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Lecture 7

Course: COMP 3868, Spring 2012
School: Hong Kong Polytechnic...
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Calculus CC2053: 7 Introduction to Calculus and Linear Algebra 2011/2012 Semester 1 INTEGRATION: Anti-Derivative and Indefinite Integrals 1 CC2053: Calculus 7 Antiderivative A function F(x) for which for every x in the domain f is said to be an antiderivative of f (x) 2 CC2053: Calculus 7 Antiderivatives and Indefinite Integrals is called the indefinite integral, represents the family of all...

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Calculus CC2053: 7 Introduction to Calculus and Linear Algebra 2011/2012 Semester 1 INTEGRATION: Anti-Derivative and Indefinite Integrals 1 CC2053: Calculus 7 Antiderivative A function F(x) for which for every x in the domain f is said to be an antiderivative of f (x) 2 CC2053: Calculus 7 Antiderivatives and Indefinite Integrals is called the indefinite integral, represents the family of all antiderivatives of if 3 CC2053: Calculus 7 Rules for Integrating Common Functions For k and C constants 1. The constant rule 2. The power rule 3. The logarithmic rule 4. The exponential rule for for all x 0 for 4 CC2053: Calculus 7 Rules for Integrating Common Functions For C constant 5. Sine function 6. Cosine function 5 CC2053: Calculus 7 Example 1 Find the following integrals: a. b. c. d. e. 6 CC2053: Calculus 7 Example 1:- solution a. b. c. d. e. 7 CC2053: Calculus 7 Algebraic Rules for Indefinite Integration 6. The constant multiple rule 7. The sum and difference rule 8 CC2053: Calculus 7 Example 2 Find the following integrals: a. c. e. b. d. f. 9 CC2053: Calculus 7 Example 2:- solution a. b. 10 CC2053: Calculus 7 Example 2:- solution c. d. 11 CC2053: Calculus 7 Example 2:- solution e. f. 12 CC2053: Calculus 7 Example 3 Find the equation of the curve that passes through (0, 3) if its slope is given by for each x. 13 CC2053: Calculus 7 Example 3:- solution The curve passes through the point (0, 3). Substitute (0, 3) into . Thus, the equation of the curve is 14 CC2053: Calculus 7 Example 4 If the marginal cost of producing x units is given by and the fixed cost is $2,000, find the cost function C(x) and the cost of producing 20 units. 15 CC2053: Calculus 7 Example 4:- solution Thus, the cost function is dollars and it costs $4200 to produce 20 units. 16 CC2053: Calculus 7 Example 5 It is estimated that x months from now the population of a certain town will be changing at the rate of people per month. The current population is 5,000. What will be the population 9 months from now? 17 CC2053: Calculus 7 Example solution Current 5:- population is 5000, i.e. P(0) = 5000 At x = 9, Thus, 9 months from now, the population is 5,126 people. 18 CC2053: Calculus 7 Integration by Substitution 19 CC2053: Calculus 7 Reversing the Chain Rule Recall: Thus, 20 CC2053: Calculus 7 Integration by Substitution 1. Introduce the letter u to stand for some expression in x that is chosen with the simplification of the integral as the goal. 2. Rewrite the integral in terms of u and du. Note: 3. Evaluate the resulting integral and then replace u by its expression in terms of x in the answer. 21 CC2053: Calculus 7 Example 6 Find 22 CC2053: Calculus 7 Example 6:- solution Let Then (Step 1) (Step 2) (Step 3) 23 CC2053: Calculus 7 Example 7 a. b. c. d. 24 CC2053: Calculus 7 Example 7a:- solution Let Then 25 CC2053: Calculus 7 Example 7b:- solution Let Then 26 CC2053: Calculus 7 Example 7c:- solution Let Then 27 CC2053: Calculus 7 Example 7d:- solution Let Then 28 CC2053: Calculus 7 Example 8 The rate of change of the unit price p (in dollars) of shampoo is given by where x is the number of bottles that supplier will make available in the market each week. a. Find the price-supply equation if the distributor of the shampoo is willing to supply 75 bottles a week at a price of $1.60 per bottle. b. How many bottles would the supplier be willing to supply at a price of $1.75 per bottle? 29 CC2053: Calculus 7 Example 8:- solution At p = 1.6, x = 75 i.e. Let Then a. Thus, the price-supply equation is b. At p = 1.75 The supplier will supply 125 bottles at price of $1.75. 30 CC2053: Calculus 7 Integration by Parts 31 CC2053: Calculus 7 Integration by Parts If u and v are differentiable functions of x, then Which can be written, in simpler form, as Integration by parts may be useful when the integrand is a product of two functions u(x) v(x). This method works if the new product on the RHS is more easily integrable than the original product on the LHS. 32 CC2053: Calculus 7 Example 9 Find 33 CC2053: Calculus 7 Example 9:- solution 34
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Hong Kong Polytechnic University - COMP - 3868
CC2053: Calculus 8Introduction to Calculus and Linear Algebra 2011/2012 Semester 1INTEGRATION: Definite Integrals1CC2053: Calculus 8Definition of a Definite IntegralLet f be a continuous defined on the closed interval a x b, and let 1. 2. 3. 4. a =
Hong Kong Polytechnic University - COMP - 3868
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Hong Kong Polytechnic University - COMP - 3868
Tutorial 1 Introduction to Calculus and Linear AlgebraTutorial 1 Introduction to Calculus and Linear AlgebraSolve Exercises 1 to 6 1. 3. 5.3x - y = 72. 4. 6.4x +3y=262x + 3y = 1 7 m + 12n = - 13 x - 11 y = - 7 y = 0.08 x5m - 3n = 7 0.2u - 0.5v
Hong Kong Polytechnic University - COMP - 3868
Tutorial 2 (Answers) Introduction to Calculus and Linear AlgebraTutorial 2 Answers Introduction to Calculus and Linear Algebra1. 1 0 - 7 3 0 12. 1 0 0 - 5 4 0 1 0 0 0 1 - 2 Infinitely many solutions. x1 = 1 s + 1 , x2 = s for all real no. s. 2 2 x1 =
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Hong Kong Polytechnic University - COMP - 3868
Tutorial 3 (Answers) Introduction to Calculus and Linear AlgebraTutorial 3 Answers Introduction to Calculus and Linear Algebra6 8 -3 12 10 - 18 4 6 24 1. 2 - 0. 2 2. 6 - 0. 6 - 0. 2 2. 2 Not defined 5 4 0 - 31 61 -3 -2 26 -2 - 11 -7 10 15 3 4 1.2.3.
