Republic of the Philippines
Carig Sur, Tuguegarao City
Cagayan State University
CARIG CAMPUS
Tel. No. (078) 304 - 0818
COLLEGE OF ENGINEERING AND ARCHITECTURE – CIVIL ENGINEERING DEPARTMENT
INTEGRAL CALCULUS
2ND SEMESTER AY 2021-2022
UNIT I
INTEGRATION CONCEPTS/FORMULAS
INTEGRAL CALCULUS
MODULE 1
INTEGRATION CONCEPTS/FORMULAS
In this chapter, we shall start our study of the inverse of differentiation and its various
applications. We have learned that differentiation is the process of finding the derivative or
differential of a given function. Clearly, we expect its inverse to be the process of finding the
function whose derivative or differential is given. The inverse of differentiation is called
integration. Many of the important applications of calculus depend on this operation.
Learning Objectives:
At the end of this lesson, the student will be able to:
1. Able to understand the concept of Integral Calculus.
2. Evaluate such indefinite integrals by applying the concepts or basic formulas.
1.1 The Indefinite Integrals
From the differential calculus, we know that by differential, a derivative or differential is
obtained from a given function. Suppose the derivative of a function is given, how can the function
be found? For instance, what function has for its derivative 3𝑥 or for its differential 3𝑥 𝑑𝑥? From
our experience with differentiation, we know that
(𝑥 ) = 3𝑥 𝑜𝑟 𝑑(𝑥 ) = 3𝑥 𝑑𝑥. Hence, our
answer to the question above is 𝑥 . But that is not the only possible answer. Recalling that the
derivative or differential of a constant is zero, we may also give the following functions as valid
answer:
𝑥 +4
𝑥 − 15
𝑥 + √2
Thus, we see that there is no unique answer to the question above. However, one thing quite
noticeable from the possible answers above is that, pairwise, the functions differ by a constant. In
fact, it can be proved that if two functions have the same derivative, their difference is a constant.
Thus, if we let C be any constant, then we may write 𝑥 + 𝐶 as our general answer.
The function 𝑥 + 𝐶 is called the antiderivative or the integral of 3𝑥 . The process of finding
the antiderivative is called antiderivative and integration instead of antidifferentiation. We may
define integration as the process of finding the function whose derivative or differential is given.
INTEGRAL CALCULUS
To denote the process of integration, the symbol ⎰ is used. Since 𝑑(𝑥 + 𝐶) = 3𝑥 𝑑𝑥, then
the statement “the integral of 3𝑥 is 𝑥 + 𝐶” can be expressed symbolically as
3𝑥 𝑑𝑥 = 𝑥 + 𝐶
Since C has no known value, the expression ∫ 3𝑥 𝑑𝑥 is called an indefinite integral. In general, the
indefinite integral of a function 𝑓(𝑥) is denoted by ∫ 𝑓(𝑥) 𝑑𝑥 and defined symbolically as follows:
𝒇(𝒙) 𝒅𝒙 = 𝑭(𝒙) + 𝑪
𝒊𝒇 𝒅𝑭(𝒙) = 𝒇(𝒙)𝒅𝒙
We call the symbol ⎰ the integral sign, 𝑓(𝑥) the integrand, C, the constant of integration and
𝐹(𝑥) + 𝐶 the value of the indefinite integral ⎰ 𝑓(𝑥)𝑑𝑥. The differential dx indicates that x is the
variable of integration.
As symbols of operation, d and ⎰ are inverse to each other. Let us see what happens if d
precedes ⎰ or when it follows ⎰. Suppose 𝑓(𝑥) and 𝐹(𝑥) are functions whose relationship is given
by
𝑑𝐹(𝑥) = 𝑓(𝑥)𝑑𝑥
Integrating both sides
𝐹(𝑥) 𝑑𝑥 =
𝑓(𝑥) 𝑑𝑥
𝐹(𝑥) 𝑑𝑥 = 𝐹(𝑥) + 𝐶
The result above tells us that integrating the differential of a function gives that function plus an
arbitrary constant.
