Download Chapter 6: Techniques of Integration

Chapter 6: Techniques of Integration
Winter 2016
Department of Mathematics
Hong Kong Baptist University
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§6.1 Integration by Parts
In §5.6, we have introduced the method of substitution.
Recall that the method of substitution can be regarded as
inverse to the Chain rule for differentiation.
In this section, we introduce another general method for
anti-differentiation, Integration by Parts. The method for
integration by parts is inverse to the Product rule for
differentiation.
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Suppose that U(x) and V (x) are two differentiable functions.
According to the Product rule,
d
dV
dU
(U(x)V (x)) = U(x)
+ V (x)
.
dx
dx
dx
Or equivalently,
U(x)
dV
d
dU
=
(U(x)V (x)) − V (x)
.
dx
dx
dx
Integrating both sides of the above equation, we obtain
Z
U(x)
dV
dx = U(x)V (x) −
dx
Z
V (x)
dU
dx.
dx
This leads to the so-called Integration by Parts,
Z
Z
UdV = UV − VdU.
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To apply the method, we first break up the given integrand
into a product of two pieces, U and V 0 , then by Integration by
Parts,
Z
Z
UV 0 dx = UV −
where
R
VU 0 dx,
VU 0 dx is usually a simpler integral than
R
UV 0 dx.
An improper choice of U and V 0 can result in making the
integral more difficult rather than easier.
As a common practice, we always look for a factor of the
integrand that is easily integrated, and include dx with that
factor to make up V 0 dx = dV . The remaining factor of the
integrand is then assigned to be U.
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Z
Example 1: Find the integral
xe x dx.
Approach 1:
Let U = e x and xdx = dV . Then dU = e x dx and V = x 2 /2. By
Integration by Parts,
Z
Z
x2
x
xe dx =
e x d( )
2
Z 2
2
x
x x
=
e −
d(e x )
2
2
Z 2
x2 x
x x
=
e −
e dx.
2
2
Z
Note that the new integral to be evaluated,
VdU, is becoming
more complicated.
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Approach 2:
Let U = x and e x dx = dV . Then dU = dx and V = e x . Using
Integration by Parts,
Z
Z
x
xe dx =
x · d(e x )
Z
x
= xe − e x dx
= xe x − e x + C .
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The following are two useful rules of thumb for choosing U and V :
(i) If the integrand involves a polynomial multiplied by an
exponential function, a sine, a cosine, or some other readily
integrable function, try U equals the polynomial an dV equals
the rest.
(ii) If the integrand involves a logarithm, an inverse trigonometric
function, or some other function that is not readily integrable
but whose derivative is readily calculated, try that function for
U and let dV equal the rest.
Of course, these “rules” come with no guarantee.
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Z
Example 2: Find the integral
x · cos(x)dx.
Solution:
Following the rule of thumb, we let U = x and cos(x)dx = dV .
Then dU = dx and V = sin(x). Using Integration by Parts,
Z
Z
x · cos(x)dx =
x · d(sin(x))
Z
= x · sin(x) − sin(x)dx
= x · sin(x) + cos(x) + C .
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Z
Example 3: Find the integral
x 2 ln(x)dx.
Solution:
Following the rule of thumb, we let U = ln(x) and x 2 dx = dV .
1
Then dU = dx and V = x 3 /3. Using Integration by Parts,
x
Z
Z
x3
2
x ln(x)dx =
ln(x)d( )
3
Z
1 3
1 3
=
x ln(x) −
x d(ln(x))
3
3
Z
1 3
1 2
=
x ln(x) −
x dx
3
3
1 3
1
=
x ln(x) − x 3 + C .
3
9
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Z
Example 4: Find the integral I =
e x sin(x)dx.
Solution: Using Integration by Parts,
Z
I
=
e x sin(x)dx
Z
e x d(− cos(x))
Z
x
= −e cos(x) + e x cos(x)dx
Z
= −e x cos(x) + e x d(sin(x))
Z
= −e x cos(x) + e x sin(x) − e x sin(x)dx
=
= −e x cos(x) + e x sin(x) − I .
1
This leads to I = (e x sin(x) − e x cos(x)) + C .
2
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Definite Integral:
If we want to evaluate a definite integral by the method of
integration by parts, we must remember to include the appropriate
evaluation symbol with the integrated term.
