MH1101 Chapter 1: Integrals
1.1
Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.2
Area Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.3
Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.4
The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . 19
1.5
The Substitution Rule . . . . . . . . . . . . . . . . . . . . . . . . . . 24
1.6
Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
1
MH1101 CHAPTER 1: INTEGRALS
1.1
2
Antiderivatives
Reference: Stewart Calculus 8E, Section 3.9
Goal: There are many antiderivatives of a given function, but they are all the same
up to adding a constant.
Z
We will introduce the notation f (x) dx of an indefinite integral. It is a convenient
notation when speaking about antidifferentiations.
When we differentiate a function, we get its (unique) derivative. When we “reverse”
the process of differentiation, we get an antiderivative.
Definition 1. (Antiderivative & Indefinite Integral)
A function F (x) is an antiderivative of f (x) on an interval (a, b) if
F 0 (x) = f (x) for all x ∈ (a, b).
Example 1.1. The following are obvious from the definition (assuming all functions are defined on an appropriate interval):
• f is an antiderivative of f 0 .
• f 0 is an antiderivative of f 00 .
Example 1.2. Let f (x) = 3x2 , where x ∈ R = (−∞, ∞). Consider the functions
F (x) = x3 , G(x) = x3 + 1, H(x) = x3 + 100.
By definition, all F (x), G(x), H(x) all antiderivatives of 3x2 on R since
F 0 (x) = G0 (x) = H 0 (x) = 3x2 = f (x).
MH1101 CHAPTER 1: INTEGRALS
3
Indeed, in the preceding example, any function of the form x3 + C, where C is a
constant, is an antiderivative of 3x2 . The question is: Are there any others? The
answer is No!
The proof below uses the Mean Value Theorem (MVT): Suppose f is continuous
on [a, b] and is differentiable on (a, b). Then there exists a point c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Theorem 1. All antiderivatives of f differ by a constant. That is, if F is an
antiderivative of f on an interval (a, b), then the most general antiderivative of f
on (a, b) is
F (x) + C
where C is an arbitrary constant.
Proof. Let F (x) and G(x) be two antiderivatives of f ; so by definition, we have
F 0 (x) = f (x), G0 (x) = f (x).
We want to show that G(x) = F (x) + C for some constant C.
Approach: Let h(x) = G(x) − F (x). We will show that h(x) = C for some
constant C. To do this: Let x1 , x2 be two arbitrary numbers inside the interval
(a, b), where x1 < x2 . We will prove that h(x1 ) = h(x2 ) using the Apply the Mean
Value Theorem.
Note that
h0 (x) = G0 (x) − F 0 (x) = f (x) − f (x) = 0 for all x ∈ (a, b).
(1.1)
Applying the Mean Value Theorem to h on the interval [x1 , x2 ], we get a number
c ∈ (x1 , x2 ) such that
h(x2 ) − h(x1 ) = h0 (c)(x2 − x1 ).
Since h0 (c) = 0 (by (1.1)), we have h(x2 ) − h(x1 ) = 0, whence h(x2 ) = h(x1 ).
Therefore, h has the same value at any two numbers x1 , x2 in (a, b). This means
that h is constant on (a, b), i.e. h(x) = C for some constant C.
MH1101 CHAPTER 1: INTEGRALS
4
We have proved that G(x) − H(x) = C, as desired.
Notation. The general antiderivative of f is also called an indefinite integral of
f , and is denoted by
Z
f (x) dx = F (x) + C,
where F is an antiderivative of f , and C is an arbitrary constant.
Table of some useful indefinite integrals
R
R n
n+1
k dx = kx + C (k is a constant)
x dx = xn+1 + C (n 6= −1)
R
R
sin x dx = − cos x + C
cos x dx = sin x + C
R
R
sec2 x dx = tan x + C
csc2 x dx = − cot x + C
R
R
sec x tan x dx = sec x + C
csc x cot x dx = − csc x + C
R x
R
x
e dx = ex + C
ax dx = lna a + C (a 6= 1)
R 1
dx = ln |x| + C
x
Some common properties of indefinite integrals (antiderivatives):
Z
•
Z
(f (x) + g(x)) dx =
Z
•
Z
f (x) dx +
g(x) dx
Z
a · f (x) dx = a ·
f (x) dx, where a is a constant.
Example 1.3. Find the most general antiderivative of the function
f (x) =
1
+ 5 sin x, x ∈ R \ {0}.