BASIC INTEGRATION FORMULA
I.1. ∫ 𝒅𝒖 = 𝒖 + 𝑪
I.2. ∫(𝒖 + 𝒗)𝒅𝒖 = ∫ 𝒖 𝒅𝒙 + ∫ 𝒗 𝒅𝒙
I.3. ∫ 𝒂𝒖𝒅𝒙 = 𝒂 ∫ 𝒖 𝒅𝒙
I.4. ∫ 𝒖𝒏 𝒅𝒙 =
I.5. ∫
𝒅𝒖
𝒖
𝒖𝒏 𝟏
𝒏 𝟏
+𝑪
= 𝐥𝐧 𝒖 + 𝑪
,𝒏 ≠ 𝟏
We wish to make the following important remarks in
using these formulas:
1. Formula I.2. can be extended to the sum of any
finite number of differentials.
2. Formula I.3. tells us that a constant may be moved
across the integral sign. (Note: You can not do this
to a variable.)
3. Formula I.4. is used for finding the integral of a
power of a function. Note that it holds for any real
number n except n = -1. Note further that if u = x,
I.4. simplifies to
𝑥
+𝐶
𝑛+1
INTEGRAL CALCULUS
PROOF of I.1: Differentiating the right member of I.1,
𝑑(𝑢 + 𝐶) = 𝑑𝑢 + 0 = 𝑑𝑢
This completes the proof. In fact, if we
replace F(x) by u from the formula
∫ 𝐹(𝑥) 𝑑𝑥 = ∫ 𝑓(𝑥) 𝑑𝑥, we get I.1.
PROOF of I.2: We must show that the differential of the right member is (u + v) dx. Hence,
differentiating the right member,
𝑑(∫ 𝑢𝑑𝑥 + ∫ 𝑣𝑑𝑥) = 𝑑 ∫ 𝑢𝑑𝑥 + 𝑑 ∫ 𝑣𝑑𝑥
= 𝑢𝑑𝑥 + 𝑣𝑑𝑥
= (𝑢 + 𝑣)𝑑𝑥
PROOF of I.3: Differentiating the right member,
𝑑(𝑎 ∫ 𝑢𝑑𝑥) = 𝑎(𝑑 ∫ 𝑢𝑑𝑥)
= 𝑎(𝑢𝑑𝑥)
PROOF of I.4: Differentiating the right member,
𝑑
+𝐶 =𝑑
=
(
+ 𝑑𝐶
)(
)
+0
=𝑢
PROOF of I.5: Differentiating the right member,
Case (1) if u > 0: 𝑑(ln 𝑢 + 𝐶) = 𝑑 ln 𝑢 + 𝑑𝑐
= 𝑑𝑢 + 0
=
Case (2) if u < 0: 𝑑(ln(−𝑢) + 𝐶) = 𝑑 ln(−𝑢) + 𝑑𝑐
= − (−𝑑𝑢) + 0
=
Since 𝑢 = |𝑢| 𝑖𝑓 𝑢 > 0 𝑎𝑛𝑑 |𝑢| = −𝑢 𝑖𝑓 𝑢 < 0, 𝑡ℎ𝑒𝑛 𝑖𝑛 𝑒𝑖𝑡ℎ𝑒𝑟 𝑐𝑎𝑠𝑒 𝐼. 5 ℎ𝑜𝑙𝑑𝑠.
