46
4
Integration and Its Applications
4.1
Antidifferentiation; Particular Antiderivatives; Integration by Substitution
A problem of the early mathematicians before is the following:
Given a function f that is continuous and nonnegative on [a, b],
find the area between the graph of f and [a, b] on the x-axis.
Not until Archimedes proposed later on that this can be solved exhaustively using areas of rectangles of varying heights and limits. Even though the rectangle method makes sense intuitively,
the resulting limits cannot be evaluated at all times. So, the area problem remained a problem
until Newton and Leibniz, in 17th century, independently found a relationship between areas and
derivatives. They showed that
if f is a nonnegative continuous function on [a, b],
and if A(x) is the area under the graph of f over the interval [a, x], where x ∈ [a, b], then
A′ (x) = f (x).
4.1.1
Antiderivative or Indefinite Integrals
The preceding equation tells us that the area under the curve is given by a function that has
f as the first derivative. In particular, the area function A is called an antiderivative of f.
Definition 36. A function F is an antiderivative of a function f on an open interval I if F ′ (x) = f (x)
for all x ∈ I.
The process of finding a function from its derivative is called antidifferentiation, and a method
for finding areas using antidifferentiation is called antiderivative method. This shows that antidifferentiation is actually a process in which one tries to “undo” a differentiation.
Remark 75. If an antiderivative of f exists, then it is not unique.
The following theorem strengthens the preceding remark.
Theorem 76. If F is an antiderivative of f on an interval I, then every antiderivative of f on I is
of the form F (x) + C, where C is an arbitrary constant.
If F is an antiderivative of f, we write
Z
f (x)dx = F (x) + C ,
where F (x) + C is the general antiderivative of f .
The following shows the immediate properties of indefinite integrals.
Z
Theorem 77.
1.
dx = x + C
2. A constant factor can be moved through an integral sign; i.e., if a is a constant, then
Z
Z
af (x)dx = a f (x)dx.
3. An antiderivative of a sum (difference) is the sum (difference) of the antiderivatives; i.e., if f
and g are defined on the same interval, then
Z
Z
Z
Z
Z
Z
[f (x) + g(x)]dx = f (x)dx + g(x)dx
[f (x) − g(x)]dx = f (x)dx − g(x)dx .
4. If n ∈ R and n ̸= −1, then
Z
MATH 21 DISCUSSION NOTES
xn dx =
xn+1
+ C.
n+1
S. C. TUASON (2023)
4.1
Antidifferentiation; Particular Antiderivatives; Integration by Substitution
47
The following enumerates the antiderivatives of trigonometric/circular functions.
Z
Z
Theorem 78.
1.
sin xdx = − cos x + C
4.
csc2 xdx = − cot x + C
Z
2.
Z
3.
4.1.2
Z
cos xdx = sin x + C
5.
sec2 xdx = tan x + C
6.
sec x tan xdx = sec x + C
Z
csc x cot xdx = − csc x + C
Particular Antiderivatives
Definition 37. If F (x) + C is the general antiderivative of f , then each antiderivative of f is called
a particular antiderivative of f .
If we wish to find a particular antiderivative F (x) of f (x), we must have first a given condition
called an initial or boundary condition, in which the particular antiderivative must satisfy.
4.1.3
Integration by Substitution
Suppose F is an antiderivative of f and g is a differentiable function. The Chain Rule tells us that
the derivative of F (g(x)) is
d
[F (g(x))] = F ′ (g(x))g ′ (x),
dx
which, in integral form, is
Z
F ′ (g(x))g ′ (x)dx = F (g(x)) + C,
and since F ′ = f as F is an antiderivative of f,
Z
f (g(x))g ′ (x)dx = F (g(x)) + C.
If we let u = g(x) so that du = g ′ (x)dx (since if viewed as differentials, then we can perform crossdu
multiplication on
= g ′ (x)), then the preceding equation yields
dx
Z
f (u)du = F (u) + C.
This shows the following theorem.
Theorem 79. If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I with antiderivative F , then
Z
Z
f (g(x))g ′ (x)dx = f (u)du = F (u) + C.
Remark 80. Easy substitutions occur when the integrand is the derivative of a known function,
except for a constant added to, subtracted from, multiplies, or divides the independent variable.
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.1
Antidifferentiation; Particular Antiderivatives; Integration by Substitution
48
Exercise 21.
1. Find the following antiderivatives.
Z Z
3
6
4 cos x12
(a)
x 4 − 5x4 − 4 dx
dx
(e)
x
x3 sin3 x12
Z
4z 2 − 6z + 3
√
(b)
dz
Z
3
z
Z
(f)
csc (2x + 1) cot(2x + 1) dx
(c)
4 csc x cot x + 2 sec2 x dx
Z p
Z
3
2
(d)
16n − 8n + 1 dn
(g)
(sin y − tan2 y) sec2 y dy
2. Solve completely: the point (3, 2) is on a curve, and at any point (x, y) on the curve, the
tangent line has a slope equal to 2x − 3. Find an equation of the curve.
