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Chapter 7
Integration
7.1
Indefinite and Definite Integration
Goals
By completing this section, students should be able to
• recall the definition of indefinite integral and apply properties/formulas of indefinite integrals;
• recall the definition of definite integral and apply properties/formulas of definite integrals;
• recall and apply the fundamental theorems of calculus.
7.1.1
Definite Integral
An object is moving along a striaght line (say the y–axis). Its velocity is recorded and shown as the graph y = v(x)
below.
Can we find the distance traveled within a time interval [a, b]?
Suppose that we divide the time interval [a, b] into equal parts of length d > 0. For the purpose of illustration, let’s
partition [a, b] into five equal parts so that d = (b − a)/5 and we obtain the following sub–intervals:
[a, a + d], [a + d, a + 2d], ⋯[a + 3d, a + 4d], [a + 4d, b].
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CHAPTER 7. INTEGRATION
Over the interval [a, a + d], we take the (approximated) velocity to be v(a) so that the distance traveled is approximately equal to v(a) ⋅ d.
We then move to the next sub–interval [a + d, a + 2d], and we take the (approximated) velocity to be v(a + d)
so that the distance traveled over [a + d, a + 2d] is approximately equal to v(a + d) ⋅ d. Further, the accumulated
distance traveled is (v(a) + v(a + d))d.
Similarly, if we take v(a + 2d) to be the approximated velocity over [a + 2d, a + 3d], then the distance traveled is
approximately v(a + 2d)d and the accumulated distance traveled is (v(a) + v(a + d) + v(a + 2d))d.
Repeat the same process over the remaining sub–intervals [a + 3d, a + 4d] and [a + 4d, b], we found that the total
distance tranveled is approximately equal to
[v(a) + v(a + d) + v(a + 2d) + v(a + 3d) + v(a + 4d)] d.
If the number of partition n is large enough, the distance traveled can be regarded as the area of the region bounded
by the graph y = v(x) and the horizontal axis.
In general, if the graph of a bounded function f is given and if R is the region enclosed by the graph of f , the
horizontal axis, and vertical lines x = a and x = b (say a < b), then one can approximate the signed area of R by
the same procedure.
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MATH1013/LN/2019 Sem 1
7.1. INDEFINITE AND DEFINITE INTEGRATION
For the purpose of illustration, let’s take a = 0, b = 2 and the number of uniform sub–intervals is N = 8. Then the
length of each sub–interval is ∆t = (2 − 0)/8 = 1/4.
Over the sub–interval [i∆t, (i+1)∆t], take f (i∆t) to be the approximated value of f so that the area of the graph
of f over this interval is f (i∆t)∆t.
Consequently, the area of the region is approximately given by
7
∑ f (a + i∆t) ⋅ ∆t.
i=0
Again, one may increase the number of partitions (N ) (equivalently decrease the length of the sub–intervals):
∆t = 1/8
∆t = 1/12
∆t → 0
When ∆t → 0, the sum seems to have a limit which can be regarded as the area of region bounded by the curve
y = f (t) over the interval [a, b] ( = [0, 2] in this example).
Remark 7.1.1. In the previous approximation, we take the functional value at the left endpoint as an approximation
of the the functional values over a sub–interval [i∆t, (i + 1)∆t]. However, why not we use the right endpoint, or
the midpoint, or any arbitrary point from the interval?
Right–end
Midpoint
Arbitrary
If f is nice, we may expect that the area should be independent from such a choice.
Definition 7.1.2. Let f be a bounded function on an interval [a, b], i.e., there is M > 0 such that
∣f (x)∣ ≤ M
for all x ∈ [a, b].
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CHAPTER 7. INTEGRATION
We say that a number L is the definite integral of f over [a, b] if
L = lim [
n→∞
b − a n−1
⋅ ∑ f (si )]
n
i=0
b
for any choice of si in the interval [ti , ti+1 ]. We shall denote the limit L by ∫
f (t) dt. In addition, f is said to
a
be (Riemann) integrable over [a, b] if this limit exists.
b
a
By convention, we define ∫
f (x) dx ∶= − ∫
b
f (x) dx.
a
Theorem 7.1.3. If f is continuous over [a, b], then it is integrable.
Theorem 7.1.4 (Properties of Definite Integral). Let f and g be continuous.
a
1. ∫
f (x) dx = 0.
a
b
2. ∫
b
(kf (x)) dx = k ∫
a
f (x) dx where k is a real.
a
b
3. ∫
b
[f (x) ± g(x)] dx = ∫
a
b
4. ∫
b
g(x) dx.
a
c
f (x) dx = ∫
a
a
b
f (x) dx ± ∫
f (x) dx + ∫
c
f (x) dx.
a
b
5. If f ≥ g over [a, b], then ∫
b
6. ∣∫
a
a
b
f (x) dx ≥ ∫
f (x) dx.
a
b
f (x) dx∣ ≤ ∫
∣f (x)∣ dx.
a
Theorem 7.1.5 (Further Properties of Definite Integrals). Let f be continuous on the intervals below.
1. If f is even (f (−x) = f (x)) on [−a, a], then
a
∫
−a
a
f (x) dx = 2 ∫
f (x) dx.
0
2. If f is odd (f (−x) = −f (x)) on [−a, a], then
a
∫
f (x) dx = 0.