INTEGRAL CALCULUS
EXAMPLE 1.1: Evaluate ∫(5𝑥 + 3𝑥 + 6 )𝑑𝑥
SOLUTION:
∫(5𝑥 + 3𝑥 + 6 )𝑑𝑥 = ∫ 5𝑥 𝑑𝑥 + ∫ 3𝑥 𝑑𝑥 + ∫ 6𝑑𝑥
= 5 ∫ 𝑥 𝑑𝑥 + 3 ∫ 𝑥 𝑑𝑥 + 6 ∫ 𝑑𝑥
=5
+𝑐 +3
=5
+ 𝑐 + 6𝑥 + 𝑐
+𝑐 +3
+ 𝑐 + 6𝑥 + 𝑐
= 𝑥 + 𝑥 + 6𝑥 + (𝑐 + 𝑐 + 𝑐 )
= 𝒙𝟓 + 𝒙𝟑 + 𝟔𝒙 + 𝑪
where 𝐶 = 𝑐1 + 𝑐2 + 𝑐3 . In practice,
the above integral is simply
evaluated
in
the following
manner:
∫(5𝑥 + 3𝑥 + 6 )𝑑𝑥
=5
+3
+ 6𝑥 + 𝐶
= 𝒙𝟓 + 𝒙𝟑 + 𝟔𝒙 + 𝑪
EXAMPLE 1.2: Evaluate ∫(3𝑥 + 4 ) 𝑑𝑥
SOLUTION:
∫(3𝑥 + 4 ) 𝑑𝑥 = ∫(9𝑥 + 24𝑥 + 16)𝑑𝑥
= 9 ∫ 𝑥 𝑑𝑥 + 24 ∫ 𝑥 𝑑𝑥 + 16 ∫ 𝑑𝑥
=9
+ 24
=9
+ 24
+ 16𝑥 + 𝐶
+ 16𝑥 + 𝐶
= 𝟑𝒙𝟓 + 𝟏𝟐𝒙𝟑 + 𝟏𝟔𝒙 + 𝑪
EXAMPLE 1.3: Evaluate ∫(
+ ) 𝑑𝑥
SOLUTION:
∫(
+ ) 𝑑𝑥 = ∫ 4𝑥
= 4∫𝑥
=4
+
𝑑𝑥
𝑑𝑥 + 2 ∫
+ 2(ln|𝑥|) + 𝐶
=4
+ 2(ln|𝑥|) + 𝐶
𝟐
= − 𝟐 + 𝟐 𝒍𝒏 |𝒙| + 𝑪
𝒙
INTEGRAL CALCULUS
1.2 Integration by Substitution
Some integrals can not be evaluated readily by direct application of the standard
integration formulas. The techniques for evaluating such integrals leans heavily on what is known
as the method of substitution. This method involves a change of variable, say from x to another
variable u. The purpose of substituting a new variable is to bring the problem to a form for which a
standard formula can be applied. This integration by substitution is justified by the so-called Chain
Rule for Integration which we shall briefly state below.
Let F(u) be a function whose derivative is f(u), that is,
𝐹(𝑢) = 𝑓(𝑢). If u is a
differentiable function of x, say u = h(x) then
𝑓(𝑢) 𝑑𝑢 =
𝑓[ℎ(𝑥)] ℎ′(𝑥) 𝑑𝑥
EXAMPLE 1.4: Evaluate ∫(3𝑥 + 4) 𝑑𝑥
Let 𝑢 = 3𝑥 + 4. Then 𝑑𝑢 = 3𝑑𝑥, and 𝑛𝑓 = . Hence,
∫(3𝑥 + 4) 𝑑𝑥 = ∫ 𝑢 𝑑𝑢
=
=
+𝐶
(𝟑𝒙 𝟒)𝟑
𝟗
+𝑪
“nf” or neutralizing factor is the
reciprocal of the number which we
wish to introduce after the integral
sign or preceding dx. That is in the
example 1.4 nf = 1/3. With this new
idea, our solution can be shortened
further. Thus, knowing that nf = 1/3,
we may integrate the given integral
right away.
