Basic calculus anti-derivative of a functions grade PDF

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BASIC CALCULUS Quarter 4 – Module 3: Antiderivative using Substitution and Problems involving Antidifferentiation

Basic Calculus Alternative Delivery Mode Quarter 4 – Module 3: Antiderivative Using Substitution and Problems involving Antidifferentiation First Edition, 2021 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

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Basic Calculus Quarter 4 – Module 3: Antiderivative Using Substitution and Problems involving Antidifferentation

Introductory Message This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson. Each SLM is composed of different parts. Each part shall guide you step-bystep as you discover and understand the lesson prepared for you. Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to monitor your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these. In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-ba sed learning. Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task. If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Thank you.

What I Need to Know

This module was designed and written to help you master the substitution technique used in taking the antiderivative of a function . Also, the scope of this module allows you to develop critical thinking skills through the guided word problems. This can be used in many different learning situations. The language used recognizes the diverse vocabulary level of students. The module is divided into two lessons, specifically: Lesson 1 - Finding the Antiderivative of a Function using Substitution Lesson 2 – Problem Solving Involving Antidifferentiation After going through this module, you are expected to: 1.

Compute the antiderivative of a function using substitution rule.

2.

Solve problems involving antidifferentiation.

1

What I Know

Part A. Directions. Find the antiderivative of the following integrals. Write your answers on a separate sheet of paper. 1. ∫(1 − 2𝑡)𝑑𝑡

2. ∫(25𝑥 2 + 30𝑥 + 9) (5𝑥 + 3)𝑑𝑥 1

3. ∫ 3 𝑒 3𝑥 𝑑𝑥

4. ∫(2𝑥√𝑥2 − 3) 𝑑𝑥 5. ∫ sin 7𝑥 𝑑𝑥 6. ∫

1

√1−𝑥

𝑑𝑥

2

7. ∫ 2𝑥+3 𝑑𝑥

8. ∫ 𝑠𝑒𝑐 5𝑥 𝑑𝑥 1

9. ∫ 𝑥 𝑙𝑛𝑥 𝑑𝑥 10. ∫ (𝑥 2

7 𝑑𝑥 −14𝑥+49)

Part B. Directions. Analyze and solve the following problems. Write your answers on a separate sheet of paper. An object is thrown vertically upward with an initial velocity of 75ft/sec from a building 40ft above the ground. (a = 32ft/sec2) 1. Determine the equation of the distance s from the ground at a given time t in seconds. 2. Calculate the distance of the object from the ground after 2 seconds.

2

Lesson

1

Antiderivative of Functions Using Substitution

In the previous lessons, different rules of antidifferentiation are presented and illustrated through examples and exercises. Let us take a few moments to recall those rules by answering the proceeding exercises.

What’s In Directions. Match Column A and B with antiderivative of the following integrands. Write the letters of your answers in a separate sheet of paper.

COLUMN A

1) ∫ 6𝑥𝑑𝑥 2) ∫(16𝑥 3 + 9𝑥 2 + 3𝑥 − 5)𝑑𝑥 3) ∫ 3𝑥 𝑑𝑥

+𝑐

d.

𝑒𝑥 3

+𝑐

f.

1 2𝑥 2

COLUMN B

b. +𝑐 c. 3 𝑠𝑖𝑛 𝑥 + 𝑐

1

5

3𝑥2

e. tan 𝑥 + 𝑐

5) ∫ 𝑑𝑥 𝑥

6) ∫ 𝑠𝑒𝑐 2𝑥 𝑑𝑥 7) ∫ 3 cos 𝑥 𝑑𝑥

+𝑥+𝑐

g. 5 𝑙𝑛|𝑥| + 𝑐

3

h. 4𝑥 4 + 3𝑥 3 + 2 𝑥 2 − 5𝑥 + 𝑐

1

8) ∫ 7 𝑑𝑥 𝑥

9) ∫ 9√𝑢 𝑑𝑢 ∫

𝑙𝑛3

a.

4) ∫ 𝑒 𝑥 𝑑𝑥 3

10)

3𝑥

i.

𝑥3 −1 𝑑𝑥 𝑥3

j.