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Z
Example 5: Find the integral
2
xe −x dx.
0
Solution: Using Integration by Parts, we have
Z 2
Z 2
−x
xe dx = −
xd(e −x )
0
0
2 Z 2
= −xe −x +
e −x dx
0
0
2
−2
= −(2e − 0) − e −x 0
= 1 − 3e
−2
.
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Z
Example 6: Find the integral
e
x 3 ln(x)dx.
1
Solution: Using Integration by Parts, we have
Z e
Z e
x4
3
x ln(x)dx =
ln(x)d( )
4
1
1
e Z e x 4
x4
=
ln(x) −
d(ln(x))
4
1
1 4
Z
e4 1 e 3
=
−
x dx
4
4 1
e 4 x 4 e
=
− 4
16 1
4
1
3e
+ .
=
16
16
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§6.3 Inverse Substitutions
We saw in §5.6 that the method of substitution can be used
to transform
Z
Z
f (g (x))g 0 (x) dx
into
f (u) du
by letting u = g (x).
In
however, the more “complicated” integral
R some cases,
f (g (x))g 0 (x) dx is actually simpler
to evaluate. Thus, it is
R
sometimes useful to evaluate f (u) du by making the inverse
substitution u = g (x).
Here, we will only consider trigonometric
substitutions
to
√
√
2
2
2
2
deal with integrands of the type a − x and a + x .
(The other substitutions in the textbook are optional.)
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Z
Example 7: Evaluate
√
dx
.
4 − x2
Solution: Letting x = 2 sin θ,
Z
Z
dx
√
=
4 − x2
Z
=
Z
=
we have dx = 2 cos θdθ, so that
2 cos θ
q
dθ
4(1 − sin2 θ)
2 cos θ
dθ
2 cos θ
1 dθ
= θ + C = arcsin
x 2
+ C.
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Z
Example 8: Evaluate
1/2 p
1 − 4x 2 dx.
1/4
Solution: Letting x =
Z
1/2 p
1
2
sin θ, we have dx =
1 − 4x 2 dx =
1/4
Z
1
2
π/2 p
1 − sin2 θ ·
π/6
1
=
2
=
1
4
Z
cos θdθ, so that
1
cos θ dθ
2
π/2
cos2 (θ) dθ
π/6
Z π/2
(1 + cos(2θ)) dθ
π/6
√
π/2
1
1
π
3
=
θ + sin(2θ)
=
−
.
4
2
12
16
π/6
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√
Integrals involving a2 − x 2 can often be simplified by letting
x = a sin(θ), or equivalently, θ = arcsin(x/a). We can rewrite
any other trigonometric functions that appear by noting that
cos(θ) =
1p 2
a − x 2.
a
√
1
Integrals involving a2 + x 2 or a2 +x
2 can often be simplified
by letting x = a tan(θ), or equivalently, θ = arctan(x/a). We
can rewrite any other trigonometric
functions that appear by
√
1
2
2
noting that sec(θ) = a a + x , so that
tan(θ)
x
=√
,
2
sec(θ)
a + x2
a
1
cos(θ) =
=√
.
2
sec(θ)
a + x2
sin(θ) =
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Z
Example 9: Evaluate
x +1
dx.
x2 + 9
Solution: Note that
Z
Z
Z
x +1
x
dx
dx
=
dx
+
.
2
2
2
x +9
x +9
x +9
For the first integral, we let u = x 2 + 9, du = 2x dx. For the
second integral, we let x = 3 tan(θ), dx = 3 sec2 (θ) dθ:
Z
Z
Z
x +1
1
du
3 sec2 (θ)
dx
=
+
dθ
x2 + 9
2
u
9(1 + tan2 (θ))
1
1
= ln |u| + θ + C
2
3
x 1
1
= ln(x 2 + 9) + arctan
+ C.
2
3
3
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§6.5 Improper Integrals
Rb
For the integral I = a f (x)dx considered previously, we have
assumed that f is continuous on the closed, finite interval
[a, b]. Since such a function is necessarily bounded, the
integral I is also necessarily a finite number. Such integrals
are called proper integrals.
In this section, we generalize the definite integral to allow for
two possibilities excluded in the situation described above:
(i) We may have a = −∞ or b = ∞ or both (referred to as
improper integrals of type I);
(ii) f may be unbounded as x approaches a or b or both
(referred to as improper integrals of type II).