EXAMPLE 1.5: Evaluate ∫(𝑥 − 1) 𝑥𝑑𝑥
Let 𝑢 = 𝑥 − 1. Then 𝑑𝑢 = 2𝑥𝑑𝑥, and 𝑛𝑓 = . Hence,
∫(𝑥 − 1) 𝑥 𝑑𝑥 = ∫ 𝑢 𝑑𝑢
=
=
+𝐶
𝒙𝟐 𝟏
𝟏𝟎
𝟓
+𝑪
EXAMPLE 1.6: Evaluate ∫(sin 4𝑥 cos 4𝑥) 𝑑𝑥
Let 𝑢 = sin 4𝑥 . Then 𝑑𝑢 = 4 𝑐𝑜𝑠 4𝑥 𝑑𝑥, and 𝑛𝑓 = . Hence,
∫(sin 4𝑥 cos 4𝑥) 𝑑𝑥 = ∫ 𝑢 𝑑𝑢
=
=
+𝐶
(𝒔𝒊𝒏 𝟒𝒙)𝟔
𝟐𝟒
+𝑪
INTEGRAL CALCULUS
EXAMPLE 1.7: Evaluate ∫(
Solution:
)𝑑𝑥
Here f(x) = 2𝑥 − 6𝑥 + 4 and g(x) = 𝑥 − 3. Carrying out the indicated division, we
get
In this example, we are considering a rational
fraction, 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑓(𝑥) ≥ 𝑑𝑒𝑔𝑟𝑒𝑒 𝑜𝑓 𝑔(𝑥).
= 2𝑥 +
( )
Where Q(x) = 2x and R(x) = 4
( )
Therefore,
∫(
= 𝑄(𝑥) +
( )
𝑔(𝑥)
Where:
)𝑑𝑥 = ∫(2𝑥)𝑑𝑥 + ∫(
=
)𝑑𝑥
+ 4 ln |𝑥 − 3| + 𝐶
Q(x) – quotient
R(x) – remainder of lower degree than g(x)
= 𝒙𝟐 + 𝟒 𝐥𝐧 |𝒙 − 𝟑| + 𝑪
1.3 Integration of Trigonometric Functions
The standard formulas for evaluating the integrals of six trigonometric functions are given
below. The first two can be verified by differentiation and remaining four may be proved directly.
TF.1. ∫ 𝐬𝐢𝐧 𝒖 𝒅𝒖 = − 𝐜𝐨𝐬 𝒖 + 𝑪
TF.2. ∫ 𝐜𝐨𝐬 𝒖 𝒅𝒖 = 𝐬𝐢𝐧 𝒖 + 𝑪
TF.3. ∫ 𝐭𝐚𝐧 𝒖 𝒅𝒖 = − 𝐥𝐧|𝒄𝒐𝒔 𝒖| + 𝑪
TF.4. ∫ 𝐜𝐨𝐭 𝒖 𝒅𝒖 = 𝐥𝐧|𝐬𝐢𝐧 𝒖| + 𝑪
TF.5. ∫ 𝐬𝐞𝐜 𝒖 𝒅𝒖 = 𝐥𝐧 | 𝐬𝐞𝐜 𝒖 + 𝐭𝐚𝐧 𝒖 | + 𝑪
TF.6. ∫ 𝐜𝐬𝐜 𝒖 𝒅𝒖 = − 𝐥𝐧 | 𝐜𝐬𝐜 𝒖 + 𝐜𝐨𝐭 𝒖 | + 𝑪
PROOF of TF.1: Since 𝑑(− cos 𝑢 + 𝐶) = sin 𝑢 𝑑𝑢, then we have verified that TF.1. is correct.
PROOF of TF.3: ∫ tan 𝑢 𝑑𝑢 = ∫
= −∫
= −∫
𝑑𝑢
(
)
(
𝑑𝑢
)
= − ln|cos 𝑢| + 𝐶
INTEGRAL CALCULUS
(
PROOF of TF.5: ∫ sec 𝑢 𝑑𝑢 = ∫
)
𝑑𝑢
=∫
=∫
𝑑𝑢
(
)
= ln|sec 𝑢 + tan 𝑢| + 𝐶
The following formulas are also consequences of their corresponding differentiation formulas. For
instances, TF.7. is correct since d(tan u + C) = sec 2 u du.