3

6𝑢 2 + 𝑐 1

− 6𝑥 6 + 𝑐

k. 𝑥 +

1 2𝑥 2

+𝑐

Notes to the Teacher However, these may not be adequate to integrate a given function. The question of which rule applies to a given situation depends mainly on one’s familiarity of various antidifferentiation techniques.

3

What’s New

This lesson will present the most basic technique in taking the antiderivative of a function which is antidifferentiation by substitution. This method is the inverse of the chain rule in differentiation.

Chain Rule for Integration The integration by substitution is justified through the Chain Rule of Integration. Let F(u) be a function whose derivative is f(u), that is, is a differentiable function of x, say 𝑢 = ℎ(𝑥) then,

𝑑

𝑑𝑥

𝐹(𝑢) = 𝑓(𝑢). If u

∫ 𝑓(𝑢) = 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥 Here is how to prove this rule. Given that: 𝑑 𝐹(𝑢) = 𝑓(𝑢) 𝑑𝑥 𝑑 𝐹 (𝑢) = 𝑓(𝑢)

Integrating both sides Since 𝑢 = ℎ(𝑥) then

𝑑𝑢

𝑑𝑥

∫ 𝑑 𝐹 (𝑢) = ∫ 𝑓(𝑢) 𝑑𝑢.

= ℎ′(𝑥), by the chain rule of differentiation we have: 𝑑𝐹(𝑢) = 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥

Or 𝑑𝐹 (𝑢) = 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥 Integrating both sides ∫ 𝑑𝐹(𝑢) = ∫ 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥 Using Transitive Property of Equality. If ∫ 𝑑 𝐹 (𝑢) = ∫ 𝑓(𝑢) 𝑑𝑢

and ∫ 𝑑𝐹(𝑢) = ∫ 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥 then, ∫ 𝑓(𝑢)𝑑𝑢 = ∫ 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥

4

What is It

Integration by Substitution This technique is a great help in dealing with integrals that cannot be evaluated readily by direct application of the standard integration formulas, particularly, if the integrand is the result of a differentiation using the chain-rule. Integration by substitution primarily simplifies an integrand by setting a part of the integrand to u and part of integrand with d(u). It is also described as a change of variables because we are replacing variables to obtain an expression wherein integration rules can be readily applied. Usually, we make a substitution for a function whose derivative occurs in the integrand. This section is dedicated to help us learn the skills of easily identifying the term/s in the integrand as a useful substitute.

Substitution with Indefinite Integrand Let 𝑢 = ℎ(𝑥), where ℎ′(𝑥) is continuous over an interval, let 𝑓(𝑥) be continuous over corresponding range of g and let 𝐹(𝑥) be and antiderivative of 𝑓(𝑥), then ∫ 𝑓[ ℎ(𝑥) ] ℎ′(𝑥) 𝑑𝑥

= ∫ 𝑓(𝑢)𝑑𝑢

= 𝐹 (𝑢) + 𝑐

Notes : 𝑢 = ℎ(𝑥) wherein u is written instead of the function h(x)

𝑑𝑢 = ℎ′(𝑥) while du is written instead of the derivative of the function h(x)

Example 1. Substitution to find the antiderivative of a function. Given

∫ 8𝑥 3 (2𝑥4 + 5)𝑑𝑥

:

To achieve the form ∫ 𝑓(𝑢)𝑑𝑢, we can rewrite the given as: ∫(2𝑥4 + 5) 8𝑥3 𝑑𝑥 To find the antiderivative: let 𝑢 = (2𝑥4 + 5) and 𝑑𝑢 = 8𝑥3 𝑑𝑥 ∫(𝑢)𝑑𝑢 =

𝑢1+1 1+1

+𝑐

=

𝑢2 2

=

(2𝑥 4 +5) 2

(use power rule of integrals)

+𝑐 2

+𝑐

(substitute the original expression)

5

To check :

2

ℎ(𝑥) =

(2𝑥 4+5) 2

ℎ′(𝑥) =

2(2𝑥 4 +5) 2

2−1

∙ 8𝑥 3 𝑑𝑥

ℎ′(𝑥) = (2𝑥4 + 5) 8𝑥3 𝑑𝑥

Same as ℎ′(𝑥) = 8𝑥3 (2𝑥4 + 5) 𝑑𝑥, in which we started with. Example 2. Substitution with modification. Given