−a
3. If f is periodic with period T , then for any a ∈ R,
a+T
∫
Example 7.1.6. Let f (x) =
a
T
f (x) dx = ∫
f (x) dx.
0
3
3
1
. Then f is even and hence ∫ f (x) dx = 2 ∫ f (x) dx.
2
1+x
−3
0
Example 7.1.7. Let f (x) = x2 sin x. Then f is odd and hence ∫
3
−3
f (x) dx = 0.
3π/2
√
Example 7.1.8. Let f (x) =
1 + sin x cos x. Then f is periodic with period 2πand hence ∫
2π
∫
f (x) dx.
0
78
−π/2
f (x) dx =
MATH1013/LN/2019 Sem 1
7.1.2
7.1. INDEFINITE AND DEFINITE INTEGRATION
Indefinite Integral
Definition 7.1.9 (Antiderivatives). A function F (x) is an antiderivative of f (x) if
F ′ (x) = f (x).
The process of finding antiderivatives is called indefinite integration.
Theorem 7.1.10 (Antiderivative).
If F and G are differentiable functions
on (a, b) and F ′ (x) = G′ (x) on (a, b), the F (x) = G(x) + k for some constant k. Consequently, if F and G are
antiderivatives of f , then they are different by some constant k.
Proof. Let H(x) = F (x) − G(x) so that H ′ (x) = 0 over (a, b). Fix c ∈ (a, b). By the mean value theorem, for
any x ∈ (a, b), there is ξ lying between x and c such that
(F (x) − G(x)) − (F (c) − G(c)) = H(x) − H(c) = H ′ (ξ)(x − c) = 0.
It follows that F (x) − G(x) is identically equal to F (c) − G(c).
Definition 7.1.11 (Indefinite Integral). The family of all antiderivatives of a function f is called the indefinite
integral of f , and it is denoted by ∫ f (x) dx.
If F (x) is an antiderivative of f (x), then we may write also
∫ f (x) dx = F (x) + C
where C is known as the integration of constant.
For instance, an antiderivative of 3x2 is x3 . Therefore, we write
2
3
∫ 3x dx = x + C.
Theorem 7.1.12 (Algebraic Properties of Indefinite Integrals).
1. ∫ (f (x)±g(x)) dx = ∫ f (x) dx±∫ g(x) dx.
2. ∫ (kf (x)) dx = k ∫ f (x) dx, k is a constant.
Theorem 7.1.13 (Integration Formulas for Elementary Functions).
−1
∫ x dx = ln ∣x∣ + C.
1. ∫
xa dx =
xa+1
+ C, a ≠ −1; and
a+1
2. ∫ ex dx = ex + C.
3. ∫ cos x dx = sin x + C.
4. ∫ sin x dx = − cos x + C.
5. ∫ sec2 x dx = tan x + C.
Example 7.1.14. Evaluate the following indefinite integrals
1. ∫ 4 dx.
Ans. 4x + C
2. ∫ (2x5 − 3x1/2 + x−e ) dx.
Ans. (1/3)x6 − 2x3/2 +
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x1−e
1−e
+C
CHAPTER 7. INTEGRATION
3. ∫ (x2 − 2)(x + 3) dx.
Ans. (1/4)x4 + x3 − x2 − 6x + C
4. ∫ (x2 + sin x) dx
5. ∫ (
Ans.
3
− sec2 x) dx
x
x3
3
− cos x + C
Ans. 3 ln ∣x∣ − tan x + C
Example 7.1.15. Suppose that the graph of an unknown function F passes through (2, 5). Find F if the slope of
the graph is given by 2x.
Solution. Given F ′ (x) = 2x so that F is an anti-derivative of 2x.
Since
2
∫ (2x) dx = x + C,
F (x) = x2 + k for some constant k.
In order to find k, we impose the condition F (2) = 5 to get k = 1. In conclusion, the curve is y = x2 + 1.
7.1.3
Fundamental Theorem of Calculus
Let f be a continuous function. In general, it is hard to compute definite integral of f over an interval [a, b] by
definition. However, it turns out that it is pretty easy once we can find an anti-derivative F of f although find
anti-derivative is itself not an easy problem.
The arguments below are optional. You can safely jump to Theorem 7.1.16.
Let c ∈ [a, b] be arbitrarily fixed, and define
x
G(x) = ∫
f (t) dt.
c
This function is well defined as the definite integral of f over the interval with endpoints c and x exists.
Let’s assume that x > c. As f is continuous, f has absolute maximum M (x) and absolute minimum m(x) over
[c, x]. By properties of definite integrals,
m(x) ≤ f (t) ≤ M (x)
x
m(x)(x − c) = ∫
c
for all t ∈ [c, x]
x
m(x) dt ≤ ∫
m(x) ≤
c
x
f (t) dt ≤ ∫
M (x) dt = M (x)(x − c)
c
G(x) − G(c)
≤ M (x)
x−c
As limx→c+ m(x) = limx→c+ M (x) = f (c),
lim+
G(x) − G(c)
= f (c).
x−c
lim−
G(x) − G(c)
= f (c).
x−c
x→c
Similarly, one can prove that
x→c
Now, let M be the absolute maximum of ∣f ∣ over [a, b]. For x > a slightly,
x
∣G(x) − G(a)∣ = ∣∫
a
x
f (t) dt∣ ≤ ∫
a
It follows that limx→a G(x) = G(a).
