Basic Calculus Fourth Quarter: Script For Radyo Eskwela Sa Isabela
Basic Calculus Fourth Quarter: Script For Radyo Eskwela Sa Isabela
Basic Calculus Fourth Quarter: Script For Radyo Eskwela Sa Isabela
Basic Calculus
Fourth Quarter
COPYRIGHT PAGE
Radio-Based Instruction Materials for Basic Calculus
(Grade 11)
Copyright © 2020
DEPARTMENT OF EDUCATION
Regional Office No. 02 (Cagayan Valley)
Regional Government Center, Carig Sur, Tuguegarao City, 3500
“No copy of this material 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.”
This material has been developed for the implementation of K to 12 Curriculum through the Curriculum and
Learning Management Division (CLMD). It can be reproduced for educational purposes and the source must be
acknowledged. Derivatives of the work including creating an edited version, an enhancement of
supplementary work are permitted provided all original works are acknowledged and the copyright is
attributed. No work may be derived from this material for commercial purposes and profit.
Consultants:
Regional Director : ESTELA L. CARIÑO, EdD, CESO IV
Assistant Regional Director : JESSIE L. AMIN, CESO V
Schools Division Superintendent : MADELYN L. MACALLING, PhD, CESO VI
Asst. Schools Division : EDNA P. ABUAN, PhD
Superintendent(s)
: DANTE J. MARCELO, PhD, CESO VI
Chief Education Supervisor, CLMD : OCTAVIO V. CABASAG, PhD
Chief Education Supervisor, CID : RODRIGO V. PASCUA, EdD
Division Focal Person for Radio-Based : JAY J. GALLEGOS
Instruction
Development Team
Scriptwriter : Mary Jane P. Gumangan
Content Editor : Leonor M. Balicao
Language Editor : Joy S. Ferrer-Lopez
Technical Specialist : Rhean Joy R. Lintao
Voice Talent: :
Episode Executive Producer : Joy S. Ferrer-Lopez
RBI Assistant Chair for Quality : Joy S. Ferrer-Lopez
Assurance
RBI Assistant Chair for Production : Michael Kevin A. Monforte
RBI Over-all Chair : Labi L. Upam Jr.
Focal Persons : Larina M. Colobong, EPS TLE/ Focal Person for SHS
Inocencio Balag, EPS Mathematics, SDO Isabela
Ma. Cristina Acosta, EPS LRMDS, SDO Isabela
Isagani R. Duruin, PhD., Regional Learning Area Supervisor
Introduction
Lesson Proper
Quiz
Checking of Quiz
Closing Billboard
Total:
Approved by:
JOY S. FERRER-LOPEZ
Executive Producer
3 RADIO TEACHER: Hello STEM students! How’s your feeling today? I hope that your
4 fine and okay. I am grateful because you are still actively listening to
5 our Radyo Eskwela sa Isabela. Today we will be having our new and
9 RADIO TEACHER: So, let’s get started! Please be in your most comfortable area in your
11 wherever you are. Get now your Learning Activity Sheets in Basic
16 RADIO TEACHER: Before we discuss our new topic for today, lets have a short recall on
SFX: YIEEEHHH
RADIO TEACHER: I will read the directions. Find the antiderivatives of the following
1. ∫ 3 x dx
1 2
2. ∫ x dx
2
3
3. ∫ ( e x + 2x ) dx
4 4. ∫¿¿
5 5. ∫ cot 2 x dx
6 I will give you three minutes to answer.
7 SFX: CLOCK TICKING
8 RADIO TEACHER: There you have it learners, let’s check if you get the correct answer or
9 if you still remember your last lesson.
10
3x
1.∫ 3 x dx= +C
11
ln 3
2
12 2.∫ dx =2 ln|x|+C
x
13
x x x 2x
3.∫ ( e +2 ) dx=e + +C
14 ln 2
15 4. ∫ ¿ ¿
18
RADIO TEACHER: Who among you got a perfect score? (PAUSE) Al right I'm hoping
19
that you still remember your last lesson. Today you are expected to
20
compute the antiderivative of a function using substitution rule.