∫( 𝑥√𝑥 2 − 8 )𝑑𝑥

:

To achieve the form ∫ 𝑓(𝑢)𝑑𝑢, we can rewrite the given as: ∫ ( √𝑥 2 − 8 ) 𝑥 𝑑𝑥

Express the radical sign to fractional exponent. ∫(𝑥 2 − 8)1/2 𝑥 𝑑𝑥 Let 𝑢 = (𝑥 2 − 8). Then, 𝑑𝑢 = 2𝑥 𝑑𝑥 However, modification is to be done so we can apply integration by substitution rule. Dividing 𝑑𝑢 = 2𝑥 𝑑𝑥 by 2, we have:

𝑑𝑢 = 2𝑥 𝑑𝑥 2 𝑑𝑢 = 𝑥 𝑑𝑥 2

1

Now, use 2 𝑑𝑢 as substitute with 𝑥 𝑑𝑥. 1

∫ 𝑢1/2

1 𝑑𝑢 2

= 2 ∫ 𝑢1/2 𝑑𝑢 1 +1

1 𝑢2

= (1 2

1

+1

2

= ( 2

=

1

2

1

3

𝑢2 3 2

2

)+𝑐

(use power rule of integrals)

) +𝑐 3

∙ 3 (𝑢 2) + 𝑐 3

= (𝑢 2) + 𝑐 3

1

3

= 3 (𝑥 2 − 8)2 + 𝑐

(substitute the original expression)

6

Example 3. Substitution with integrals of exponential functions. Given

∫ 6𝑒 3𝑥 𝑑𝑥

:

1

Take the constant out of the integrand. Let 𝑢 = 3𝑥 ; 𝑑𝑢 = 3𝑑𝑥 ; 3 𝑑𝑢 = 𝑑𝑥 ∫ 6𝑒 3𝑥 𝑑𝑥

1

= 6 ∫ 𝑒 𝑢 𝑑𝑢 3

(constant rule)

= 2 ∫ 𝑒 𝑢 𝑑𝑢

(simplify)

1

= 6 ∙ 3 ∫ 𝑒 𝑢 𝑑𝑢

= 2𝑒 𝑢 + 𝑐

(used of common integral)

= 2𝑒 3𝑥 + 𝑐

(substitute the value of u)

Example 4. Antiderivative of Trigonometric Functions using Substitution Given

∫ cot 𝑥 𝑑𝑥

:

Use trigonometric identities to replace cot x as shown. ∫ Let 𝑢 = sin 𝑥 and 𝑑𝑢 = 𝑐𝑜𝑠𝑥 𝑑𝑥.

𝑐𝑜𝑠 𝑥 𝑑𝑥 𝑠𝑖𝑛 𝑥

Rewrite the integrand as follows. 1

∫ 𝑠𝑖𝑛 𝑥 cos 𝑥 𝑑𝑥 1

=∫ 𝑢 𝑑𝑢

= 𝑙𝑛|𝑢| + 𝑐

= 𝑙𝑛|sin 𝑥| + 𝑐

(substitute value of u)

7

What’s More

Use substitution to evaluate the following indefinite integrals. Write your answers on a separate sheet of paper. 1. ∫ 𝑥 2 (𝑥 3 − 1)5 𝑑𝑥

2. ∫(6𝑥 2 √2𝑥 3 − 9)𝑑𝑥

3. ∫ 𝑒 𝑐𝑜𝑠 𝑥 sin 𝑥 𝑑𝑥 4. ∫

2𝑥

𝑥 2 −5

dx

3𝑥 2

5. ∫ 3 𝑑𝑥 1+𝑥

What I Have Learned

Directions. Fill in the appropriate words to summarize integration by substitution. Write your answers on a separate sheet of paper. ➢ Inspect the integral and select an expression ℎ(𝑥) within the integrand to set equal to (1) _________. ➢ Make sure to select ℎ(𝑥) such that ℎ′(𝑥) is also part of the (2) _________. ➢ Substitute 𝑢 = (3) _________ and 𝑑𝑢 = ℎ′(𝑥)𝑑𝑥 into the integral. ➢ Evaluate the (4) _________ in terms of u.