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.2
Rectilinear Motion Revisited; Integrals Yielding Logarithmic and Exponential Functions
4.2
49
Rectilinear Motion Revisited; Integrals Yielding Logarithmic and Exponential Functions
In this set of notes, we resume the study of rectilinear motion using the tools of integration.
4.2.1
Rectilinear Motion Revisited
Recall that if a particle in rectilinear motion has position function s(t), then its instantaneous
velocity and acceleration are given by
v(t) = s′ (t)
and a(t) = v ′ (t).
In view of integrals, s(t) is an antiderivative (particular) of v(t) and v(t) is an antiderivative (particular) of a(t). That is,
Z
Z
s(t) = v(t) dt and v(t) = a(t) dt.
This shows that if v(t) is known, then integrating v(t) produces a family of position functions with
that velocity function. If, in addition, the position s0 of the particle at any time t0 is given, then
one can find the constant of integration and determine a unique position function.
Similarly, if a(t) is known, then integrating a(t) yields a family of velocity functions with that
acceleration function. If, in addition, the velocity v0 of the particle at any time t0 is given, then one
can find the constant of integration and determine a unique velocity function.
4.2.2
Antiderivatives (continued)
Z
1
du = ln |u| + C
Theorem 81.
1.
u
Z
2.
tan x dx = ln | sec x| + C
Z
cot x dx = ln | sin x| + C
3.
Z
sec x dx = ln | sec x + tan x| + C
4.
Z
csc x dx = ln | csc x − cot x| + C
5.
Z
6.
Z
7.
ax dx =
ax
+C
ln a
(a > 0 and a ̸= 1)
ex dx = ex + C
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.2
Rectilinear Motion Revisited; Integrals Yielding Logarithmic and Exponential Functions
Exercise 22.
50
1. Solve the following items completely.
(a) A ball is thrown vertically upwards from a platform 8 feet above the ground. After 2
seconds, the ball is 2 feet below the platform. Find the initial velocity of the ball. (Assume
acceleration due to gravity is -32 ft/s2 .)
(b) If the brakes on a car can give the car a constant negative acceleration of 8 m/s2 , what
is the maximum speed it can go to be able to stop within 25 meters after the brake is
applied?
2. Evaluate the following integrals.
Z
x2 − x
dx
(a)
1 + 3x2 − 2x3
Z
3x5 − 2x3 + 5x2 − 2
(b)
dx
x3 + 1
Z
log4 (x2 )
(c)
dx
x
Z
(d)
cot x cot(ln csc x) dx
Z
1
(e)
1 dx
2
x ex
Z √x 5√x
5 e
√
dx
(f)
x
Z
106x + 1
(g)
dx
103x
Z
3 − 2x
(h)
dx
3x
Z x
3 −7
(i)
dx
e2x
Z
e3x
(j)
dx
ex − 5
Z x
3 [cos(3x ) − sin(3x )]
dx
(k)
sin2 (3x )
Z log (cos x)
3 2
+ sec x
(l)
dx
cot x
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.3
Integrals Yielding Inverse Trig Fcns; Integrals of Hyperbolic Fcns; The Definite Integral
4.3
51
Integrals Yielding Inverse Trig Fcns; Integrals of Hyperbolic Fcns;
The Definite Integral
In this section, we continue exploring the antiderivatives of other classes of functions. We will also
introduce a concept that links the area (of a figure) to other concepts such as length, volume, density,
etc.: the definite integral.
4.3.1
Antiderivatives Yielding the Inverse Circular Functions
The following theorem shows the antiderivatives that yield the inverse circular functions.
Theorem 82. Suppose a > 0. Then
Z
u
du
√
+C
= sin−1
1.
a
a 2 − u2
Z
1
du
−1 u
2.
=
+C
tan
a2 + u2
a
a
Z
u
du
1
√
3.
= sec−1
+C
a
a
u u2 − a2
4.3.2
Antiderivatives of Hyperbolic Functions
The next theorem presents the antiderivatives of hyperbolic functions.
Z
Theorem 83.
1.
cosh x dx = sinh x + C
Z
2.
sinh x dx = cosh x + C
Z
sech2 x dx = tanh x + C
Z
csch2 x dx = − coth x + C
3.
4.
Z
sech x tanh x dx = −sech x + C
5.
Z
csch x coth x dx = −csch x + C
6.
Z
7.
tanh x dx = ln(cosh x) + C
Z
coth x dx = ln | sinh x| + C
8.