2x
MH1101 CHAPTER 1: INTEGRALS
5
Solution. Using the table of indefinite integrals, we have
Z
Z
Z
1
1
+ 5 sin x dx =
dx + 5 sin x dx
2x
2x
Z
Z
1
1
=
dx + 5 sin x dx
2
x
1
ln |x| − 5 cos x + C.
=
2
We cannot apply Theorem 1 directly to the domain R \ {0} since it is not an interval
of the form (a, b). The function is defined on R \ {0} = (−∞, 0) ∪ (0, ∞). We can
apply Theorem 1 to each domain (−∞, 0) and (0, ∞) separately:
( 1
Z
ln |x| − 5 cos x + C1 x ∈ (−∞, 0)
1
2
+ 5 sin x dx =
1
2x
ln |x| − 5 cos x + C2 x ∈ (0, ∞)
2
where C1 and C2 are arbitrary constants.
MH1101 CHAPTER 1: INTEGRALS
1.2
6
Area Problem
Reference: Stewart Calculus 8E, Section 4.1
Goal: To calculate the area under a curve as the limit of Riemann sums. This
motivates the definition of a definite integral in the next section.
Consider the area under the parabola y = f (x) = x2 from 0 to 1:
We approximate the area as follows:
• Divide the interval [0, 1] into n equal subintervals of equal size 4x:
[x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
• Use rectangles with these subintervals as base to approximate the area between
the graph and the x-axis.
• Take the values of the function at the right endpoints or left endpoints of
the subintervals as the height of the rectangles.
(i) With right endpoints, the sum of the areas of the rectangles, denoted by
Rn , is
n
X
Rn =
f (xi )4x.
i=1
(ii) With left endpoints, the sum of the areas of the rectangles, denoted by
Ln is
n
X
Ln =
f (xi−1 )4x.
i=1
MH1101 CHAPTER 1: INTEGRALS
7
Approximation by n = 4 subintervals using right endpoints:
Approximation by n = 4 subintervals using left endpoints:
Let A be the actual area between the graph of the function and the x-axis.
Let Rn be the approximation of the area using n subintervals with right endpoints.
Let Ln be the approximation of the area using n subintervals with left endpoints.
We have
Ln < A < R n .
MH1101 CHAPTER 1: INTEGRALS
8
Approximation using right endpoints:
Approximation using left endpoints:
Intuitively, the approximations get better and better as n tends to ∞:
A = lim Rn = lim (f (x1 )4x + f (x2 )4x + · · · + f (xn )4x) .
n→∞
n→∞
A = lim Ln = lim (f (x0 )4x + f (x1 )4x + · · · + f (xn−1 )4x) .
n→∞
n→∞
In fact, instead of left endpoints or right endpoints, we could take the height of
the i-th rectangle to be the value of f at any number x∗i in the i-th subinterval
[xi−1 , xi ]. We call the numbers x∗1 , x∗2 , . . ., x∗n the sample points. So a more
general expression for the area is
A =
=
lim (f (x∗1 )4x + f (x∗2 )4x + · · · + f (x∗n )4x)
n→∞
lim
n→∞
n
X
i=1
f (x∗i )4x.
MH1101 CHAPTER 1: INTEGRALS
1.3
9
Definite Integral
Reference: Stewart Calculus 8E, Section 4.2
Goal: Motivated by the area problem in the preceding section, we define the notion
of a definite integral of a function f which occurs in a wide variety of situations even
when f is not necessarily a positive function.
Definition 2. (Definite Integral)
Suppose f is a function defined for a ≤ x ≤ b, and we divide the interval [a, b] into
n subintervals of equal width 4x = (b − a)/n:
[x0 , x1 ], [x1 , x2 ], . . . , [xn−1 , xn ].
Let x∗i ∈ [xi−1 , xi ] be any sample points in these subintervals. The definite integral
of f from a to b is
Z
b
f (x) dx = lim
a
n→∞
n
X
f (x∗i )4x,
i=1
provided that this limit exists and gives the same value for all possible choices of
the sample points.
If the definite integral
Rb
a
f (x) dx exists, we say that f is integrable on [a, b].
Pn
∗
The sum Sn =
i=1 f (xi )4x is called a Riemann sum of f . So the definite
Rb
integral a f (x) dx is the limit of the Riemmann sums Sn as n → ∞.
Z b
In the notation
f (x) dx:
a
• f (x) is called the integrand.
