Basic- Calculus-Quarter-4-part-1 PDF

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Basic CalculusModule 3####### Quarter 4Learner’s PacketBasic Calculus Quarter 4 First Edition, 2020Published by: Department of Education SDO—Rizal Schools Division Superintendent: Cherrylou D. Repia Assistant Schools Division Superintendents: Gloria C. Roque and Babylyn M. PambidBasic CalculusModule...

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Basic Calculus Module 3 Quarter 4

Learner’s Packet

Basic Calculus Module 3 Grade 11 Quarter 4

Schools Division Office Management Team: : Rosemarie C. Blando, August Jamora, Merle D. Lopez Writer/s: Rhenelee S. Ramos Illustrator: Julius Burdeos Evaluators: Al Jesse M. Arroyo, Florabel F. Hilario

Basic Calculus Quarter 4 First Edition, 2020 Published by: Department of Education SDO—Rizal Schools Division Superintendent: Cherrylou D. Repia Assistant Schools Division Superintendents: Gloria C. Roque and Babylyn M. Pambid

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WEEK

Illustrating Antiderivative of a Function

I

Lesson

After going through this material, you are expected to: 1) illustrate antiderivative of a function; and 2) apply the definition and rules of the antiderivative of a function; Learning Task 1: Complete the if –then statement below using rules in derivatives. 1. If g(x) = 5, then g’ (x) = _________________ 2. If g(x) = 5x, then g’ (x) = _________________ 3. If g(x) = 2x3—5x2 then g’ (x) = _________________ 4. If g(x) = 10 + 5t—t2 then, g’(x) = _________________ 5. If g(x) = then, g’(x) = _________________

D Learning Task 2: Complete the table below to illustrate the antiderivative of a function. GIVEN

Antiderivative

1. g(x) = 35 2. g(x) = 40x 3. g(x) = 10x + 17 4. g(x) = 8x5

5. g(x) = -15x3

Guide Questions: 1. How did you find the antiderivative of each given function? 2. What did you notice on how the antiderivative of a function was obtained? What process did you do? 3. Upon observing the process, how are you going to define antiderivative of a function?

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DISCUSSION In the previous lesson, you learned about finding ways on how to illustrate derivative of a function. Supposed that the process is reversed, this process is called antiderivative or simply integration. An antiderivative of a function f(x) is a function of F(x). Consider the illustration below: Derivatives

Antiderivatives

xn x3 Example 1: Illustrate the antiderivatives of the following functions. 1.

2.

We can find the antiderivative of the following functions by using the rule illustrated above. Solutions: 1.

2.

THE INDEFINTE INTEGRAL Consider the function f(x) = 3x2,, its antiderivative is F((x) = x3. Observe that

Consequently, x3 + 1, x3 - 4 and x3 + 2π are also antiderivative of f(x) = 3x 2., where C represents any real number. This family of all antiderivatives is called indefinite integral. Since F(x) is an antiderivative f(x)m then we can say that F(x) + C is the indefinite integral of f(x) and is denoted by

The arbitrary constant C is called the constant of integration. Function f(x) is called the integrand while the symbol ∫ is just an elongated S meaning sum and denotes the operation of integration.

Example 1: Evaluate the integral of ∫8x7 dx. Solution: The integrand is

Applying the power rule of differentiation, note that

Thus,

Example 2: Evaluate the integral of ∫5u4 du. The integrand is

Applying the power rule of differentiation, note that

Thus,

Note that: If F’ (x) = f(x), then F(x) is an antiderivative of f(x). If F’ (x) = f(x), then, ∫f(x)d = F(x) + C, for any real number C. Integral

2x

x2

Derivative

Integral Rule

Constant Rule

Integration is a way of adding slices to find the whole. Example: What is an integral of 2x? We know that the derivative of x2 is 2x, so an antiderivative or integral of 2x is x 2. The integral of many functions are well known and there are useful rules to work out the antiderivative or integral of more complicated functions. The following are the basic rules.

Illustration

Example

∫ adx = ax + C

∫3dx = 3x + C

∫ andx = axn + C

∫6x2dx = 6∫x2dx, then use the power rule on x2, hence;

Power Rule Multiplication by Constant

Sum and Difference Rule ∫ [f(x) ±g(x)]dx

= ∫ f(x)dx ± ∫g(x)dx

∫ew ± 3dw

= ∫ewdw ± ∫3dw =ew ± 3w + C

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E Learning Task 3: Apply the rules of integration to the following integral. 1.