Z
9.
sech x dx = 2 tan−1 (ex ) + C = tan−1 (sinh x) + C
Z
10.
csch x dx = ln |csch x − coth x| + C
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.3
Integrals Yielding Inverse Trig Fcns; Integrals of Hyperbolic Fcns; The Definite Integral
4.3.3
52
The Definite Integral
Definition 38. Suppose f is defined on [a, b]. The definite integral of f from a to b is
Z
n
X
b
f (x) dx =
a
lim
max ∆xi →0
f (x∗i ) · ∆xi
i=1
if the limit exists and does not depend on the choice of partitions or the choice of numbers x∗i in the
subintervals. If the said limit exists, then f is integrable on [a, b].
Remark 84. The value of the definite integral is preserved when x is changed into a different
Z b
Z b
variable. That is,
f (x) dx =
f (t) dt.
a
a
Theorem 85. Suppose f is continuous on [a, b]. Then
1. f is integrable on [a, b], and
2. the net signed area A between the graph of f and [a, b] is
Z
A=
b
f (x) dx.
a
Example. Polynomial, sine, and cosine functions are integrable on any closed interval.
Exercise 23. Evaluate the following integrals.
Z
1
√
1.
dt
(t + 3) t2 + 6t − 7
Z
sec2 t
2.
dt
1 + 16 tan2 t
√
Z
csc2 ( 3 x)
√
dx
3.
√
3
x2 (4 + cot2 ( 3 x))
Z
x
√
4.
dx
8x − x2
Z
5.
sinh x tanh x dx
Z
csch (cot−1 x)
dx
1 + x2
Z
sech x(2 + coth2 x) dx
6.
7.
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.4
Properties of the Definite Integral; First and Second Fundamental Theorems of Calculus
4.4
53
Properties of the Definite Integral; First and Second Fundamental
Theorems of Calculus
In this set of notes, we continue the discussion of the definite integral. We will also establish the
connections between definite and indefinite integrals that we will call the Fundamental Theorem of
Calculus. This relates to rectangle and integration methods for calculating areas and also provides
a convenient method for evaluating definite integrals using antiderivatives.
4.4.1
Properties of the Definite Integral
The following are the properties of the definite integral.
Theorem 86. Let f and g be integrable on [a, b]. Then
Z b
Z a
1.
f (x) dx = −
f (x) dx
a
Z
b
a
f (x) dx = 0
2.
a
Z
b
c dx = c(b − a)
3.
a
Z
b
4.
Z
cf (x) dx = c
a
Z
b
f (x) dx
a
b
Z
[f (x) ± g(x)] dx =
5.
a
b
Z
f (x) dx ±
a
b
g(x) dx
a
6. If f is integrable on a closed interval containing a, b and c, then
Z b
Z c
Z b
f (x) dx =
f (x)dx +
f (x)dx
a
a
c
regardless of the order of a, b and c.
4.4.2
The Fundamental Theorem of Calculus
Theorem 87 (The First Fundamental
Theorem of Calculus). If f is continuous on [a, b] and x ∈
Z x
[a, b], then the function F (x) =
f (t) dt is an antiderivative of f , that is, F ′ (x) = f (x).
a
Suppose f is nonnegative and continuous on [a, b]. By Theorem 85.2., the area under the graph
of f is given by
Z b
A=
f (x) dx.
a
Suppose x ∈ [a, b]. If A(x) is the area under the graph of f from a to x, then
Z x
A(x) =
f (x) dx
a
′
and A (x) = f (x). So A(x) is an antiderivative of f (x) and every other antiderivative F (x) of f (x)
is obtained by adding a constant to A(x), i.e.,
F (x) = A(x) + C.
Note that the area under the curve from a to a is the area above the point a, that is,
Z a
A(a) =
f (x) dx,
a
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.4
Properties of the Definite Integral; First and Second Fundamental Theorems of Calculus
54
hence it is zero by Theorem 86.2. Also, note that the area under the curve from a to b is given by
Z
A(b) =
b
f (x) dx = A.
a
Observe that
Z
F (b) − F (a) = [A(b) + C] − [A(a) + C] = A(b) − A(a) = A − 0 = A =
b
f (x) dx.
a
This shows the second part of the Fundamental Theorem of Calculus.
Theorem 88 (The Second Fundamental Theorem of Calculus). If f is continuous on [a, b] with F
as its any antiderivative on [a, b], then
b
Z
f (x) dx = F (b) − F (a).
a
The preceding theorem states that the definite integral can be evaluated by finding an antiderivative of the integrand and subtracting its value at the lower limit of integration from its value at the
upper limit of integration.
Exercise 24.
Z
(a) If
1. Do as indicated.
Z 6
Z
f (x) dx = 3 and
g(x) dx = 8, find
6
−1
Z 2
(b) If
−1
Z 1
g(x) dx = 4 and
−1
6
[3f (x) − 2g(x)] dx.