Similarly, one can prove that limx→b G(x) = G(b).
80
x
∣f (t)∣ dt ≤ ∫
M dt = M (x − a).
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MATH1013/LN/2019 Sem 1
7.1. INDEFINITE AND DEFINITE INTEGRATION
Theorem 7.1.16 (Fundamental Theorem of Calculus (I)). Let f be continuous on [a, b] and c ∈ [a, b]. Then the
function
t
G(t) ∶= ∫
f (s) ds,
a ≤ t ≤ b,
(7.1)
c
is continuous over [a, b] and is differentiable over (a, b) and
G′ = f.
In other words, every continuous function over an interval has an antiderivative which is given by (7.1).
Example 7.1.17. Since f (x) =
1
1+x2
is continuous on R,
x
1
d
1
(∫
dt) =
2
dx 0 1 + t
1 + x2
by the fundamental theorem of calculus (I).
Example 7.1.18. Let c be a real number and f be a continuous function f such that
x
∫
f (t) dt = cos x −
c
1
2
for all x ∈ R.
Find f .
Solution. As f is continuous, the fundamental theorem of calculus (I) asserts that
x
d
d
(∫ f (t) dt) =
(cos x − 1/2) =
dx c
dx
=
.
x
Remark 7.1.19. Let f be continuous on an interval I. The function G(x) = ∫
of c. Suppose that c1 , c2 ∈ I be distinct. Then
x
∫
c1
f (t) dt depends on the choice
c
x
f (t) dt − ∫
.
f (t) dt =
c2
Example 7.1.20. Suppose that f ∶ I → R and φ ∶ J → I be continuous, where I and J are intervals. Suppose that
x
G(x) = ∫
f (t) dt for some c ∈ I. Then
c
.
(G ○ φ)(s) =
If φ is differentiable, then by the chain rule,
(G ○ φ)′ (s) =
.
Theorem 7.1.21 (Fundamental Theorem of Calculus (II)). Let f be continuous on [a, b]. If F is continuous on
[a, b] and it is an antiderivative of f over (a, b), then
b
∫
f (t) dt = F (b) − F (a).
a
b
We sometimes use [F (t)]a to denote the difference on the RHS.
Proof. Let G be defined in (7.1). As both F and G are antiderivative of f , there is a constant k such that G(x) =
F (x) + k.
As a result,
b
∫
a
b
f (t) dt = ∫
c
a
f (t) dt − ∫
f (t) dt = G(b) − G(a)
c
= (F (b) + k) − (F (a) + k) = F (b) − F (a)
as expected.
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CHAPTER 7. INTEGRATION
If f and g are continuous and f (x) ≥ g(x) over the interval [a, b], then the (absolute) area bounded by y = f (x)
and y = g(x) over [a, b] is given by
b
∫
(f (x) − g(x)) dx.
a
Example 7.1.22. Find the area of the region enclosed by the curves y = 4 − x2 and y = 2 − x.
Solution. Solving 4 − x2 = 2 − x, these curves intersect when x = −1 or x = 2.
Further, as 4 − x2 ≥ 2 − x over [−1, 2], the area is
2
∫
−1
9
[(4 − x2 ) − (2 − x)] dx = .
2
Example 7.1.23. Find the area bounded by the curves y = f (x) = x2 − x and y = g(x) = 2x for −2 ≤ x ≤ 3.
Solution. Solving f (x) = g(x) over [−2, 3], there is a single solution x = 0. Check that g(x) ≥ f (x) over [0, 3]
while f (x) ≥ g(x) over [−2, 0].
It follows that the area of the region bounded by these curves over [−2, 3] is given by
0
∫
= ∫
−2
0
−2
((x2 − x) − (2x)) dx + ∫
3x − x2 dx + ∫
3
3
((2x) − (x2 − x)) dx
0
3x − x2 dx = ⋯ =
0
82
79
.
6
MATH1013/LN/2019 Sem 1
7.2
7.2. TECHNIQUES OF INTEGRATION
Techniques of Integration
Goals
By completing this section, students should be able to
• understand and apply the method of substitution;
• understand and apply the method of integration by parts.
• find partial fraction decomposition of rational functions;
• evaluate definite and indefinite integral of rational functions;
• understand and evaluate improper integrals.
7.2.1
Method of Substitution
Recall that the chain rule states that
d
[F (g(x))] = F ′ (g(x)) ⋅ g ′ (x)
dx
when F and g are differentiable. Therefore,
′
′
∫ F (g(x)) ⋅ g (x) dx = F (g(x)) + C.
If f (u) = F ′ (u), then
′
∫ f (g(x)) ⋅ g (x) dx = F (g(x)) + C.
(7.2)
Integration by Substitution
Let f be continuous.
Procedure.
1. Select a substitution that appears to simplify the integrand. In particular, try to select u so that
du is a factor in the integrand.
2. Express the integrand entirely in terms of u and du, completely eliminating the original variable
x and dx.
3. Evaluate the new integral (which should be easy by the choice of the substitution).
4. Express the antiderivative found in Step 3 in terms of original variable.
Example 7.2.1. Evaluate the following integrals
1. ∫ (x2 + 2x + 5)5 (2x + 2) dx.
2
2. ∫ ex
+2x+5
Ans.