4.

2.

5.

3.

Learning Task 4: The Blind Item Trivia: Who is the German philosopher, mathematician and political adviser who is distinguished because of his independent invention of the differential and integral calculus? Directions: To answer the trivia, apply the rules in antiderivatives to the following given and write the letter of your answer to the decoder below. Z ∫6dx

I ∫3x8dx

L ∫7x-4 dx

E ∫(5x4 + 3x2 + 6) dx

B ∫x6 dx

N ∫(3x + 4)2dx

Answer Box

A Learning Task 4: Illustrate the antiderivative of the following functions and step by step, explain your answer.

1. 2. 3.

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Computing the General Antiderivative of Polynomial, Radical, Exponential and Trigonometric Functions

I

Lesson

After going through this material, you are expected to compute the general antiderivative of 1) polynomial; 2) radical; 3) exponential ; and 4) trigonometric functions.

Learning Task 1: For a given pair of functions F and f, validate whether F is an antiderivative of f. If not, give its correct antiderivatives. Function

Right Antiderivative

1. f(x) = 6 ; F(x) = 6x + 3 2. f(x) = 30x9 ; F(x) = 3x10 3. f(x) = 15x3 –6x; F(x) = 3x 5 + 3x2 - 1 4. f(x) = 6x2 –8x—5; F(x) = 2x3 - 4x2 - 5x + 5 5.

D Definition 1 (Antiderivative) . If F’(x) = f(x) we call F an antiderivative of f.

Definition 2 (Indefinite Integral) . If F is an antiderivative of f, then ∫f(x)dx = F(x) + c I is called the (general) Indefinite Integral of f, where c is an arbitrary constant. The general antiderivatives is the family of functions that have a derivative equal to f (x). To find the antiderivatives we need to compute the indefinite integral of f(x).

THE ANTIDERIVATIVE OF POLYNOMIAL FUNCTIONS

Example1: Compute the integral of The above given only consists of terms added together, you can integrate them separately and the results, giving us:

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WEEK

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Notice that each term can be integrated using the Power Rule for integration, which is:

Plugging our 3 terms into this formula, we have:

Now we have our final answer by adding these together, remembering to add our constants C in the end.

Remember that the antiderivative of a polynomial functions can be express by letting An antiderivative F(x) of f(x) can be found by

Example 2: Consider the function general antiderivative.

,compute for the

Solutions:

THE ANTIDERIVATIVE OF RADICAL FUNCTIONS To integrate radicals or function having square roots, you just need to simply express the root as an exponent. Example: Compute for the antiderivative of Solutions: Express the square root as an exponent The integral therefore becomes: Apply the rule in integration The answer is

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Sometimes, you may have more than one term under the radical sign, as in tgis example:

At this case, you can use u– substitution to proceed. You need to set u equal to the quantity in the denominator. Solve this for x by squaring both sides and subtracting. This allows you to get dx in terms of u by taking the derivative of x dx = (2u)du Substituting back into the original integrals gives

Now, you can integrate this using the basic formula and expressing u in terms of x.

THE ANTIDERIVATIVE OF EXPONENTIAL FUNCTIONS The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential function, y = e x, is its own derivative and its own integral. Exponential functions can be integrated using the following formulas: a.

b.

Examples: Find the antiderivative of the following exponential functions. 1.

Solution: Let u equal the exponent e.

2. Solutions: First rewrite the problem using a rational exponent Using substitution, choose u = 1 + ex. Then du = ex. We have, Then,

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THE ANTIDERIVATIVE OF TRIGONOMETRIC FUNCTIONS Recall from the definition of an antiderivative that, if That is, every time there is a differentiation formula, we get an integration formula for nothing. Here is a list of some of them.

Question: 1. What can you say about the formulas for the antiderivatives of sin x and cos x?

2. What about the other four? Examples: Compute the following. 1.

2.

,This given requires a substitution

Then we have,

3. Write

and use the same substitution.

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Moreover, the illustration below shows the antiderivatives of all six trigonometric functions which you may use in getting the integral of trigonometric functions.

Integral Rule

General Rule

E Learning Task 3: Compute for the general antiderivatives of the functions.

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A Learning Task 4: Connect the following function on the left side to its general derivative on the right side.

anti-

1. 2. 3. 4. 5.