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Example 10:
Find the area of the region A lying under the curve y = 1/x 2 and
above the x-axis to the right of x = 1.
Solution: The area can be expressed as
Z ∞
1
A=
dx,
2
x
1
which is an improper integral of type I. To evaluate this improper
integral, we interpret it as a limit of proper integrals over intervals
[1, R] as R → ∞. Then
Z
A=
1
∞
1
dx
x2
Z
=
=
R
lim
R→∞ 1
lim (−
R→∞
1
1 R
dx = lim (− )
R→∞
x2
x 1
1
+ 1) = 1.
R
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Example 11:
Find the area of the region under y = 1/x, above y = 0, and to
the right of x = 1.
Solution: The area is given by the improper integral of type I,
A =
=
∞
Z R
1
1
dx = lim
dx
R→∞ 1 x
x
1
R
lim ln(x) = lim ln(R) = ∞.
Z
R→∞
1
R→∞
Note that this improper integral diverges to infinity. Comparing to
y = 1/x 2 , the “spike” of y = 1/x for x > 1 is much thicker. Such
a difference makes the region has infinite area.
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Definition (Improper integrals of type I)
If f is continuous on [a, ∞), we define the improper integral of f
over [a, ∞) as a limit of proper integrals:
∞
Z
Z
R
f (x)dx = lim
f (x)dx.
R→∞ a
a
Similarly, if f is continuous on (−∞, b], we define
Z
b
Z
f (x)dx =
−∞
lim
b
R→−∞ R
f (x)dx.
In either case, if the limit exists and is a finite number, we say that
the improper integral converges; if the limit does not exist, we say
that the improper integral diverges. If the limit is ∞ (or −∞), we
say that the improper integral diverges to infinity (or diverges to
negative infinity).
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Z
∞
Example 12: Evaluate the integral
e −|x| dx.
−∞
Solution: Note that this is an improper integral of type I at both
endpoints. We usually break it into two separate integrals for this
kind of integrals.
Z ∞
Z 0
Z ∞
−|x|
−|x|
e
dx =
e
dx +
e −|x| dx
−∞
−∞
0
Z
R
e −x dx
R
= 2 lim (−e −x )
= 2 lim
R→∞ 0
R→∞
= 2 lim (1 − e
R→∞
0
−R
)
= 2.
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Z
∞
cos(x)dx.
Example 13: Evaluate the integral
0
Solution: For this improper integral, we have
Z
∞
Z
cos(x)dx
=
0
=
lim
R→∞ 0
R
cos(x)dx
lim sin(R).
R→∞
Note that the limit does not exist (and it is not ∞ or −∞). So all
we can say is that the given integral diverges.
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Definition (Improper Integrals of Type II)
If f is continuous on the interval (a, b] and is possibly unbounded
near a, we define the improper integral
Z
b
Z
f (x)dx = lim
a
c→a+ c
b
f (x)dx.
Similarly, if f is continuous on [a, b) and is possibly unbounded
near b, we define
Z b
Z c
f (x)dx = lim
f (x)dx.
a
c→b− a
These improper integrals may converge, diverge, diverge to infinity,
or diverge to negative infinity.
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Example 14:
√
Find the area of the region S lying under y = 1/ x, above the
x-axis, between x = 0 and x = 1.
Solution: The area A is given by
Z 1
1
√ dx,
A=
x
0
which is an improper integral of type II since y is unbounded near
x = 0. By definition, we have
Z 1
1
A = lim
x −1/2 dx = lim 2x 1/2 c→0+ c
c→0+
c
√
= lim (2 − 2 c) = 2.
c→0+
This shows that the integral converges and S has a finite area of 2
square units.
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Z
Example 15: Evaluate the integral
f (x) =
√
1/ x
x −1
2
f (x)dx, where
0
if 0 < x ≤ 1
if 1 < x ≤ 2.
Solution: Note that f (x) is a piecewise continuous function as in
§5.4. We have
Z 2
Z 1
Z 2
1
√ dx +
f (x)dx =
(x − 1)dx
x
0
0
1
Z 1
2
1
1
√ dx + ( x 2 − x)
= lim
c→0+ c
2
1
x
1
= 2 + (2 − 2 − + 1)
2
5
=
.
2
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