TF.7. ∫ 𝐬𝐞𝐜 𝟐 𝒖 𝒅𝒖 = 𝐭𝐚𝐧 𝒖 + 𝑪
TF.8. ∫ 𝐜𝐬𝐜 𝟐 𝒖 𝒅𝒖 = − 𝐜𝐨𝐭 𝒖 + 𝑪
TF.9. ∫ 𝐬𝐞𝐜 𝒖 𝐭𝐚𝐧 𝒖 𝒅𝒖 = 𝐬𝐞𝐜 𝒖 + 𝑪
TF.10. ∫ 𝐜𝐬𝐜 𝐮 𝐜𝐨𝐭 𝒖 𝒅𝒖 = − 𝐜𝐬𝐜 𝒖 + 𝑪
EXAMPLE 1.8: Evaluate ∫ sin 4𝑥 𝑑𝑥
Solution: This takes the form of TF.1. with 𝑢 = 4𝑥. Then 𝑑𝑢 = 4𝑑𝑥 and 𝑛𝑓 = . Hence
applying TF.1., we have,
𝟏
∫ sin 4𝑥 𝑑𝑥 = ∫ sin 𝑢 𝑑𝑢 = (− cos 4𝑥) + 𝐶 = − 𝟒 𝐜𝐨𝐬 𝒖 + 𝑪
EXAMPLE 1.9: Evaluate ∫(
)𝑑𝑥
Solution: This takes the different from integration of trigonometric functions. with 𝑢 = 2𝑥.
Then 𝑑𝑢 = 2𝑑𝑥 and 𝑛𝑓 = . Hence applying integration of trigonometric functions
formulas, we have,
∫(
)𝑑𝑥 = ∫(
)𝑑𝑥 + ∫(
)𝑑𝑥
= ∫(sec 2𝑥)𝑑𝑥 + ∫(tan 2𝑥)𝑑𝑥
𝟏
𝟏
𝟐
𝟐
= 𝐥𝐧|𝐬𝐞𝐜 𝟐𝒙 + 𝐭𝐚𝐧 𝟐𝒙| − 𝐥𝐧|𝐜𝐨𝐬 𝟐𝒙| + 𝑪
EXAMPLE 1.10: Evaluate ∫(1 + sec 𝑥) 𝑑𝑥
∫(1 + sec 𝑥) 𝑑𝑥 = ∫(1 + 2 sec 𝑥 + sec 𝑥)𝑑𝑥
= ∫ 𝑥 𝑑𝑥 + ∫ 2 sec 𝑥 𝑑𝑥 + ∫ sec 𝑥 𝑑𝑥
= 𝒙 + 𝟐 𝒍𝒏 |𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏 𝒙| + 𝒕𝒂𝒏 𝒙 + 𝑪
INTEGRAL CALCULUS
1.4 Integration of Exponential Functions
The following formulas for evaluating the integrals of exponential functions can be proved
by differentiation.
EF.1. ∫ 𝒆𝒖 𝒅𝒖 = 𝒆𝒖 + 𝑪
EF.2. ∫ 𝒂𝒖 𝒅𝒖 =
𝒂𝒖
𝐥𝐧 𝒂
+𝑪
EXAMPLE 1.11: Evaluate ∫ 𝑒
𝒂 > 𝟎, 𝒂 ≠ 𝟏
𝑑𝑥
Solution: hence 𝑢 = 4𝑥. Then 𝑑𝑢 = 4𝑑𝑥 and 𝑛𝑓 = . applying EF.1, we have,
∫𝑒
𝟏
𝑑𝑥 = ∫ 𝑒 𝑑𝑢 = 𝒆𝟒𝒙 + 𝑪
𝟒
EXAMPLE 1.12: Evaluate ∫ 4
𝑑𝑥
Solution: this takes the 𝑎 with 𝑎 = 4 𝑎𝑛𝑑 𝑢 = 3𝑥. Then 𝑑𝑢 = 3𝑑𝑥 and 𝑛𝑓 = . applying
EF.2, we have,
∫4
𝑑𝑥 = ∫ 𝑎 𝑑𝑢 = ∗
+𝐶 =
+𝐶 =
𝟒𝟑𝒙
𝐥𝐧 𝟔𝟒
+𝑪
1.5 Integration of Hyperbolic Functions
The following formulas are used for evaluating the integrals of hyperbolic functions.