➢ Write the (5) _________. in terms of x and the expression ℎ (𝑥 ).

8

What I Can Do

Antiderivative using u-substitution Given

∫

:

2𝑥 𝑑𝑥 √𝑥−1

Let u = x – 1 then x = u +1 and du = dx. Use this substitution values to rewrite the given as shown: 2∫

= 2∫

𝑥 𝑑𝑥 √𝑥−1

𝑢+1 √𝑢

we have: = 2∫(

𝑑𝑢

𝑢

(substitute)

𝑢+1 √𝑢

2∫

√𝑢

(Take out the constant)

+

1

1

𝑑𝑥

√𝑢

) 𝑑𝑥 1

= 2 ∫ (𝑢1− 2 + 𝑢− 2) 𝑑𝑥 1

1

= 2 ∫ (𝑢 2 + 𝑢− 2) 𝑑𝑥 = 2( = 2(

1 𝑢 2+1 1 +1 2 3

− 𝑢 2+1

+

𝑢2

+

2

3

3 2

1

(use fractional exponents)

1

𝑢2 1 2

1 −2+1

)+𝑐

(use common integration)

)+𝑐 1

= 2 ( 𝑢 2 + 2 𝑢2) + 𝑐 3

=

4 3 𝑢2 3 1

1

+ 4 𝑢2 + 𝑐 3

1

= 4 (3 𝑢 2 + 𝑢 2) + 𝑐 1

(simplify)

3

(take out common factor 4) 1

= 4 ( (𝑥 − 1) 2 + (𝑥 − 1)2) + 𝑐 3

(substitute original value of u)

Now, it’s your turn. Directions. Evaluate the following using u-substitution. Write your answers on a separate sheet of paper.

1. ∫

2. ∫

𝑥

√𝑥−2 1

𝑑𝑥

√2𝑥+3

𝑑𝑥

9

Lesson

2

Solve Problems involving Antidifferentiation

What’s In

Directions. Evaluate the following integrals. Choose your answer from the box. Write your answers on a separate sheet of paper.

1) ∫ 𝑥 (𝑥 2 + 5)3 𝑑𝑥

8(8𝑥 − 3) + 𝑐

1 (𝑥 2 8

2) ∫ √8𝑥 − 3 𝑑𝑥

1

+ 5)4 + 𝑐

− 8 𝑒 (−8𝑥+11) + 𝑐

𝑥

3) ∫ 2 𝑑𝑥 𝑥 −4

1 2

[2𝑠𝑒𝑐 2 (𝑡𝑎𝑛2𝑥 )𝑠𝑒𝑐 22𝑥 ] 𝑑𝑥

4) ∫

ln |𝑥2 − 4| + 𝑐

1 (8𝑥 12

5) ∫ 𝑒 −8𝑥+11𝑑𝑥

3

− 3) 2 + 𝑐

𝑡𝑎𝑛(𝑡𝑎𝑛(2𝑥)) + 𝑐

Notes to the Teacher This lesson will give you insights of some practical applications of the integrals or antiderivatives.

10

What’s New

Read and Learn!

In a Physics class, Ms. Riza Fuentes asked the students which of the feather and the coin will fall first when dropped at the same time when air resistance is taken out of the variable. She provided a vacuum pump with tube and end caps. Following the procedure illustrated below, the students were asked to list down their observations.

chamber assembly

with chamber lying horizontally

with air removed

11

The students observed that both the coin and the feather freely fell at the same speed. This was because the air was removed from the chamber, thus creating a vacuum where air resistance did not affect the objects as they fall. Given time (t), we calculate the distance (s) by the function: 𝑠(𝑡) = −𝑎𝑡 2 + 𝑣𝑖 𝑡 + 𝑐,

where c is constant; a is acceleration due to gravity and 𝑣𝑖 is initial velocity. The students recorded the time when the coin and the feather fell as 2seconds. They were asked to calculate the initial velocity of the objects with a constant acceleration due to gravity of 32𝑓𝑡/𝑠𝑒𝑐2 . Note that the length of the chamber is 2 ft. Solution : From the function 𝑠(𝑡) = −𝑎𝑡 2 + 𝑣𝑖 𝑡 + 𝑐, substitute the data provided in the problem and solve for 𝑣𝑖 as shown. Since the objects came from the other end of the chamber, let c = 0 𝑠(𝑡) = −𝑎𝑡 2 + 𝑣𝑖 𝑡 + 𝑐 𝑓𝑡