21
Antidifferentiation is more challenging than differentiation. To find
what I’ve told you that substitution for integrals corresponds to the
∫ f ( u ) du=F (u )+ c
1 Assuming that u = u(x) is differentiable function and using the chain
2
d
3 rule, we have F ( u ( x ) )=F ' ( u ( x ) ) u ' ( x )=f ( u ( x ) ) u ' (x)
dx
4 Integrating both sides gives ∫ f ( u( x)) u' ( x ) dx=F ( u ( x ) ) + c .
5
Hence
6 ∫ f ( u( x)) u' ( x ) dx=f ( u ) du , where u=u( x )
7
This is the substitution rule formula for indefinite integrals. Note that
8
the integral on the left is expressed in terms of the variable x . The
9
integral on the right is in terms of u .
10
This change of variable is one of the most important tools available to
11
us. This technique is called integration by substitution. It is often
12
important to guess what will be the useful substitution. The
13
substitution method (also called u−substitution) is used when an
14
integral contains some function and its derivative. In this case, we can
15
set u equal to the function and rewrite the integral in terms of the new
16
variable u . This makes the integral easier to solve. Do not forget to
17
express the final answer in terms of the original variable x. (PAUSE)
18
Let’s try this example by following the process in your Learning
19
Activity Sheet.
20
Step 1: Idetifying u.
6 ∫¿¿
7 Step 4: Integrate resulting integral
8
∫ (7 x+5)4 dx=∫ ¿ ¿ ¿
9
Step 5: Return to the initial variable x.
10 5
u5 (7 x +5)
+C= +C
11 35 35
13 SFX: YIEEEHHH
14 RADIO TEACHER: Very Good. Let’s have another example. Get your paper and pencil,
15 and let’s try to solve this. Evaluate ∫ ¿ ¿ using the process on how to
16
find the antiderivatives with the substitution rule. (Read twice)
17
(PAUSE) What is the first step?
18
SFX: CORRECT ANSWER
19
RADIO TEACHER: Correct. The first step is to identifying u. Then u = x+4. How about
20
the second step?
21
SFX: CORRECT ANSWER
du
RADIO TEACHER: Very Good! Determine the value of dx. Now, since u = x + 4, =1
dx
1 RADIO TEACHER: Very Good! Make the substitution. Then, substituting (x+4) and dx.
2 5 5
We have ∫ (x+ 4) dx=∫ u du. How about the fourth step?
3
SFX: CORRECT ANSWER
4
RADIO TEACHER: Exactly! Integrate resulting integral. We have,
5
∫ (x+ 4)5 dx=∫ u5 du
6
=
u6
7 +C
6
8 Get it? (PAUSE) how about the last step?
10 u6 (x +4 )6
RADIO TEACHER: Return to the initial variable x. Therefore, +C = +C
6 6
11
(PAUSE) Did you understand learners?
12
SFX: YIEEEHHH
13
RADIO TEACHER: Now, let’s try another example by answering the Activity 1 on page
14
11 – 13 . Are you ready?
15
SFX: YIEEEHHH
16
RADIO TEACHER: Let’s answer all item number 2 and 4 only per activity. And the other
17
item will serve as your activity after the session. Directions: Find the
18
the antiderivatives of the following polynomials function using U –
19
substitution.
20
21 2. ∫ 16 x √ 80−2 x 2 dx
u=
du/dx =
dx =
ANSWER:
1 6X
4. ∫ dx
√ x2 +2
2
u=
3
du/dx =
4
dx =
5
ANSWER:
6
I will give you two minutes to answer.
7
SFX: CLOCK TICKING
8
RADIO TEACHER: Let’s check if your answer is correct. Number 2 we have,
9 ∫ 16 x √ 80−2 x 2 dx, let u = 80−2 x 2, it follows that du/dx =
10
11 16 x du
=¿ -4x , then dx = . Therefore, by substitution the answer
−4 −4 x
12
13 −1 6X
is ¿. Let’s have number 4, ∫ 2
dx , let u ¿ x 2+ 2, it follows
6 √X +2
14
15
6x du
du/dx = =¿ 2x, then dx = . Therefore, by substitution the
16 3 2x
17 answer is 6 ¿ ¿
18 RADIO TEACHER: Now, let’s proceed to the next, exponential and logarithmic function.
20 1 x+3
2. ∫ e dx
4
u=
du/dx =
dx = du
Answer:
4
4. ∫ dx
1 e 3 x+ 4
2 u=
3 du/dx =
4 dx = du
5 Answer:
8 1 x+3
RADIO TEACHER: Let’s check if your answer is correct. For number 2, ∫ e dx, let u
4
9
10 e x+3
= x + 3, then du/dx = 1 and dx = du. Therefore, the answer is +C
4
11
12
4
. For number 4, ∫ 3 x+ 4
dx , let u = 3x +4, then du/dx =3 and dx = du .