0
x2
+2x+5
+C
2
Ans. sin(x3 /3 + x) + C
dx
.
4x + 7
5
5. ∫
+C
Ans. ex
(2x + 2) dx.
3. ∫ (x2 + 1) cos(x3 /3 + x) dx.
4. ∫
(x2 +2x+5)6
6
Ans.
x
dx.
+ 10
1
4
ln ∣4x + 7∣ + C
5
Ans. [ 12 ln(x2 + 10)]0 =
83
1
2
ln 35
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CHAPTER 7. INTEGRATION
In the previous discussion, we always manage to represent the integrand as a product
F ′ (g(x)) ⋅ g ′ (x)
for some suitable functions F and g.
However, sometimes it’s more convenient to set x = φ(θ) for some suitable function φ. To be more precise, we
have
Theorem 7.2.2. Let f be continuous and φ be continuously differentiable. Suppose that φ has inverse function
and φ′ is never zero. If
F ′ (θ) = f (φ(θ)) ⋅ φ′ (θ),
then
−1
∫ f (x) dx = F (φ (x)) + C.
The proof is included for the sake of completeness but you can safely skip it.
Proof. Since φ has inverse function and φ′ is never zero, the inverse function theorem asserts that if x = φ(θ),
then
′
(φ−1 ) (x) = 1/φ′ (θ).
By the chain rule,
d
′
F (φ−1 (x)) = F ′ (φ−1 (x)) ⋅ ((φ−1 ) ) (x)
dx
= F ′ (θ)/φ′ (θ) = f (φ(θ)) = f (x).
Example 7.2.3. Show that for a ≠ 0,
∫
dx
1
x
= arctan + C.
x2 + a2 a
a
Solution. The motivation for the following substitution is the identity
1 + tan2 θ = sec2 θ and the formula (tan θ)′ = sec2 θ.
Let x = a tan θ. Then
dx =
dθ.
Using the previously stated identity, we get
dx
=∫
x2 + a 2
√
Example 7.2.5. Evaluate ∫
1 − x2 dx.
∫
+ a2
dθ =
1
1
1
x
∫ dθ = θ + C = arctan + C.
a
a
a
a
Solution. The motivation for the following substitution is the identity
1 − sin2 θ = cos2 θ and the formula (sin θ)′ = cos θ.
Let x = sin θ. Then dx =
dθ, and thus
√
√
1 − x2 dx = ∫
1−
∫
⋅
dθ = ∫ cos2 θ dθ.
To evaluate the last integral, we use the double–angle formula:
1
1
1
(1 +
) dθ = [θ + sin(2θ)] + C.
2
2
2
It remains to express the last expression in terms of x. Clearly, θ =
. Meanwhile,
√
1
sin(2θ) = sin θ ⋅ cos θ = x ⋅ 1 − x2
2
2
∫ cos θ dθ = ∫
because
.
84
(7.3)
MATH1013/LN/2019 Sem 1
7.2.2
7.2. TECHNIQUES OF INTEGRATION
Integration by Parts
Recall that the product rule states that if u and v are differentiable functions in x, then
dv
du
d
uv = u
+v .
dx
dx
dx
Integrate to yield the integration–by–parts formula
′
′
∫ uv dx = uv − ∫ vu dx.
(7.4)
If we use the differential notation du = u′ dx and dv = v ′ dx, (7.4) can be written as
∫ u dv = uv − ∫ v du.
b
If we evaluate definite integral ∫
(uv) dx instead, then we get
a
b
∫
a
uv ′ dx = u(b)v(b) − u(a)v(a) − ∫
b
vu′ dx.
(7.5)
a
Technical Notes
1. This formula allows one to express the integral ∫
which may be easier to find.
u dv in terms of another integral ∫
2. To apply this formula to an integral ∫ f dx, we must write
f dx = u dv
for some suitable parts u and v.
Example 7.2.10. Evaluate ∫ xex dx and ∫
2
xex dx.
1
Solution. Note that
xex dx = x dex .
Apply the integration–by–parts formula (7.4),
x
∫ xe dx
= xex − ∫ ex dx
= xex − ex + C = ex (x − 1) + C.
To find the definite integral, apply fundamental theorem of calculus,
2
∫
1
2
xex dx = [ex (x − 1)]0 = e2 .
2e
Example 7.2.12. Evaluate ∫ ln y dy and ∫
ln y dy.
e
Solution. We may simply take u = ln y and v = y. Apply the integration–by–parts formula (7.4),
∫ ln y dy = y ln y − ∫ y d(ln y)
= y ln y − ∫ y ⋅
1
dy = y ln y − ∫ dy = y ln y − y + C.
y
To find the definite integral, apply fundamental theorem of calculus,
2e
∫
e
2e
y ln y dy = [y ln y − y]e = e ln 4.
85
v du
CHAPTER 7. INTEGRATION
Example 7.2.13. Evaluate ∫ x2 ex dx.
Solution. Note that x2 ex dx = x2 d(ex ). Apply the integration–by–parts formula (7.4),
2 x
2 x
x
∫ x e dx = x e − 2 ∫ xe dx.
Apply the formula again to evaluate the integral on the RHS, we have
x
x
∫ xe dx = e (x − 1) + C.
Finally, ∫ x2 ex dx = ex (x2 − 2x + 2) + C.