2 𝑓𝑡 = −32 𝑠𝑒𝑐 2 (2𝑠𝑒𝑐)2 + 𝑣𝑖 (2𝑠𝑒𝑐) + 0

2 𝑓𝑡 = −128 𝑓𝑡 + 𝑣𝑖 (2𝑠𝑒𝑐)

𝑣𝑖 𝑣𝑖

=

=

(128+2)𝑓𝑡 2 𝑠𝑒𝑐

130𝑓𝑡 2 𝑠𝑒𝑐

𝑣𝑖 = 65

𝑓𝑡

𝑠𝑒𝑐

Do you know that the formula 𝑠(𝑡) = −𝑎𝑡 2 + 𝑣𝑖 𝑡 + 𝑐 is derived through antidifferentiation?

12

What is It

In solving problems involving integrals, student’s critical thinking skills allows them to analyze the appropriate formula as well the applicable technique or rules in given situations. Since integration or antidifferentiation is a reverse of differentiation, familiarity on the process of solving differential calculus problems will help you in dealing problems on integral calculus. Most of the practical applications of integration involves how one quantity changes with respect to another quantity. Thus, finding their rate of change and relations.

Rectilinear Motion This application deals with a value which changes with respect to a certain value of another variable. In the study of rectilinear motion, the velocity v of a moving object is defined as the time rate of change of the distance s and the acceleration a as the time rate of change of the velocity. Symbolically, it is written as: 𝑣=

𝑑𝑠 𝑑𝑡

and

𝑎=

𝑑𝑣 𝑑𝑡

Illustrative Example 1. A ball is thrown vertically upward from the initial height of 1 meter with an initial velocity of 30 m/sec. Find the maximum height attained by the ball.

1m

13

Given from the problem Constant value of gravitational acceleration (

𝑚

𝑠𝑒𝑐 𝑠

)

: 𝑣𝑖 = 30𝑚/𝑠𝑒𝑐

: −9.8 (

𝑚

𝑠𝑒𝑐 𝑠

)

Solution : Since 𝑎 =

𝑑𝑣

𝑑𝑣 𝑑𝑡

= −𝑎. Therefore, 𝑑𝑣 = (−9.8)𝑑𝑡. To solve for the integral of 𝑑𝑣 = (−9.8)𝑑𝑡, we have : 𝑑𝑡

and a is negative then,

𝑑𝑣 𝑑𝑡

= −9.8. That is,

∫(−9.8)𝑑𝑡

𝑣 = −9.8𝑡 + 𝐶

(velocity of falling object)

However, we know that the initial velocity 𝑣 = 30𝑚/𝑠𝑒𝑐 when 𝑡 = 0. To determine the value of constant c, substitute the given values to the above equation. 𝑣 = −9.8𝑡 + 𝐶

30 = −9.8(0) + 𝐶 𝐶 = 30

Then, 𝑣 = −9.8𝑡 + 𝐶 becomes 𝑣 = −9.8𝑡 + 30. This equation represents the velocity of the ball at any given time. Use these two equations, 𝑣 = −9.8𝑡 + 30 and 𝑣 =

shown.

𝑑𝑠 𝑑𝑡

𝑑𝑠 𝑑𝑡

to find s (distance) as

= −9.8𝑡 + 30

𝑑𝑠 = (−9.8𝑡 + 30)𝑑𝑡

To find s, take the integral of ds, we have: ∫(−9.8𝑡 + 30)𝑑𝑡 𝑠=−

9.8𝑡 2 2

+ 30𝑡 + 𝐶

𝑠 = −4.9𝑡2 + 30𝑡 + 𝐶 From the given, ...