13 e 3
14
−4
15 Therefore, the answer is +C . (PAUSE) I’m hoping that you
3 e3 x+4
16 really understand our lesson today. (PAULet’s try trigonometric
17 function, again you will be given two minutes to answer items 2 and
18 4.
19
2. ∫ 𝑠𝑖𝑛6 𝑥 cos 𝑥 𝑑x
20
u=
du/dx =
dx = du
Answer:
4. ∫ 6 sin(6𝑥) 𝑑x
u=
1
du/dx =
2
dx = du
3
Answer:
4
SFX: CLOCK TICKING
5
RADIO TEACHER: Let’s check. Number 2 ∫ 𝑠𝑖𝑛6 𝑥 cos 𝑥 𝑑x, let u = sin x, then du/dx = - cos
6
7 du −sin7 x
x and dx = . Therefore, the answer is +C. Let’s have
−cos x 7
8
9
du
number 4, ∫ 6 sin(6𝑥) 𝑑x, let u = 6x., then du/dx = 6 and dx = .
10 6
11 Therefore, the answer is −𝑐𝑜𝑠(6𝑥) + c.
12 RADIO TEACHER: Do you understand our lesson? (PAUSE) I believe that you can now
18 RADIO TEACHER: Exactly! The Substitution Rule. What is the first step on how to
20 RADIO TEACHER: Very Good! Indetifying u. How about the second step?
RADIO TEACHER: Correct determine the value of dx. How about the third step?
RADIO TEACHER: That’s right! Make the substitution. How about the fourth step?
1 RADIO TEACHER: Very Good! Integrate resulting integral. How about the last step?
3 RADIO TEACHER: Exactly! Return to the initial variable x. What are the different
6 trigonometric functions. (PAUSE) I know and I believe that you really understand our
7 lesson today. Give yourself a ten clap for a job well done. (PAUSE) For now, get your
8 ballpen and paper and let’s see if you really understand our lesson. Be ready because we
10 RADIO TEACHER: Direction: Find the antiderivatives of the following functions using
be honest all the time, even if others may not. Are you ready?
du
1
RADIO TEACHER: Very Good! ∫ e3 x dx , we let u = 3x. Then du = 3 dx. Hence, dx = 3
.
2 So,
3x
du 1 1 1
3
∫ e3 x dx=∫ eu = ∫ e u du= e u= e 3 x +C or e +C
3 3 3 3 3
4
5
RADIO TEACHER: Number 2, we have ∫ 2 4 x dx ,we let u = 4x and so du = 4dx.
6
du 4x u du 1 u 1 2u 1 4x
Thus, dx = . Hence, we have ∫ 2 dx=∫ 2 = ∫ 2 du= +C= 2 +C
7 4 4 4 4 ln 2 4 ln 2
8
24 x
9 or +C
4 ln2
10
1
Number 3, we have ∫ 2 x−1 dx , suppose we let u = 2x – 1. The du = 2 dx. Hence,
11
12
13 du
dx= . We have
2
14
15 1 1 du 1 1 1 1
∫ 2 x−1 dx=∫ u 2
=¿ ∫ du= ln ∨u∨¿+ C= ln ¿2 x−1∨¿+C ¿ ¿ ¿
2 u 2 2
16
1 1
4 dx and dx = du .So, ∫ cos ( 4 x +3 ) dx=∫ cos u ∙ du
4 4
1
cos u du
4∫
=
1
= sin u+C
4
1
= sin ( 4 x +3 ) +C
4
4
Number 5, we have ∫ x sec ¿ ¿ ¿
du
Let u = x 5. Then du = 5 x4 dx . Thus, x4 dx = . We have
5
∫ x 4 sec ¿ ¿ ¿
1
sec u du
5∫
¿
1
= ln ¿ sec u+ tan u∨+C
5
1
= ln ∨ sec x 5 + tan x 5∨+C
5
RADIO TEACHER: Congratulations for a job well done learners. Who among you got a
understand and enjoy our lessons today. Please try to answer some
activities in your LAS and practice more. If you have any questions
okay?
CLOSING BILLBOARD
-END-