Formulas H1 to H6 may be verified by differentiation. For example, H1 is correct since
𝑑(sinh 𝑢 + 𝐶) = cosh 𝑢 𝑑𝑢. The student may give the proof of H7 and H8.
HF.1. ∫ 𝒄𝒐𝒔𝒉 𝒖 𝒅𝒖 = 𝐬𝐢𝐧𝐡 𝒖 + 𝑪
HF.2. ∫ 𝒔𝒊𝒏𝒉 𝒖 𝒅𝒖 = 𝒄𝒐𝒔𝒉 𝒖 + 𝑪
The hyperbolic functions are defined in terms of the
exponential functions:
sinh 𝑥 =
cosh 𝑥 =
HF.4. ∫ 𝒄𝒔𝒄𝒉𝟐 𝒖 𝒅𝒖 = −𝐜𝐨𝐭𝐡 𝒖 + 𝑪
tanh 𝑥 =
coth 𝑥 =
HF.5. ∫ 𝒔𝒆𝒄𝒉 𝒖 𝒕𝒂𝒏𝒉 𝒖 𝒅𝒖 = − 𝐬𝐞𝐜𝐡 𝒖 + 𝑪
sech 𝑥 =
csch 𝑥 =
HF.6. ∫ 𝒄𝒔𝒄𝒉 𝒖 𝒄𝒐𝒕𝒉 𝒖 𝒅𝒖 = − 𝐜𝐬𝐜𝐡 𝒖 + 𝑪
The hyperbolic functions have identities that are
similar to those of trigonometric functions:
HF.3. ∫ 𝒔𝒆𝒄𝒉𝟐 𝒖 𝒅𝒖 = 𝐭𝐚𝐧𝐡 𝒖 + 𝑪
HF.7. ∫ 𝒕𝒂𝒏𝒉 𝒖 𝒅𝒖 = 𝐥𝐧 |𝐜𝐨𝐬𝐡 𝒖| + 𝑪
HF.8. ∫ 𝒄𝒐𝒕𝒉 𝒖 𝒅𝒖 = 𝒍𝒏 |𝐬𝐢𝐧𝐡 𝒖| + 𝑪
cosh 𝑥 − sinh 𝑥 = 1
sinh 2𝑥 = 2 sinh 𝑥 cosh 𝑥
1 − tanh 𝑥 = sech 𝑥
cosh 2𝑥 = cosh 𝑥 + sinh 𝑥
coth 𝑥 − 1 = csch 𝑥
INTEGRAL CALCULUS
EXAMPLE 1.13: Evaluate ∫ cosh(3𝑥 + 4) 𝑑𝑥
Solution: hence 𝑢 = 3𝑥 + 4. Then 𝑑𝑢 = 3𝑑𝑥 and 𝑛𝑓 = . Then,
𝟏
∫ cosh(3𝑥 + 4) 𝑑𝑥 = ∫ cosh 𝑢 𝑑𝑢 = 𝟑 𝒔𝒊𝒏𝒉 (𝟑𝒙 + 𝟒) + 𝑪
EXAMPLE 1.14: Evaluate ∫ 𝑥 tanh 𝑥 𝑑𝑥
Solution: hence 𝑢 = 𝑥 . Then 𝑑𝑢 = 2𝑥𝑑𝑥 and 𝑛𝑓 = . Then,
𝟏
∫ 𝑥 tanh 𝑥 𝑑𝑥 = ∫ tanh 𝑢 𝑑𝑢 = 𝟐 𝒍𝒏 𝒄𝒐𝒔𝒉 𝒙𝟐 + 𝑪
EXAMPLE 1.15: Evaluate ∫ 𝑒 sinh 𝑥 𝑑𝑥
Solution: since sinh 𝑥 =
∫𝑒
𝑒𝑥 −𝑒−𝑥
2
, we obtain
𝑑𝑥 = ∫(
)𝑑𝑥 = ∫(𝑒
=
=
𝑒
𝒆𝟐𝒙
𝟒
− 1)𝑑𝑥 = ∫ 𝑒 𝑑𝑥 − ∫ 𝑑𝑥
−𝑥 +𝐶
𝒙
− +𝑪
𝟐
EXAMPLE 1.16: Evaluate ∫ sinh 𝑥 𝑑𝑥
Solution: sinh 𝑥 = sinh 𝑥 sinh 𝑥, and sinh 𝑥 = cosh 𝑥 − 1, then
∫ sinh 𝑥 𝑑𝑥 = ∫ sinh 𝑥 sinh 𝑥 𝑑𝑥 = ∫(cosh 𝑥 − 1) sinh 𝑥 𝑑𝑥
= ∫(𝑢 − 1)𝑑𝑢
=
=
−𝑢+𝐶
𝐜𝐨𝐬𝐡𝟑 𝒙
𝟑
− 𝐜𝐨𝐬𝐡 𝒙 + 𝑪
𝑢 = 𝑐𝑜𝑠ℎ 𝑥 , 𝑑𝑢 = sinh 𝑥 𝑑𝑥
INTEGRAL CALCULUS
1.6 Applications of Indefinite Integration
Indefinite integration finds applications in some geometrical and physical problems in
physics, chemistry, mathematicians and engineering. We shall illustrate these with some examples
after our discussion of some important concepts which we have to understand before solving such
problems.
An equation which involves derivatives or differentiation called a differential equation.
From the previous topics, the function F(x) and f(x) are related by the equation
dF(x)=f(x)dx
if we let y = F(x), then we have,
dy = dF(x)=f(x)dx
= f(x)
The above equations are the simplest type of differential equation where f(x) is known function