• a and b are called the limits of integration; a is the lower limit and b is
the upper limit.
• The term dx simply indicates that the independent variable is x. It is also a
‘dummy’ variable in the notation since replacing x by any other variable will
not change the value. E.g:
Z b
Z b
Z b
f (x) dx =
f (t) dt =
f (u) du.
a
a
a
MH1101 CHAPTER 1: INTEGRALS
10
• The procedure of calculating an integral is called integration.
With right endpoints as sample points, we have
Z
n
X
b
f (x) dx = lim
n→∞
a
f (xi )4x,
i=1
whereas, with left endpoints as sample points, we have
Z
b
f (x) dx = lim
n→∞
a
n
X
f (xi−1 )4x,
i=1
where
xi = a + i4x (i = 0, 1, 2, . . . , n) x0 = a, xn = b.
Z
Example 1.4. With right endpoints as sample points, compute
uating the corresponding limit of Riemann sums.
1
x2 dx by eval-
0
Solution. It is known (by mathematical induction) that
n
X
i2 =
i=1
n(n + 1)(2n + 1)
.
6
Let f (x) = x2 . With a = 0, b = 1, we have
4x =
b−a
1−0
1
i
=
= , xi = 0 + i4x = , i = 0, . . . , n.
n
n
n
n
With right endpoints as sample points, we have
n
X
Sn =
f (xi )4x
i=1
n
n 2
X
i 1 X i
1
=
f
=
n n
n n
i=1
i=1
=
n
1 X 2
i
n3 i=1
=
1 n(n + 1)(2n + 1)
(n + 1)(2n + 1)
·
=
.
n3
6
6n2
MH1101 CHAPTER 1: INTEGRALS
11
Hence
Z
1
x2 dx =
0
=
=
=
=
=
lim Sn
n→∞
(n + 1)(2n + 1)
n→∞
6n2
1
n + 1 2n + 1
lim
·
6 n→∞ n
n
1
1
1
lim 1 +
2+
6 n→∞
n
n
1
·1·2
6
1
.
3
lim
If f takes on both positive and negative values, then the Riemann sum is the sum
of the areas of the rectangles that lie above the x-axis and the negatives of the areas
of the rectangles that lie below the x-axis (the areas of the blue rectangles minus
the areas of the gold rectangles).
When we take the limit of such Riemann sums, the definite integral can be interpreted as a net area, that is, a difference of areas:
Z b
f (x) dx = A1 − A2
a
where
• A1 is the area of the region above the x-axis and below the graph of f (between
x = 1 and x = b) ,
MH1101 CHAPTER 1: INTEGRALS
12
• A2 is the area of the region below the x-axis and above the graph of f (between
x = a and x = b).
Properties of Definite Integrals
Rb
When we defined the definite integral a f (x) dx, we implicitly assumes that a < b.
In fact, the definition as the limit of Riemann sums still works even if a > b, in
which case, we would have
4x =
a−b
b−a
=−
.
n
n
Therefore, if we were to switch the limits of integration, we only have to multiply
the resulting integral by a ‘−1’.
Z
b
Z
f (x) dx = −
a
a
f (x) dx.
b
Also, if a = b, then 4x = 0 and so
Z
a
f (x) dx = 0.
a
Assuming f and g are continuous and c is a constant, the following basic properties
of definite integrals make our life easier:
MH1101 CHAPTER 1: INTEGRALS
13
Basic rules of definite integrals:
Z
b
c dx = c(b − a).
(P1)
a
Z
b
Z
(f (x) ± g(x)) dx =
(P2)
a
Z
b
b
g(x) dx.
a
f (x) dx.
a
a
c
Z
f (x) dx +
a
f (x) dx ±
b
Z
cf (x) dx = c
(P4)
Z
a
(P3)
Z
b
b
b
Z
f (x) dx =
f (x) dx
c
a
Comparison rules of definite integrals:
b
Z
(P5) If f (x) ≥ 0 on [a, b] then
f (x) dx ≥ 0.
a
Z
b
Z
f (x) dx ≥
(P6) If f (x) ≥ g(x) on [a, b] then
b
g(x) dx.
a
a
(P7) If m ≤ f (x) ≤ M on [a, b] then
Z
b
f (x) dx ≤ M (b − a).
m(b − a) ≤
a
The proofs of the above rules are quite straight forward using the definition of
definite integrals. Here we only provide the proofs of Property (P6) and (P7).