−1
Z 1
3g(x) dx = 6, find
−1
g(x) dx.
2
2. Find the derivative of the following functions.
Z 2
t3
dt
(a) F (x) =
3
1−3x 1 + t
Z x1 p
1 + m3 dm
(b) F (x) =
1
Z cos x
(c) F (x) =
y 2 dy
sin x
3. Evaluate the following definite integrals.
Z 4
(a)
1 + 3t − t2 dt
0
π
4
1 + cos2 θ
dθ
cos2 θ
0
Z 2 p
(c)
y y − 1 dy
Z
(b)
1
Z
4
(d)
√
1
Z
1
√
3 dw
w ( w + 2)
2
|x + 1| dx
(e)
−2
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.5
Area of Plane Regions; Arc Length of Plane Curves
4.5
55
Area of Plane Regions; Arc Length of Plane Curves
Previously, we were able to show how to find the area between a curve y = f (x) and an interval on
the x-axis. Now, in this section, we wish to find the area bounded by arbitrary curves and express
it as a definite integral. We will also characterize a smooth function and obtain an expression for
its length.
4.5.1
Area of Plane Regions
The following shows the area of a plane region using the vertical approach (bounded above and below).
Theorem 89. If f and g are continuous on [a, b] and f (x) ≥ g(x) for all x ∈ [a, b], then the area
of the region R bounded above by y = f (x) and below by y = g(x), and the vertical lines x = a and
x = b is
Z b
AR =
(f (x) − g(x)) dx.
a
In the preceding, one can view f (x) − g(x) as the height and dx as the width of the rectangle at
an arbitrary point x in [a, b].
The following gives the area of a plane region using the horizontal approach (bounded on the
left and right).
Theorem 90. If u and v are continuous on [c, d] and v(y) ≥ u(y) for all y ∈ [c, d], then the area of
the region R bounded on the left by x = u(y), and on the right by x = v(y), and the horizontal lines
y = c and y = d is
Z d
AR =
(v(y) − u(y)) dy.
c
In the preceding, one can view v(y) − u(y) as the length and dy as the width of the rectangle at
an arbitrary point x in [c, d].
4.5.2
Arc Length of Plane Curves
We define the following.
Definition 39. If the derivative f ′ of f is continuous on [a, b], then the curve with the equation
y = f (x) is smooth on [a, b].
The following gives the arc length of smooth plane curves.
Theorem 91. If y = f (x) is a smooth curve on [a, b], then the arc length L of this curve from x = a
to x = b is
s
2
Z bp
Z b
dy
1+
1 + [f ′ (x)]2 dx.
L=
dx =
dx
a
a
If x = u(y) is a smooth curve on [c, d], then the arc length L of this curve from y = c to y = d is
s
2
Z d
Z dp
dx
L=
dy =
1+
1 + [u′ (y)]2 dy.
dy
c
c
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
4.5
Area of Plane Regions; Arc Length of Plane Curves
Exercise 25.
56
1. Find the area of the following regions.
(a) Triangle whose vertices are the points (−1, 4), (2, −2), and (5, 1)
(b) Trapezoid whose vertices are the points A(−1, −1), B(2, 2), C(6, 2), and D(7, −1)
(c) Region bounded by the upper branch of the parabola x = y 2 , the tangent line to the
parabola at (1, 1), and the x-axis
(d) Region bounded by y = 5 − x2 , y = 2x + 2, y = −x − 1
2. Find the arc length of the following curves in the specified interval.
3
(a) y = 32 (x − 5) 2 from the point where x = 6 to the point where x = 8
(b) 2x − 4y + 6 = 0 between y = 0 and y = 2
π
(c) y = ln(cos x) from x = 0 to x =
4
√
2
3
(d) 9y = 4x from the origin to the point (3, 2 3)
3. Set up a definite integral for the following:
(a) length of the arcsin function throughout its domain
√
(b) arc length of y = 3 x − 1 from x = 0 to x = 1
MATH 21 DISCUSSION NOTES
S. C. TUASON (2023)
Name:
Student Number:
Passport 4.1
Duration of completion:
mins
General Directions: Provide what is asked for in each item. Show complete solutions and box your final
answers. Work independently. Use blue or black ink only. Make sure to indicate above how many minutes
you finished this activity. Submit your answers on any blank sheet of paper (preferably 1/2 A4 paper,
crosswise, write in portrait) during official class hours.
Passport 3.6 (Seatwork)
(0.4 pts)
1. Evaluate the following integral.
Z
(1 − y)2
√
dy
1+y
“Any form of cheating in examinations or any act of dishonesty in relation to studies, such as plagiarism,
shall be subject to disciplinary action.”
Highest Possible Score: 0.4 pts