and y is an unknown function of x. If we integrate both sides of dy = f(x)dx, we have
𝑦 + 𝐶 = 𝐹(𝑥) + 𝐶
Or
𝑦 = 𝐹(𝑥) + 𝐶 − 𝐶
If we let 𝐶 = 𝐶 − 𝐶 , then it simplifies to
𝒚 = 𝑭(𝒙) + 𝑪
A more general type of differential equation is given by the form
𝑑𝑦 𝑓(𝑥)
=
𝑑𝑥 𝑔(𝑦)
where f(x) and g(y) are known functions of x and y respectively and y is an unknown function of x.
The equation above may be written in the form
𝒈(𝒚)𝒅𝒚 = 𝒇(𝒙)𝒅𝒙
INTEGRAL CALCULUS
so that no x’s appear on one side and no y’s appear on the other side of the equation. The variables
are said to be separated. For example, the variables of the differential equation
=
can be
separated by writing as 3ydy = 2xdx. The later form may now be integrated. Thus
3𝑦𝑑𝑦 =
2𝑥𝑑𝑥
3𝑦
=𝑥 +𝐶
2
An equation of the form 𝑦 = 𝐹(𝑥) + 𝐶 may represent the equation of a family of curves having a
common property, say, the same slope at any point (x, y). Through a specific point 𝑃 (𝑥 , 𝑦 ), there
passes just one member of the family. The value of C for the member passing through 𝑃 can thus
be obtained by substituting the coordinates of 𝑃 in the equation of the family. This is illustrated in
the following example.
EXAMPLE 1.17: Find the equation of the family of curves whose slope at any point is 2x. Find also
the equation of the member which passes through the point (2, 1).
Solution: Since the slope at any point (x, y) is 2x, then we have
𝑑𝑦
= 2𝑥
𝑑𝑥
𝑑𝑦 = 2𝑥 𝑑𝑥
Integrating both sides
𝑑𝑦 =
2𝑥𝑑𝑥
𝑦 =𝑥 +𝐶
The equation 𝑦 = 𝑥 + 𝐶 is the equation of the family of curves with slope 2x. To
find the equation of the member through (2, 1), we substitute x = 2 and y = 1 then
solve for the value of C. Thus
𝑦 =𝑥 +𝐶
1=2 +𝐶
𝐶 = −3
Therefore, the required equation of the specific member of the family of curves is
𝒚 = 𝒙𝟐 − 𝟑
INTEGRAL CALCULUS
From the differential calculus of rectilinear motion, we have defined the velocity v of a moving
particle as the time rate of change of the distance s and the acceleration a as the time rate of
change of the velocity v. In symbol, we write
𝒗=
𝒅𝒔
and
𝒅𝒕
𝒂=
𝒅𝒗
𝒅𝒕
We have solved problems dealing with such quantities by using the process of differentiation. This
time, we shall solve such problems by use of integration.