Proof of Property (P6). Since f (x) ≥ g(x) on [a, b], we have
f (x∗i ) ≥ g(x∗i ) for all i =⇒ f (x∗i )4x ≥ g(x∗i )4x for all i
=⇒
n
X
i=1
f (x∗i )4x
≥
n
X
i=1
g(x∗i )4x
MH1101 CHAPTER 1: INTEGRALS
14
Taking limit as n → ∞,
lim
n→∞
n
X
f (x∗i )4x
≥ lim
i=1
i.e.
Z
b
n→∞
Z
f (x) dx ≥
n
X
g(x∗i )4x
i=1
b
g(x) dx.
a
a
Proof of Property (P7). Since m ≤ f (x) ≤ M , Property (P6) gives
Z b
Z b
Z b
m dx ≤
f (x) dx ≤
M dx.
a
a
a
Using Property (P1) to evaluate the integrals on the left and right sides, we have
Z b
f (x) dx ≤ M (b − a).
m(b − a) ≤
a
Though it’s quite easy to write down the limit of Riemann sums given a definite
integral, the reverse, i.e. converting a limit to a definite integral, may not be so
straightforward.
Example 1.5. Express the following limit as a definite integral.
(i) lim
n→∞
n
X
π2i
i=1
n2
sin
iπ
n
4
n 3X
3i
(ii) lim
2+
n→∞ n
n
i=1
Solution. We will make use of right-endpoints approximations. The idea is to
identify a suitable choice of 4x, xi = a + i4x, f (x) (these choices may not be
unique) which will determine the rest:
Z b
n
X
a = x0 , b = xn , lim
f (xi )4x =
f (x) dx.
n→∞
i=1
a
MH1101 CHAPTER 1: INTEGRALS
15
(i) Consider the i-th term:
π2i
sin
n2
Choose
4x =
iπ
n
iπ
=
sin
n
iπ
n
·
π
.
n
π
iπ
, xi = , f (x) = x sin x.
n
n
Then
a = x0 = 0, b = xn = π.
So
lim
n→∞
n
X
π2i
i=1
n2
sin
iπ
n
n
X
iπ
=
=
=
lim
n→∞
lim
n→∞
lim
n→∞
Z
i=1
n
X
i=1
n
X
n
sin
f (xi )4x
i=1
f (x) dx
Z0 π
=
x sin x dx
0
(ii) Consider the i-th term:
Choose
4x =
3i
2+
n
4
·
xi sin xi 4x
π
=
iπ
n
3
.
n
3
3i
, xi = 2 + , f (x) = x4 .
n
n
Then
a = x0 = 2, b = xn = 5.
·
π
n
MH1101 CHAPTER 1: INTEGRALS
16
So
4
n 3X
3i
lim
2+
=
n→∞ n
n
i=1
4
n X
3i
3
lim
2+
·
n→∞
n
n
i=1
=
lim
n→∞
=
lim
n→∞
Z
n
X
(xi )4 4x
i=1
n
X
f (xi )4x
i=1
5
=
f (x) dx
2
Z
=
5
x4 dx.
2
Can you find a different choice of 4x, xi and f (x)?
Example 1.6. Suppose f is continuous, and
Z 3
Z 4
f (x) dx = 3,
f (x) dx = 7.
0
Z
Find
0
3
f (x) dx.
4
Solution. By Property (P4),
Z
Z 4
f (x) dx =
3
Z
f (x) dx +
0
0
4
f (x), dx,
3
4
Z
7=3+
f (x) dx,
3
Z
4
f (x) dx = 4.
3
Hence,
Z
3
Z
f (x) dx = −
4
4
f (x) dx = −4.
3
MH1101 CHAPTER 1: INTEGRALS
17
Example 1.7. Use the properties of integrals to verify that
π
≤
12
√
π/3
Z
3π
.
12
sin x dx ≤
π/6
Solution. Note that sin x is increasing on [π/6, π/3], so for
π
6
≤x≤
π
3
we have
√
3
1
= sin(π/6) ≤ sin x ≤ sin(π/3) =
.
2
2
By Property P7, we have
1
(π/3 − π/6) ≤
2
Z
π
≤
12
Z
π/3
π/6
√
3
sin x dx ≤
(π/3 − π/6).
2
√
π/3
3π
.
2
sin x dx ≤
π/6
Example 1.8. Which of the following integrals has the largest value? Why?