Hong Kong Polytechnic University - COMP - 3868
Tutorial 3 Introduction to Calculus and Linear AlgebraTutorial 3 Introduction to Calculus and Linear AlgebraGiven the following matrices: 2 A= 0-13 - 3 B= 4 - 2 20 -1 1 C = 4 -3 5 - 2 3 2 3 - 2 1 D = 0 - 1 1 5 2 Perform the indicated operations, i
Hong Kong Polytechnic University - COMP - 3868
Tutorial 4 (Answers) Introduction to Calculus and Linear AlgebraTutorial 4 Answers Introduction to Calculus and Linear Algebra1.3. 5. 1 0 - 3 - 2 0 - 1 - 1 - 9 1 - 7 - 5 - 1 - 1 - 1 - 1 - 1 0 2. 4. Does not Exists - 5 - 2 1 26.( A) ( B)x1 = -20, x2
Hong Kong Polytechnic University - COMP - 3868
Tutorial 4 Introduction to Calculus and Linear AlgebraTutorial 4 Introduction to Calculus and Linear AlgebraIn Exercises 1 to 6, given A, find A1, if it exists. You may try row operations or determinant and cofactor approaches. Check each inverse by sho
Hong Kong Polytechnic University - COMP - 3868
Tutorial 5 (Answers) Introduction to Calculus and Linear AlgebraTutorial 5 Answers Introduction to Calculus and Linear Algebra1. 2. 3. 4. 5. 6. 2 7 5/3 5 1/3 (i) Yes (ii) No (iii) No (i) 1 (ii) 4 (iii) No A=37.8.Page 1 of 1
Hong Kong Polytechnic University - COMP - 3868
Tutorial 5 Introduction to Calculus and Linear AlgebraTutorial 5 Introduction to Calculus and Linear AlgebraIn Exercises 1 to 5, find the indicated limit if it exists. 1. 2. 3. 4. 5. 6.limx 2 -1 x 1 x - 1 x 2 - 3 x - 10 lim x 5 x-5 ( x + 1)( x - 4) li
Hong Kong Polytechnic University - COMP - 3868
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Hong Kong Polytechnic University - COMP - 3868
Tutorial 6 Introduction to Calculus and Linear AlgebraTutorial 6 Introduction to Calculus and Linear AlgebraIn Exercises 1 to 8, compute the derivative of the given function by first principle and then find the slope and equation of the line that is tan
Hong Kong Polytechnic University - COMP - 3868
Tutorial 7 (Answers) Introduction to Calculus and Linear AlgebraTutorial 7 Answers Introduction to Calculus and Linear Algebra1. 3. 5. 7.9.1 1 f ' ( x ) = 6 x 5 - 12 x 3 + 4 x + 1 + 2 3 x 2 - (t + 2) f ' (t ) = 2 (t - 2) 22. 4. 6. 8.f ' ( x) =-3 (
Hong Kong Polytechnic University - COMP - 3868
Hong Kong Polytechnic University - COMP - 3868
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Hong Kong Polytechnic University - COMP - 3868
Tutorial 8 Introduction to Calculus and Linear AlgebraTutorial 8 Introduction to Calculus and Linear AlgebraIn Exercises 1 to 8, use the chain rule to compute the derivative and simplify your answer. 1. 3.5. 7.y = u 2 + 1; u = 3 x - 2 1 y = 2 ; u = x2
Hong Kong Polytechnic University - COMP - 3868
Tutorial 9 (Answers) Introduction to Calculus and Linear AlgebraTutorial 9 Answers Introduction to Calculus and Linear Algebra1. 3. 5. 7.f ' ( x) = (6 x 2 + 20 x + 33)e 6 x 3 f ' ( x) = e 3x 2 3x -2 f ' ( x) = ( x + 1)( x - 1)2. 4. 6. 8.f ' ( x ) = -
Hong Kong Polytechnic University - COMP - 3868
Tutorial 9 Introduction to Calculus and Linear AlgebraTutorial 9 Introduction to Calculus and Linear AlgebraIn Exercises 1 to 8, differentiate the given function. 1. 3. 5. 7.f ( x) = ( x 2 + 3 x + 5)e 6 x2. 4. 6. 8.f ( x) = (1 - 3e x ) 2 f ( x) = x 2
Hong Kong Polytechnic University - COMP - 3868
Hong Kong Polytechnic University - COMP - 3868
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Hong Kong Polytechnic University - COMP - 3868
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Hong Kong Polytechnic University - COMP - 3868
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