Example 7.2.14. Let a and b be nonzero real numbers. Evaluate
Ic = ∫ eax cos bx dx
and Is = ∫ eax sin bx dx.
Solution. Integrate-by-parts to get
Ic
sin bx
)=
b
1 ax
a
e sin(bx) − ∫
b
b
= ∫ eax d (
=
1 ax
1
e sin(bx) − ∫ sin(bx) d (eax )
b
b
1
a
ax
e sin(bx) dx = eax sin(bx) − Is .
b
b
Similarly, one gets (check it)
a
eax
cos bx + Ic .
b
b
Solving this system of equations for Ic and Is , we have
Is = −
Ic =
eax
eax
(a
cos
bx
+
b
sin
bx)
and
I
=
(−b cos bx + a sin bx) .
s
a2 + b2
a2 + b2
Note that these formulas are valid even a or b (but not both) is zero.
Reduction Formula
We have evaluated ∫ xex dx in Example 7.2.10 and then evaluated ∫ x2 ex dx in Example 7.2.13. In fact, the
formula for the second integral is deduced from the first one.
If we define a sequence of integrals by
In = ∫ xn ex dx,
we may guess that the formula of In can be deduced from In−1 and may be as well as the predecessors In−2 , In−3 , . . .
In general, suppose that we have a sequence of integrals
In = ∫ fn (x) dx.
If each term In can be defined in terms of its preceding terms, then we say that there is a recurrence relation among
the terms of the sequence. Evaluating In ’s by establishing such a recurrence relation is called the technique of
integration by reduction formula.
Example 7.2.15. If Jn = ∫ cosn x dx where n ≥ 2, show that
Jn =
1
n−1
cosn−1 x ⋅ sin x +
Jn−2 .
n
n
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MATH1013/LN/2019 Sem 1
7.2. TECHNIQUES OF INTEGRATION
Solution. As n ≥ 2, we write
Jn = ∫ cosn x dx = ∫ cosn−1 x cos x dx = ∫ cosn−1 x d(sin x).
Integration–by–parts to get
Jn = cosn−1 x ⋅ sin x + (n − 1) ∫ cosn−2 x ⋅ sin2 x dx.
As sin2 x = 1 − cos2 x, a little algebra leads us to the desired reduction formula.
π/2
Remark 7.2.16. Let In = ∫
cosn x dx for n ≥ 2. With the help of Example 7.2.15, it is readily seen that
0
In =
n−1
In−2 .
n
It follows that
2k − 1 2k − 3 1 π
2k
2k − 2 2
⋅
⋯ ⋅ and I2k+1 =
⋅
⋅ .
2k
2k − 2 2 2
2k + 1 2k − 1 3
For the sake of completeness, we include the following
dx
, where n > 1 and a > 0. Then
Example 7.2.18. Let In = ∫
(x2 + a2 )n
I2k =
In
a2 In
In
=
1
∫
a2
(x2 + a2 ) − x2
1
dx = 2 (In−1 − ∫
(x2 + a2 )n
a
(x2
x2
dx)
+ a2 )n
1
x
dx
( 2
−∫
)
2
n−1
2
2(n − 1) (x + a )
(x + a2 )n−1
2n − 3
1
x
In−1 + 2
⋅
.
2a2 (n − 1)
2a (n − 1) (x2 + a2 )n−1
= In−1 +
=
(7.6)
Recall that I1 is given in (7.3).
7.2.3
Integration of Rational Functions
We shall introduce the integration of rational functions in this subsection. Recall that a rational function is any
function that can be written in the form
n(x)
f (x) =
d(x)
where n(x) and d(x) are polynomials. Domain of f (x) is the set of all the reals x such that d(x) ≠ 0.
Example 7.2.19. Using the method of substitution,
⎧
1
1
⎪
⎪
⋅
+ C,
if n ∈ N ∖ {1},
dx
⎪ −
n
−
1
(x
+
a)n−1
=
⎨
(7.7)
n
∫
⎪
(x + a)
⎪
⎪
if n = 1.
⎩ ln ∣x + a∣ + C,
Example 7.2.20. Using the method of substitution,
1
⎧
⎪
ln (x2 + a2 ) + C
⎪
⎪
⎪
x
⎪ 2
n dx = ⎨
∫
1
1
⎪
(x2 + a2 )
⎪
⎪
⋅
⎪
n−1
⎪
2
2(1
−
n)
⎩
(x + a2 )
if n = 1,
(7.8)
if n ∈ N ∖ {1}.
Indeed, if u = x2 + a2 , then du = 2x dx whereby
∫
xdx
1
n =
∫
2
(x2 + a2 )
du
.
un
Recall also the equation (7.3) that
∫
and the general ∫
dx
1
x
= arctan + C
2
+a
a
a
x2
dx
can be obtained by the reduction formula (7.6).
(x2 + a2 )n
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CHAPTER 7. INTEGRATION
Partial Fraction Decomposition
We say that a rational function n(x)/d(x) is proper if deg n < deg d; otherwise it is improper. An improper
rational function can always be expressed as the sum of a polynomial and a proper rational function as
n(x)
f (x)
= d(x)q(x) + r(x),
d(x)q(x) + r(x)
r(x)
=
= q(x) +
d(x)
d(x)
where r(x) is the remainder so that deg r < deg d. As a result,
∫ f (x) dx = ∫ p(x) dx + ∫
Example 7.2.21. Evaluate ∫
r(x)
dx.
d(x)
4x − 2
dx.