2
Z
√
x dx,
2
Z
p
1/x dx,
1
1
Solution. Note that
we have
Z
2
q
√
x dx.
1
√
√
x ≤ x for 1 ≤ x ≤ 2. Since · is an increasing function,
q
p
√
√
1/x <
x < x, 1 ≤ x ≤ 2.
1
x
≤
Z
2
So
1
p
Z
1/x dx <
R2√
Thus, 1 x dx has the largest value.
1
2
q
√
Z
x dx <
1
2
√
x dx.
MH1101 CHAPTER 1: INTEGRALS
18
Definition 3. (Average value)
We define the average value of f on [a, b] as
fave
1
=
b−a
Z
b
f (x) dx.
a
Theorem 2. (The Mean Value Theorem for Integrals)
If f is continuous on [a, b], then there exists a number c in [a, b] such that
1
=
b−a
f (c) = fave
Z
b
f (x) dx.
a
Proof. Since f is continuous on [a, b], by the Extreme Value Theorem, there
exist u, v ∈ [a, b] such that
f (u) ≤ f (x) ≤ f (v), for all x ∈ [a, b].
By Property (P1) and (P6), we have
Z
b
Z
b
a
b
f (v) dx = (b − a)f (v).
f (x) dx ≤
f (u) dx ≤
(b − a)f (u) =
Z
a
a
Dividing the terms by (b − a), we have
1
f (u) ≤
b−a
Z
b
f (x) dx ≤ f (v).
a
Finally, the Intermediate Value Theorem says that there must exist c ∈ [a, b]
such that
Z b
1
f (c) =
f (x) dx.
b−a a
MH1101 CHAPTER 1: INTEGRALS
1.4
19
The Fundamental Theorem of Calculus
Reference: Stewart Calculus 8E, Section 4.3
Goal: To connect integration and differentiation, by showing that we can use antiderivative to calculate a definite integral (so that we can avoid using the limit of
Riemann sums to evaluate a definite integral!)
Theorem 3. (Fundamental Theorem of Calculus Part 1 (FTC1))
Suppose f is continuous on [a, b]. Define
Z x
F (x) =
f (t) dt, a ≤ x ≤ b.
a
Then F (x) is differentiable on (a, b) with
Z x
d
0
f (t) dt = f (x), a < x < b.
F (x) =
dx a
In short: FTC1 says that “integration” followed by “differentiation” returns the
original function!
Proof.
Idea: Use definition of differentiability to expand F 0 (x). Then apply the Mean
Value Theorem for Integrals (Theorem 2) to evaluate the resulting term to show
that F 0 (x) = f (x).
Step 1: Expand F (x).
Let x ∈ (a, b). Using the definition of differentiability, we have
F 0 (x) =
By the definition of F (x), we have
F (x + 4x) − F (x)
.
4x→0
4x
lim
(1.2)
MH1101 CHAPTER 1: INTEGRALS
20
x+4x
Z
Z
x
f (t) dt −
0
F (x) =
lim
a
f (t) dt
a
4x
4x→0
1
= lim
·
4x→0 4x
x+4x
Z
f (t) dt
(1.3)
x
Step 2: Use the Mean Value Theorem for Integrals.
Applying the Mean Value Theorem for Integrals (Theorem 2) to the function f (t)
on the interval [x, x + 4x], there exists a number c, x ≤ c ≤ x + 4x such that
1
f (c) =
·
4x
Z
x+4x
f (t) dt.
(1.4)
x
Substituting (1.4) into (1.3):
F 0 (x) =
lim f (c).
4x→0
(1.5)
Now, notice that c → x as 4x → 0 (this is because c is sandwiched between x and
x + 4x).
It follows from (1.5) that
F 0 (x) = lim f (c) = f (x),
c→x
as desired.
MH1101 CHAPTER 1: INTEGRALS
dy
for y =
Example 1.9. Find
dx
x
Z
21
√
1 + t2 dt.
1
Solution. By FTC1, we have
dy
d
=
dx
dx
d
Example 1.10. Find
dx
Z
Z
x
√
√
1 + t2 dt = 1 + x2 .
1
x4
sec t dt
1
Solution. We use Chain Rule in conjunction with FTC1. Let u = x4 . Then
d
dx
Z
x4
sec t dt =
1
=
=
=
Z u
d
sec t dt
dx 1
Z u
d
du
sec t dt ·
(by Chain Rule)
du
dx
1
du
(by FTC1)
sec u ·
dx
sec(x4 ) · 4x3 .