Learning Task 4: After thorough understanding of the lesson, you need to complete the statements below for your reflection. I have learned that _______________________________________________________________ ___________________________________________________________________________________ I realized that _____________________________________________________________________ ___________________________________________________________________________________

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WEEK

Computing the Antiderivative of a Function Using Substitution Rule

I

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Lesson

After going through this material, you are expected to compute the antiderivative of a function using substitution rule. Learning Task 1: Match the function to its corresponding antiderivative by writing the letter of the correct answer on the space provided before each number.

_____1.

A.

_____2.

B.

_____3.

C.

_____4.

D.

_____5.

E.

D Learning Task 2: Observe how the function below was solved and then swer the questions that follow.

an-

1. Let f(x) = (x2 + 3x –5)10 f’(x) = 10(x2 + 3x –5)9 ● (2x + 3) = (20x + 30) (x2 + 3x - 5)9 Now consider this: 2. What is ∫ (20x + 30) (x2 + 3x - 5)9 dx? The answer would be (x2 + 3x –5)10 + C. Questions: 1. What process is used in the first given? 2. How would we have evaluated this indefinite integral without starting with f (x) as we did? 3. What do you think is the process used? DISCUSSION The second example above explores integration by substitution. It allows us to "undo the Chain Rule." Substitution allows us to evaluate the above integral without knowing the original function first. The underlying principle is to rewrite a "complicated" integral of the form ∫fx dx as a not--so--complicated integral ∫h(u) du . Let us consider again ∫ (20x + 30) (x 2 + 3x - 5)9 dx, you will see that the

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The most complicated part of the integrand is (x2 + 3x - 5)9. To make it simpler , used substitution method rule. Solutions: Let u = x2+3x−5u. Thus, (x2+3x−5)9=u9 Then, established u as a function of x, so now consider the differential of u: du=(2x+3)dx . Keep in mind that (2x+3)(2x+3) and dx are multiplied. Return to the original integral and do some substitutions through algebra:

This method works when the integrand contains a function and the derivative of the function’s argument — in other words, when it contains that extra thing produced by the chain rule — or something just like it except for a constant. And the integrand must not contain anything else. STEPS IN SOLVING ANTIDERIVATIVE OF A FUNCTION USING SUBSTITUTION RULE

1. Set u equal to the argument of the main function. 2. Take the derivative of u with respect to x. 3. Solve for dx. 4. Make the substitutions. 5. Anti differentiate by using the simple reverse rule. 6. Substitute x-squared back in for u — coming full circle. Example: Compute the antiderivative of ∫x sin(x2+5)dx using substitution. Solutions: Let u be the "inside" function of sin (x 2+5) u = x2+5; du=2xdx The integrand has an x dx term, but not a 2x dx term. (Recall that multiplication is commutative, so the x does not physically have to be next to dx for there to be an x dx term.) We can divide both sides of the du expression by 2:

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Now, proceed to substitution method:

Thus,

.We can check our work by evaluating the derivative

of the right hand side.

E Learning Task 3: Use substitution rule to find the antiderivative of the following functions. Show your complete solutions.

1.

6.

2.

7.

3.

8.

4.

9.

5.

10.

A Learning Task 4: Evaluate the following functions. Use substitution method.

1. 2. 3. 4.

5.

15

WEEK

Solving Problems Involving Antidifferentiation

I

Lesson

After going through this material, you are expected to solve problems involving antidifferentiation.

Learning Task 1: Determine the antiderivative of the following. 1.

4.

2.

5.

3. Guide Questions: 1. What method did you use to obtain the antiderivative of each given? 2. What kind of functions did you encounter in the given?

3. How did you determine the antiderivatives of the given?

D Read and analyze the problems below and then answer the questions that follow. 1. A rock is dropped from the top of the of a 400 foot cliff. It’s velocity at time t seconds is v(t) = -32t feet per second. a. Find s(t), the height of the rock above the ground at time t b. How long will it take the rock to reach the ground? c. What will be its velocity when it hits the ground? Solutions: a. Let s(t) = v(t) dt , in order to find s(t) m we must take the antiderivative of our v(t), so,

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Since the rock was dropped off a 400 foot cliff, this means that when we started,, the rock was at a position of 400. In other words, when t = 0, s(t) = 400. Therefore, we can put 0 in for t and 400 in for s(t) and solve for C. Therefore,

Thus, the final answer is: s(t) = -16t + 400 b. When the rock hits the ground, it is 0 feet off the ground. Therefore,, s(t) = 0. So we set s(t) equal to zero and solve for t:

c. In order to find the velocity, all you need is to plug the t =25 into the equation for velocity.