2x + 3
Solution. Note that the integrand is an improper rational function and
(2x + 3) +
4x − 2
=
2x + 3
2x + 3
It follows that
∫
= +
2x + 3
4x − 2
dx =
2x + 3
.
+ C.
In view of our previous observation, it suffices to study the integral of proper rational functions.
Recall that the fundamental Theorem of Algebra asserts that every polynomial over C has at least one (complex)
zero. Consequently, every degree n complex polynomial is the product of n degree one complex polynomial.
We say that a real polynomial (i.e., polynomial with real coefficients) is irreducible if it cannot be factorized as a
product of two real polynomials with lower degrees.
√
√
For instance, x2 + 1 is irreducible – although it can be factorized as (x − −1)(x + −1), these ‘factors’ are not
real polynomial.
Corollary 7.2.22. Every real polynomial is the product of degree one real polynomials and degree two irreducible
real polynomials.
Its proof is optional.
Proof. Let p(x) = ∑nk=0 ak xk be a real polynomial and p(α) = 0 for some α ∈ C ∖ R. Take complex conjugation
to get
n
n
k=0
k=0
k
∑ ak (α) = ∑ ak αk = 0
whereby α is also a zero. As a result,
(x − α)(x − α) = x2 − 2R(α) + ∣α∣2
is an irreducible degree two factor of p. As zeroes of p are either real or appear in complex conjugate pair, we are
done.
For instance, 2x3 − 3x2 − 3x + 2 = (x − 2)(x + 1)(2x − 1), x3 − x2 − x − 1 = (x − 1)2 (x + 1) and 5x3 − 6x2 + 5x − 6 =
(5x − 6)(x2 + 1).
Accordingly, the denominator polynomial of n(x)/d(x) may be decomposed into
d(x) = (a1 x + b1 )m1 (ax2 + b2 )m2 ⋯(axk + bk )mk (A1 x2 + B1 x + C1 )n1 ⋯(Al x2 + Bl x + Cl )nl
88
MATH1013/LN/2019 Sem 1
7.2. TECHNIQUES OF INTEGRATION
where m1 , m2 , . . . , mk , n1 , . . . , nl are natural numbers; and the quadratic polynomials are irreducible. We shall
investigate firstly the situation when there is no repeated factors. It can be proved that
l
β j x + γj
αi
n(x) k
=∑
+∑
d(x) i=1 ai x + bi j=1 Aj x2 + Bj x + Cj
for some real numbers αi ’s, βj ’s, and γj ’s.
Denominator Polynomial has Simple Factors Only
Procedure – Integration by Partial Fractions I (Simple Factors)
Suppose that
n(x)
is proper and d(x) has distinct factors:
d(x)
d(x) = (a1 x + b1 )⋯(axk + bk )(A1 x2 + B1 x + C1 )⋯(Al x2 + Bl x + Cl ).
1. To each linear factor ax + b of d(x), introduce a term
α
in the partial fraction decomposition.
ax + b
βx + γ
2. To each irreducible factor ax2 + bx + c of d(x), introduce a term
in the partial fraction
2
ax + bx + c
decomposition.
3. Solve the undetermined coefficients.
Example 7.2.23. Suppose a ≠ 0. Evaluate ∫
1
dx.
x2 − a2
Solution. Note that x2 − a2 = (x − a)(x + a). Suppose that
1
A
B
(A + B)x + a(A − B)
=
+
=
.
x2 − a 2 x − a x + a
x2 − a2
Comparing the coefficients in the numerator, we have A = −B =
dx
1
=
(∫
x2 − a2 2a
∫
Example 7.2.24. Evaluate I = ∫
dx
−∫
x−a
1
.
2a
As a result,
dx
1
x−a
)=
ln ∣
∣ + C.
x+a
2a
x+a
9
dx.
2x3 − 3x2 − 3x + 2
Solution. Factorize the denominator to get 2x3 − 3x2 − 3x + 2 = (x − 2)(x + 1)(2x − 1). Suppose that
=
A
B
C
9
=
+
+
2x3 − 3x2 − 3x + 2 x − 2 x + 1 2x − 1
A(x + 1)(2x − 1) + B(x − 2)(2x − 1) + C(x − 2)(x + 1)
.
(x + 1)(x − 2)(2x − 1)
Equating the numerators, we have
A(x + 1)(2x − 1) + B(x − 2)(2x − 1) + C(x − 2)(x + 1) = 9.
Substitute x = 2, −1 and 1/2 respectively to get A = 1, B = 1 and C = −4.
Consequently,
I =∫
1
dx + ∫
x−2
1
dx − ∫
x+1
89
4
∣(x − 2)(x + 1)∣
dx = ln
+ C.
2x − 1
(2x − 1)2
CHAPTER 7. INTEGRATION
Example 7.2.25. Evaluate I = ∫
x2 + 4x − 16
dx.
− 6x2 + 5x − 6
5x3
Solution. Factorize the denominator to get 5x3 − 6x2 + 5x − 6 = (5x − 6)(x2 + 1). Suppose that
Ax + B
C
(Ax + B)(5x − 6) + C(x2 + 1)
x2 + 4x − 16
= 2
+
=
.