Theorem 4. (Fundamental Theorem of Calculus Part 2 (FTC2))
If f is continuous on [a, b], then
Z
b
f (x) dx = F (b) − F (a) = [F (x)]ba
a
where F is any antiderivative of f .
Proof.
Z
Idea: Let G(x) =
x
f (t) dt. Both G(x) and F (x) are antiderivatives of f (x), and
a
so they differ by a constant. We will exploit this fact.
MH1101 CHAPTER 1: INTEGRALS
22
From FTC1, G0 (x) = f (x). So G is an antiderivative of f .
Given than F is any antiderivative of f . By Theorem 1,
F (x) − G(x) = C
where C is some constant.
Substituting x = a and x = b, we have
Z a
Z b
G(a) =
f (t) dt = 0, G(b) =
f (t) dt.
a
a
Thus,
(F (b) − G(b)) − (F (a) − G(a)) = C − C = 0
Z b
F (b) −
f (t) dt − (F (a) − 0) = 0
a
b
Z
f (t) dt = F (b) − F (a).
a
Z
1
x3 dx
Example 1.11. Evaluate the integral
−2
Solution.
Z
1
3
x dx =
−2
x4
4
1
−2
1 (−2)4
=
−
4
4
15
= − .
4
Example 1.12. Find the area of the region bounded between the cosine curve and
the x-axis, from x = 0 to x = π/2.
Solution. The area is
Z
0
π/2
cos x dx = [sin x]π/2
= 1.
0
MH1101 CHAPTER 1: INTEGRALS
23
Example 1.13. Use the properties of integrals to verify that
Z 3√
26
x4 + 1 dx ≥ .
3
1
Solution. For 1 ≤ x ≤ 3, we have x4 + 1 ≥ x4 , so
√
√
x4 + 1 ≥ x4 = x2 .
Let f (x) = x2 . Using right endpoints approximations, we have
3 3
Z 3√
Z 3
Z 3√
x
26
2
4
4
x dx =
x + 1 dx ≥
x dx =
= .
3 1
3
1
1
1
Example 1.14. Calculate the derivative
Z x
d
sin u
du,
√
dx x u
where x ∈ (0, ∞).
Solution. Note that
Z
x
√
x
sin u
du =
u
Z
a
√
x
sin u
du +
u
Z
a
x
sin u
du
u
for any a ∈ (0, ∞).
√
We choose a = 2. Then the function sinx x is continuous on [ x, 2] and on [2, x]. By
FTC1 and Chain Rule, we have
Z x
Z 2
Z x
sin u
d
sin u
d
sin u
d
du
=
du +
du
√
√
dx
dx
dx
u
x u
x u
2
!
Z x
Z √x
d
d
sin u
sin u
= −
du +
du
dx
u
dx
u
2
2
Z w
Z x
√
d
sin u
dw
d
sin u
= −
du
+
du
(where w = x)
dw
u
dx dx
u
2
2
sin w 1
sin x
√ +
= −
w 2 x
x
√
sin x 1
sin x
√ +
= − √
x
x 2 x
√
sin x sin x
=
−
.
x
2x
MH1101 CHAPTER 1: INTEGRALS
1.5
24
The Substitution Rule
Reference: Stewart Calculus 8E, Section 4.5
Goal: By changing the underlying variable x to a new variable u, we can convert a
difficult integral with respect to x into one with respect to u that we already know
how to compute. This change of variable takes the form
u = g(x),
where g is some function of x.
Theorem 5. (Substitution Rule for Definite Integral)
Suppose g 0 is continuous on [a, b] and f is continuous on the range of u = g(x).
Then
Z g(b)
Z b
0
f (u) du.
f (g(x))g (x) dx =
g(a)
a
d
Proof. Let F be an antiderivative of f . Notice that dx
(F (g(x)) = F 0 (g(x))g 0 (x) =
0
f (g(x))g (x) by Chain Rule. This means that F (g(x)) is an antiderivative of f (g(x))g 0 (x).
Thus, by FTC2,
Z b
f (g(x))g 0 (x) dx = F (g(b)) − F (g(a)).