EXAMPLE 1.18: A body is thrown vertically upward from the ground with an initial velocity 96
ft/sec. Find the maximum heigh attained by the body.
Solution: Let:
s – distance of the body from the ground
v – velocity at any time t
Note:
v > 0 since the body is directed upward
a < 0 since the body is thrown vertically upward and gravity imparts an
acceleration downward
Given: t = 0; v = 96 ft/sec; s = 0 ft.
Since that a < 0, then we start with
a = -32 ft/sq.sec
But by definition, a = dv/dt. Hence,
𝑑𝑣
= −32 𝑓𝑡/𝑠
𝑑𝑡
Or
𝑑𝑣 = −32 𝑑𝑡
Integrating, we have
𝑣 = −32𝑡 + 𝐶
But v = 96 ft/sec when t = 0. Substituting these in the equation above and solving for
C, we get
INTEGRAL CALCULUS
96 = -32(0) + C
C = 96
Then the equation becomes,
𝑣 = −32𝑡 + 96
This equation expresses the velocity of the body at any time t. Since v = ds/dt, then
𝑑𝑠
= −32𝑡 + 96
𝑑𝑡
Or
𝑑𝑠 = −32𝑡𝑑𝑡 + 96𝑑𝑡
Integrating, we have
𝑠 = −16𝑡 + 96𝑡 + 𝐶′
Again, when t = 0, we have s = 0. Hence
0 = −16(0) + 96(0) + 𝐶′
𝐶 =0
Then the equation becomes
𝑠 = −16𝑡 + 96𝑡
Form equation above which is expresses the distance of the body from the ground at
any time t. To find the maximum height attained by the body, we must know the
time required to reach that height. Since v = 0 at the maximum height, then
𝑣 = −32𝑡 + 96
0 = −32𝑡 + 96
32𝑡 = 96
𝑡 = 3 𝑠𝑒𝑐
Then solve for the maximum height
𝑠 = −16𝑡 + 96𝑡
𝑠 = −16(3) + 96(3)
𝒔 = 𝟏𝟒𝟒 𝒇𝒕.
INTEGRAL CALCULUS
For activity: Assess yourself base on your understanding from this module.
A. Evaluate the following and simply the answer to its simplest form.
1. ∫
2. ∫
√
(
−
+
)
𝑑𝑥
𝑑𝑥
√
3. ∫ √1 + 2 sin 3𝑥 cos 3𝑥 𝑑𝑥
4. ∫
𝑑𝑥
5. ∫
𝑑𝑥
6. ∫
𝑑𝑥
7. ∫(𝑒
+ sec 2𝑡 tan 2𝑡)𝑑𝑥
8. ∫(𝑥 cos(𝑥 + 1) +
9. ∫
)𝑑𝑥
𝑑𝑥
10. ∫ 𝑒 sech 𝑥 𝑑𝑥
B. Find the equation of the family of curves whose slope at any point (x, y) is
1. 3𝑥 + 4
2.
3.
𝑦; 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (1, 1)
4. 𝑒 ; 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (0, 1)
5. 𝑥
; 𝑡ℎ𝑟𝑜𝑢𝑔ℎ (1, 2)
C. With what velocity must an object be thrown vertically upward so that it will attain a
maximum height of 900 ft from its starting point?
D. From the top of a building 256 ft high, an object is thrown vertically upward with an initial
velocity of 96 ft/sec. When will it strike the ground?
INTEGRAL CALCULUS
THANK
YOU!!!