
written in the form
3x + 7
A
B
A(x + 3) + B(x + 2)
=
+
=
(x + 2)(x + 3)
x+2 x+3
(x + 2)(x + 3)
for some constants A and B. So, we require
3x + 7 = A(x + 3) + B(x + 2) for ALL x.
In this case we can find A and B by taking x equal to the two roots of the original
denominator in turn. x = ¡2 gives 1 = A, while x = ¡3 gives ¡2 = ¡B, i.e. B = 2.
Substituting back, we see that
¶
Z 1µ
1
2
+
dx = [ ln jx + 2j + 2 ln jx + 3j ]10
I =
x+2 x+3
0
= (ln 3 + 2 ln 4) ¡ (ln 2 + 2 ln 3) = 3 ln 2 ¡ ln 3 = ln(8/3).
Example
I=
Solution
I=
Z
Z
x2 + 5
dx
x3 ¡ 3x + 2
x2 + 5
dx,
(x ¡ 1)2 (x + 2)
where the factorization of the cubic denominator is easy once it is spotted that 1 is a
root. Since it is in fact a repeated root, the appropriate resolution of the integrand into
partial fractions here is
x2 + 5
C
A
B
+
=
+
,
2
2
(x ¡ 1) (x + 2)
x ¡ 1 (x ¡ 1)
x+2
where A, B and C are constants and thus
x2 + 5 = A(x ¡ 1)(x + 2) + B(x + 2) + C(x ¡ 1)2
66
for ALL x.
Putting x = 1, we see that 6 = 3B, B = 2. Putting x = ¡2, we see that 9 = 9C, C = 1.
To find A, equate coefficients of x2 , to get 1 = A + C = A + 1, A = 0. Substituting back
now gives
¸
Z ·
2
2
1
dx
=
¡
+ ln jx + 2j + const.
+
I=
(x ¡ 1)2 x + 2
x¡1
Example
I=
Z
x2 ¡ 14x ¡ 5
dx .
(x ¡ 4)(x2 + 4x + 13)
Solution
The quadratic factor in the denominator cannot be expressed as the product of two
real linear factors, because its roots
p
p
p
p
¡4 § ¡36
¡4 § 16 ¡ 52
=
= ¡2 § ¡9 = ¡2 § 3 ¡1 = ¡2 § 3i
2
2
are complex numbers. The partial fraction corresponding to this factor therefore has
the original quadratic as denominator and has a numerator which is linear rather than
constant. Thus
x2 ¡ 14x ¡ 5
A
Bx + C
=
+ 2
2
(x ¡ 4)(x + 4x + 13)
x ¡ 4 x + 4x + 13
where A, B and C are real constants, and so
x2 ¡ 14x ¡ 5 = A(x2 + 4x + 13) + (Bx + C)(x ¡ 4)
for ALL x.
Putting x = 4 gives 16 ¡ 56 ¡ 5 = A(16 + 16 + 13), i.e. ¡45 = 45A, so A = ¡1.
Equating coefficients of x2 gives 1 = A + B = ¡1 + B, B = 2.
Equating constant terms (or, equivalently, putting x = 0) gives ¡5 = 13A ¡ 4C =
¡13 ¡ 4C, 4C = ¡8, C = ¡2.
Substituting back, we deduce that
¸
¶
Z ·
Z µ
2x + 4
1
2x ¡ 2
6
¡
dx
+
dx = ¡ ln jx ¡ 4j +
¡
I =
x ¡ 4 x2 + 4x + 13
x2 + 4x + 13 (x + 2)2 + 32
µ
¶
x+2
= ¡ ln jx ¡ 4j + ln(x2 + 4x + 13) ¡ 2 tan−1
+ const,
3
where the first term in the square brackets was integrated by “version 1” substitution
using the fact that 2x + 4 is the derivative of x2 + 4x + 13, while the second term was
integrated by (I.7) of Appendix I (and a trivial change of variable from x to x + 2) alternatively, do it directly via the substitution x + 2 = 3 tan θ.
In general, the coefficients for partial fractions are best determined by a judicious combination of setting x equal to any real roots possessed by the original denominator and
equating coefficients of various powers of x.
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