(1.6)
a
On there other hand, since F is an antiderivative of f , we have (by FTC2)
g(b)
Z
f (u) du = F (g(b)) − F (g(a)).
g(a)
Combining (1.6) and (1.7), we have
Z
b
0
Z
g(b)
f (g(x))g (x) dx =
a
as desired.
f (u) du,
g(a)
(1.7)
MH1101 CHAPTER 1: INTEGRALS
25
Z
b
h(x) dx, we do the
In practice, when applying the Substitution Rule to evaluate
a
following:
• Think of a function u = g(x). Then
dx = k(u) du for some function k of u.
du
= g 0 (x). From here, try to write
dx
Z
b
• Think of a function f (u). Then try to convert the integral
respect to x to an integral with respect to u.
h(x) dx with
a
• Replace the lower limit a by g(a), and the upper limit b by g(b) when computing the resulting integral with respect to u.
Z
Example 1.15. Evaluate
1
2
1
dx.
(3 − 5x)2
Solution.
• Let u = g(x) = 3 − 5x. Then
• Let f (u) =
du
1
= g 0 (x) = −5 =⇒ dx =
du
dx
(−5)
1
. Then
u2
Z
Z
Z
1
1
1
1
1
dx =
·
du =
du
2
2
(3 − 5x)
u (−5)
(−5)
u2
Substituting the lower and upper limits:
Z 2
Z g(2)
1
1
1
dx =
du
2
(−5) g(1) u2
1 (3 − 5x)
Z −7
1
1
=
du
(−5) −2 u2
−7
1
= − −u−1 −2
5
1 1 1
= −
−
5 7 2
1
=
.
14
MH1101 CHAPTER 1: INTEGRALS
4
Z
Example 1.16. Evaluate
26
√
2x + 1 dx.
0
Solution.
• Let u = g(x) =
√
2x + 1. Then du = 12 (2x + 1)−1/2 (2) dx =⇒ dx = u du.
• Let f (u) = u. Then
Z
√
Z
2x + 1 dx =
Z
u(u du) =
u2 du.
Substituting the lower and upper limits:
Z
4
√
2x + 1 dx =
Z
0
=
g(4)=3
u2 du
g(0)=1
3 3
u
3
1
1
= 9−
3
26
=
.
3
MH1101 CHAPTER 1: INTEGRALS
1.6
27
Improper Integrals
Reference: Stewart Calculus 8E, Section 7.8
Rb
Goal: Previously, in defining a definite integral a f (x) dx we dealt with function
f which is continuous on closed and bounded interval [a, b]. In this section, we wish
to extend the notion of integrals to cases where
• the underlying interval is infinite, e.g. [a, ∞), (−∞, b], (−∞, ∞)
• the integrand f has an infinite discontinuity on a finite interval.
Z
∞
1
dx, which is the area under the
x2
1
curve y = x12 , above the x-axis, and on the right-hand side of the line x = 1:
Motivation: Suppose we want to compute
Idea: We can first calculate the area between x = 1 and x = t (for some t). As t
gets larger and larger, the area seems to converge to the area we want.
This brings us to the following definition of improper integrals.
MH1101 CHAPTER 1: INTEGRALS
28
Definition 4. (Improper Integral of Type 1)
(a) If
Rt
a
f (x) dx exists for every number t ≥ a, then
∞
Z
t
Z
f (x) dx = lim
t→∞
a
(b) If
Rb
t
f (x) dx.
a
f (x) dx exists for every number t ≤ b, then
b
Z
Z
f (x) dx = lim
t→−∞
−∞
b
f (x) dx.
t
In each case, the improper integrals are called convergent if the corresponding
limit exists, and divergent if the limit does not exist.
(c) If both
R∞
a
f (x) f x and
Z
Ra
−∞
f (x) dx are convergent, then we define
∞
a
Z
Z
−∞
−∞
∞
f (x) dx.
f (x) dx +
f (x) dx =
a
Note: In part (c), any real number a can be used. This will be justified in the
tutorial (Tutorial 3).
∞
Z
Example 1.17. Determine whether
1
1
dx is convergent or divergent.
x
Solution. Let t ≥ 1. Then
Z
1
exists. Now,
Z
1
∞
t
1
dx = [ln |x|]t1 = ln t,
x
1
dx = lim
t→∞
x
Z
Thus, the improper limit is divergent.
1
t
1
dx = lim ln t = ∞.
t→∞
x
MH1101 CHAPTER 1: INTEGRALS
∞
Z
Example 1.18. Evaluate
−∞
29
1
dx.