2. A ball is thrown vertically upward from the ground with an initial velocity of 39.2 m/sec. If the only force considered is that attributed to the acceleration due to gravity, find: a. how long will it take for the ball to strike the ground? (b) the speed with which will it strike the ground? and (c) how high the ball will rise?

Solutions: Let the initial velocity = 39.2 m/sec, By differentiating "distance" we get "velocity", by differentiating "velocity" we get "acceleration". By doing the process vice versa, we get "velocity" by integrating "acceleration", integrating "velocity" we get "distance". Hence, a. When a ball is thrown vertically upward, it reaches the maximum height then it will strike the ground. Acceleration = 9.8 m/sec2 When we throw a ball in upward direction, it moves against the gravity, so we have to consider the acceleration as -9.8. Velocity = ∫a dt a = -9.8 velocity = ∫(-9.8) dt 17

v(t) = -9.8 t + c1 ------(1) When an object reaches its maximum height, its velocity will become zero at the point. Initial velocity = 39.2 at t = 0 39.2 = -9.8 (0) + c1 c1 = 39.2 v(t) = -9.8 t + 39.2 By applying v(t) = 0, we may find the time taken by the ball to reach its maximum height.

0 = -9.8 t + 39.2 -9.8 t = -39.2 ==> t = 39.2/9.8 = 4 Hence it takes 4 seconds to reach maximum height. Distance = ∫ v(t) = ∫(-9.8 t + 39.2) dt Distance = -9.8t2/2 + 39.2 t + c2 D(t) = -4.9 t2 + 39.2 t + c2 When t = 0, D(t) = 0, then c2 = 0

Hence D(t) = -4.9 t2 + 39.2 t So let us answer the questions earlier, a. how long will it take for the ball to strike the ground? Time taken by the ball to strike the ground = 8 seconds [It takes 4 seconds to reach its maximum height and 4 seconds to strike the ground.] b. the speed with which will it strike the ground? v(t) = -9.8 t + 39.2

when t = 8 v(8) = -9.8 (8) + 39.2 = 39.2 m /sec c. how high the ball will rise? When t = 4 D(4) = -4.9 (4)2 + 39.2 (4) = -4.9(16) + 156.8

= -78.4 + 156.8 = 78.4 m Hence the maximum height reached by the ball is 78.4 m.

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E Learning Task 2: Directions: Read and analyze the problems carefully and then solve Problem

Solution

1.The growth of a population bacteria is proportional to the cube root of t, where P is the population size and t is the time in days. Thus, ,the initial size of the population is 800, and after one day the population grows to 1000. Estimate the population P after 1 week. 2. A particle, starting out at rest, moves along the x– axis such that its acceleration is a(t) = 2cos(t), t in seconds. At the time t = 0, its position is p = 3. a. Find the velocity v(t) of the particle at time t. b. Find when the particle is at rest. c.Find the position of the particle p(t) at any time t.

3. An automobile maker claims that their car can accelerate at a constant rate, and the car can accelerate from a dead stop to a velocity of 75 ft/ sec in a distance of 150 feet. What is the acceleration of the car?

4. At the moment of the traffic light turns green, Albert, who has been waiting at the intersection starts with a constant acceleration of 10 feet/s2. At the same moment, Albert’s friend Alreign, who is travelling at a constant velocity of 35 feet per second passes Albert. Albert passes Alreign sometime after this. How fast will Albert be travelling when he passes Alreign? 5. The speed of a car travelling in a straight line is reduced from 45 to 30 miles per hour in a distance of 264 feet. Find the distance in which the x=car can be brought to rest from 30 miles per

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A Learning Task 3: Directions: Read and analyze the problems carefully and then solve it. 1. An airplane taking off from a runaway travels 3600 feet before lifting off. The airplane starts from rest, moves with constant acceleration, and makes run in 30 seconds. With what speed does it lift off? 2. A ball drop from the top of a tall building and has a velocity of v(t) = -10t - 5 m/s. Find the height of the building given that the ball strikes the ground after t = 4

Learning Task 4: After thorough understanding of the lesson, you need to make a poem on how antiderivative applications are essential to real life situations. You may recite the poem through recorded video and submit it to your teacher as your performance task. (Note that the teacher shall create their on rubrics in rating the each student).

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Answers

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Answers

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References Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. Gregory Hartman (Virginia Military Institute). Contributions were m...