2
− 6x + 5x − 6
x +1
5x − 6
(5x − 6)(x2 + 1)
5x3
Equating the numerators to get
(Ax + B)(5x − 6) + C(x2 + 1) = x2 + 4x − 16.
Substitute x = 6/5 to get C = −4, and thus
x2 + 4x − 16 = (5A − 4)x2 + (5B − 6A)x − 6B − 4.
By comparing the coefficient of x2 and the constant term, we get A = 1 and B = 2. It turns out that
I =∫
xdx
+ 2∫
x2 + 1
dx
− 4∫
x2 + 1
dx
(x2 + 1)1/2
= ln
+ 2 arctan x + C.
5x − 6
(5x − 6)4/5
Denominator Polynomial has Repeated Factor
Procedure – Integration by Partial Fractions II (Repeated Factors)
Suppose that
n(x)
is proper and d(x) has repeated factors:
d(x)
1. To each repeated linear factor (ax + b)m of d(x), introduce a term
α1
α2
αm
+
+⋯+
.
ax + b (ax + b)2
(ax + b)m
2. To each repeated irreducible factor (ax2 + bx + c)n of d(x), introduce a term
β1 x + γ 1
β2 x + γ 2
β n x + γn
+
+⋯+
.
ax2 + bx + c (ax2 + bx + c)2
(ax2 + bx + c)n
3. Solve the undetermined coefficients.
To understand why higher degree numerator polynomial is not used, let’s illustrate with the core idea with a couple
of simple cases.
Consider the rational function
ax2 + bx + c
. Note that
(x − α)3
ax2 + bx + c = a(x − α)2 + (2aα + b)(x − α) + aα2 + bα + c.
As a result,
Next, consider
ax2 + bx + c
a
(b + 2a)α aα2 + bα + c
=
+
+
.
(x − α)3
x − α (x − α)2
(x − α)3
ax3 + bx2 + cx + d
. By long division,
(x2 + x + 1)2
ax3 + bx2 + cx + d = (x2 + x + 1)(αx + β) + γ
for some α, β and γ whereby
ax3 + bx2 + cx + d
αx + β
γ
= 2
+ 2
.
2
2
(x + x + 1)
x + x + 1 (x + x + 1)2
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MATH1013/LN/2019 Sem 1
7.3. IMPROPER INTEGRALS
x
dx.
(x − 3)3
Example 7.2.26. Evaluate ∫
Solution. Suppose that the partial fraction decomposition of the integrand is
x
A
B
C
A(x − 3)2 + B(x − 3) + C
=
+
+
=
.
3
2
3
(x − 3)
x − 3 (x − 3)
(x − 3)
(x − 3)2
Equating the numerator to get
x = A(x − 3)2 + B(x − 3) + C.
It’s readily seen that A = 0, B = 1 and C = 3. Thus,
x
1
3
dx = −
−
+ C.
(x − 3)2
x − 3 2(x − 3)2
∫
Example 7.2.27 ((Optional)). Find the partial fraction decomposition of f (x) =
x−1
.
x2 (x2 + 1)2
α1 α2 β1 x + γ1 β2 x + γ2
+ 2
. Express the RHS as a single fraction and equate
+
+ 2
x x2
x +1
(x + 1)2
Solution. Suppose that f (x) =
the numerators, we have
⎧
⎪
⎪
⎪
⎪
⎪
α1
⎪
⎪
⎪
⎪
⎪
⎨
2α1
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩ α1
Solve the system to get f (x) =
7.3
α2
2α2
+ γ1
+ β1
α2
+ γ2
+ β2
+ γ1
+ β1
= −1
=
1
=
0
=
0
=
0
=
0
1
1
1−x
1−x
−
+
+
.
x x2 x2 + 1 (x2 + 1)2
Improper Integrals
b
Recall that definite integral ∫ f (x) dx exists when f is continuous over [a, b]. Is it possible to define a certain
a
kind of definite integral when this assumption is no longer valid.
For instance, suppose that c ∈ [0, 1] and
f (x) = {
1
1/2
if x ∈ [0, 1] ∖ {c}
if x = c
1
This function is bounded but it is discontinuous at c. It should be quite convincing that ∫0 f (x) dx = 1 after
redefining f (c) = 1. It is by no means a rigourus treatment.
In this section, we are going to investigate two ‘less trivial’ situations:
1. f diverges to infinity at one of the endpoint.
2. The domain of integration is not bounded.
Definition 7.3.1. Let f be continuous on [a, b) and x = b is a vertical asymptote of the graph of f . If
c
lim ∫
c→b−
f (x) dx
a
exists, then we define the improper integral of f over [a, b] to be
b
∫
a
c
f (x) dx = lim− ∫
c→b
91
f (x) dx.
a
CHAPTER 7. INTEGRATION
Similarly, we can define
b
b
f (x) dx = lim+ ∫
c→a
∫
a
f (x) dx
c
when f is continuous on (a, b] and x = a is a vertical asymptote of the graph of f .
Example 7.3.2. Consider the function f (x) = x−1/2 , x > 0. Note that 0 ∉ dom (f ). However, f is continuous on
any interval [c, 1] where 0 < c < 1. Thus
1
∫
Since lim+
c→0
√
c
1
f (t) dt = [2x1/2 ]c = 2(1 −
√
c).
c = 0, we conclude that
1
f (t) dt = lim+ [2(1 −
∫
√
c→0
0
c)] = 2
which is understood to be an improper integral.