1 + x2
Solution. Let a = 0 in part (c) of Definition (4). Then
Z 0
0
1
dx = tan−1 x t = tan−1 0 − tan−1 t = − tan−1 t.
2
t 1+x
So
Z
0
π π
1
−1
= .
dx
=
lim
(−
tan
t)
=
−
−
t→−∞
1 + x2
2
2
lim
t→−∞
t
On the other hand,
Z t
0
−1 t
1
dx
=
tan x 0 = tan−1 t − tan−1 0 = tan−1 t.
1 + x2
So
Z
t
lim
t→∞
0
1
π
dx = lim tan−1 t = .
2
t→∞
1+x
2
Combining, we have
Z ∞
Z 0
Z t
1
1
1
π π
dx = lim
dx + lim
dx = + = π.
2
2
2
t→−∞
t→∞
2
2
−∞ 1 + x
t 1+x
0 1+x
WARNING: The following calculations are WRONG!
(i)
∞
∞
= 0.
−∞
t
x2
x dx = lim
x dx = lim
t→∞ −t
t→∞
2
−∞
Z
(ii)
∞
x2
x dx =
2
−∞
Z
Z
Z
t
= lim
−t
t→∞
∞
What is the correct way to determine
x dx?
−∞
t2 (−t)2
−
2
2
= lim 0 = 0.
t→∞
MH1101 CHAPTER 1: INTEGRALS
30
An integral can also be ‘improper’ on a finite interval. For example, this happens
when the function has an asymptote at the endpoint of the interval, or even at a
point in the middle of the interval.
Definition 5. (Improper Integral of Type 2)
(a) If f is continues on [a, b) and is discontinuous at b, then
Z
b
f (x) dx = lim−
t→b
a
t
Z
f (x) dx
a
if this limit exists (as a finite number).
(b) If f is continuous on (a, b] and is discontinuous at a, then
Z
a
b
Z
f (x) dx = lim+
t→a
b
f (x) dx
t
if this limit exists (as a finite number).
In each case, the improper integrals are called convergent if the corresponding
limit exists, and divergent if the limit does not exist.
(c) If f has a discontinuity at c, where a < c < b, and both
Rb
f (x) dx are convergent, then we define
c
Z
b
Z
f (x) dx =
a
c
Z
f (x) dx +
a
b
f (x) dx.
c
Rc
a
f (x) dx and
MH1101 CHAPTER 1: INTEGRALS
5
Z
Example 1.19. Find
2
31
1
√
dx.
x−2
Solution. The function f (x) = √
1
has an infinite discontinuity at the left
x−2
endpoint of (2, 5]. By Definition 5,
Z 5
1
√
dx =
x−2
2
Z
lim+
x→2
5
√
t
1
dx
x−2
Substituting u = x − 2, we have
Z
Z
√
√
1
1
√
√ du = 2 u + C = 2 x − 2 + C.
dx =
u
x−2
Thus,
Z
2
5
5
√
lim+ 2 x − 2 t
t→2
√
√
lim+ 2( 3 − t − 2)
t→2
√
= 2 3.
1
√
dx =
x−2
=
Z
3
Example 1.20. Evaluate
0
1
dx if possible.
x−1
1
has a discontinuity at x = 1 inside the interval
x−1
[0, 3]. By part (c) of Definition 5, the integral (if it exists) is equal to
Z 3
Z 1
Z 3
1
1
1
dx =
dx +
dx,
0 x−1
0 x−1
1 x−1
Z 1
Z 3
1
1
where both
dx and
dx must exist.
0 x−1
1 x−1
Solution. The function f (x) =
MH1101 CHAPTER 1: INTEGRALS
Z
1
Unfortunately,
0
32
1
dx does not exist because
x−1
Z 1
Z t
1
1
dx = lim−
dx
t→1
0 x−1
0 x−1
= lim− [ln |x − 1|]t0
t→1
=
=
=
lim (ln |t − 1| − ln |0 − 1|)
t→1−
lim ln |t − 1|
t→1−
lim ln(1 − t)
t→1−
= −∞.
Z
Hence, the integral
0
3
1
dx is divergent.
x−1
WARNING: The following calculation is WRONG!
Z
0
3
1
dx = [ln |x − 1|]30 = ln 2 − ln 1 = ln 2.
x−1
This is wrong because the integral is improper and must be calculated in terms of
limits as given in Definition 5.