Definition 7.3.3. Let f be continuous on [a, c) ∪ (c, b] and x = c be a vertical asymptote of the graph of f . If both
the improper integrals
t
lim ∫
t→c−
b
f (x) dx
and
a
lim ∫
t→c+
f (x) dx
t
exist, then we define the improper integral of f over [a, b] to be
b
∫
c
f (x) dx = ∫
a
b
f (x) dx + ∫
a
f (x) dx.
c
Example 7.3.4. Evaluate the improper integral of f (x) = ∣x∣−1/2 over [−1, 1].
Solution. Clearly f is continuous over [−1, 1] ∖ {0} and x = 0 is a vertical asymptote of f .
Note that f (x) = x−1/2 over (0, 1], and we have seen that the improper integral ∫
0
(Example 7.3.2).
1
x−1/2 dx exists and is 2
On the other hand, for any d ∈ (−1, 0), f (x) = (−x)−1/2 is continuous on [−1, d]. By the substitution u = −x,
d
∫
−1
∣x∣−1/2 dx = ∫
1
−d
u−1/2 du = 2(1 −
√
−d) Ð→ 2 as d → 0− .
1
Hence the improper integral ∫
−1
f (x) dx exists and
1
∫
−1
0
f (x) dx = ∫
−1
1
f (x) dx + ∫
f (x) dx = 4.
0
Now we turn to integration over infinite interval.
Definition 7.3.5. Let f ∶ [a, +∞) → R be continuous. If
b
lim ∫
b→+∞
f (x) dx
a
exists, then the improper integral of f over [a, +∞) is the limit
+∞
∫
a
b
f (x) dx = lim ∫
b→+∞
f (x) dx.
a
Similarly, one can define
b
b
∫
−∞
f (x) dx = lim ∫
a→−∞
when f ∶ (−∞, b] → R is continuous.
92
f (x) dx
a
MATH1013/LN/2019 Sem 1
7.4. REFERENCES
Example 7.3.6. Let f (x) = xe−x . Integrate–by–parts to get
−x
−x
−x
−x
∫ xe dx = −xe + ∫ e dx = −e (1 + x) + C.
It follows that
b
∫
f (x) dx = − [e−b (1 + b) − e0 (1 + 0)]
0
+∞
∫
0
b
f (x) dx = 1 = lim ∫
b→+∞
f (x) dx = 1 − lim
b→+∞
0
1
= 1.
eb
However,
0
lim ∫
a→−∞
f (x) dx = −∞
a
0
and thus ∫
−∞
f (x) dx diverges.
Definition 7.3.7. Let f ∶ R → R be continuous. If there is c ∈ R so that both the improper integrals
+∞
c
∫
exist, then we define
+∞
∫
−∞
−∞
f (x) dx
and
∫
f (x) dx
c
c
f (x) dx = ∫
−∞
f (x) dx + ∫
+∞
f (x) dx.
c
Remark 7.3.8. One can easily prove that if the two improper integrals exist for some c, then one can replace c by
any real number.
Example 7.3.9. Let f (x) = (1 + x2 )−1 . We have seen that ∫ f (x) dx = arctan x + C. It follows that
b
lim ∫
b→+∞
f (x) dx
=
f (x) dx
=
lim (arctan b − arctan 0) = π/2
b→+∞
0
0
lim ∫
a→−∞
a
As a result,
+∞
∫
7.4
−∞
lim (arctan 0 − arctan a) = π/2.
a→−∞
f (x) dx =
π π
+ = π.
2 2
References
Rohde, Ulrich L. ; Jain, G. C. ; Poddar, Ajay K. ; Ghosh, A. K., Introduction to Integral Calculus: Systematic
Studies with Engineering Applications for Beginners, John Wiley & Sons, Inc., Hoboken, NJ, USA.
Section 1
• Chapter 5: § 3–5 – (on definite integral). You may skip the proof of the integrability theorem. These sections
contain also demonstrating how to obtain definite integral by computing the limits of Riemann sums. You
may study them if you’re interesting in.
• Chapter 7: § a.3–3.1, § b.2 (on properties of definite integrals).
• Chapter 1: § 1–5 – (on indefinite integral).
• Chapter 6a: § 1–5.1 (on Fundamental Theorem of Calculus). You may safely skip the proofs of the theorems.
• Chapter 8. You may study this chapter on application of definite integrations to compute area of regions
enclosed by graphs of functions.
93
CHAPTER 7. INTEGRATION
Rohde, Ulrich L. ; Jain, G. C. ; Poddar, Ajay K. ; Ghosh, A. K., Introduction to Integral Calculus: Systematic
Studies with Engineering Applications for Beginners, John Wiley & Sons, Inc., Hoboken, NJ, USA.
Sections 2
• Chapter 3a+b. Method of substitution.
• Chapter 4a+b. Integration by Parts.
• Chapter 6+7. Applications of these techniques in evaluating definite integrals.
Certainly, the integration techniques introduced in these chapters are not that elementary and you are not required
to be familiar with all of them. Also, you can skip some examples in your first reading whenever it involves
integration techniques you haven’t seen before.
Sections 3
• Chapter 9a: § 7. Improper Integral.
94