Integration

Chapter 1 : Integration QS025 1 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 1 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.1 INTEGRATION OF FUNCTIONS LEARNING OUTCOMES : At the end of the lesson, students should be able to: (a) relate integration and differentiation (b) use the basic rules of integration Integration as Anti-differentiation Integration is the reverse process of differentiation. In differentiation, we start with an expression and then proceed to find its derivative. In integration, we start with the derivative and then work backward to find the function from which it is derived. The process of finding anti-derivatives is called anti-differentiation or integration. If F(x) f (x) dx d , then the functions of the form F(x)C are the anti-derivatives of f (x) . We denote this by writing f (x) dx F(x) C . “ ” is called an integral sign and C is the constant of integration. Basic Rules of Integration. (i) k dx kx c (ii) , 1 1 1 C n n x x dx n n (iii) k f(x) dx k f(x) dx (iv) [ f(x) g(x)] dx f(x) dx g(x) dx (v) , 1 ( 1) 1 c n a n (ax b) (ax b) dx n n

Chapter 1 : Integration QS025 2 Mathematics Unit KMK 06010 Changlun Kedah Example 1: Evaluate the following integrals: a) 3 dx c) dx x 1 b) x dx 6 Solution: Example 2: Evaluate the following integrals: a) 2 x dx b) dx x x 3 2 4 1 Solution:

Chapter 1 : Integration QS025 3 Mathematics Unit KMK 06010 Changlun Kedah Example 3: Evaluate the following integrals a) x x dx 3 2 6 2 b) x dx 2 2 5 c) 2x 1 3x 2 dx d) x dx 9 5 7 Solution:

Chapter 1 : Integration QS025 4 Mathematics Unit KMK 06010 Changlun Kedah Exercises 1 Find the following integrals. 1. z dz 4 3 7 2. dx x x 3 5 4 3. x (x 2) dx 2 4. dx x x 2 1 2 5. x x dx 2 3 4 5 6. x dx 5 3 9 Answer: 1. z c 4 7 4 2. c x x 2 5 2 3. c x x 3 2 4 4 3 4. c x x x 1 4 3 4 3 5. c x x 4 5 3 4 3 4 6. x c 6 3 9 54 1

Chapter 1 : Integration QS025 5 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 2 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.1 INTEGRATION OF FUNCTIONS LEARNING OUTCOMES : At the end of the lesson, students should be able to: (a) determine integration of x 1 , x e and x a (b) determine the integral of the forms: (i) dx f x f x ( ) '( ) (ii) f x e dx f x ( ) '( ) (iii) f x f x dx n '( ) ( ) (iv) a dx px q The Integral of x 1 (special case of x dx n , when n 1 ) Remembered that x x dx d 1 (ln ) . Therefore, reversing the process gives dx x C x ln 1 Example 1: Find the following integrals. a) dx x 2 b) dx 3x 1 Solution:

Chapter 1 : Integration QS025 6 Mathematics Unit KMK 06010 Changlun Kedah The integral of ( ) ( ) f x f x In general, ( ) ( ) ln ( ) f x f x f x dx d dx f x c f x x ln f Example 2: Find the following integrals. a) dx x x 1 2 2 b) dx 2x 1 10 c) dx x x x x 3 2 2 2 3 Solution:

Chapter 1 : Integration QS025 7 Mathematics Unit KMK 06010 Changlun Kedah The integral of ( ) '( ) f x f x e e e e dx e c dx d x x x x e f x e f x e dx e c dx d f x f x f x f x ( ) ( ) ( ) ( ) '( ) '( ) Example 3: Find the following integrals. a) e dx 3x 3 b) x e dx 2 x x 2 3 3 1 c) e dx 5x Solution:

Chapter 1 : Integration QS025 8 Mathematics Unit KMK 06010 Changlun Kedah The integral of n f'(x) f (x) If f (x) is a function of x and f (x) f '(x), dx d n n f x f x n f x dx d '( ) ( ) 1 ( ) 1 . Reversing the process, C n f x f x f x dx n n 1 ( ) '( ) ( ) 1 Example 4: Find the following integrals. a) x x dx 2 4 2 3 b) dx x ln x c) x x 5 dx 2 3 Solution:

Chapter 1 : Integration QS025 9 Mathematics Unit KMK 06010 Changlun Kedah The Integral of px q a Example 5: Find dx x 2 Solution: NOTE: c a a a a a a dx dx d x x x x ln ( ) ln

Chapter 1 : Integration QS025 10 Mathematics Unit KMK 06010 Changlun Kedah Example 6: Find dx 2x 1 3 Solution: NOTE: c p a a a a p a a dx dx d p x q p x q p x q p x q ln ( ) ln

Chapter 1 : Integration QS025 11 Mathematics Unit KMK 06010 Changlun Kedah Exercises 2 Questions : Answer: 1. dx x 8x 3 7 2 x c x ln 8 3 7 2. dx x x x 2 3 4 3 2 3 x x c 2 ln 3 4 3 2 1 3. x x 1 dx 2 x c 2 3 2 1 3 1 4. dx x x 1 3 2 x c 2 1 3 1 3 2 5. dx e e x 4x 2 e e c x x 3 3 1 2 6. dx x x 2 2 1 c x x ln 2 2 ln 2 2 1

Chapter 1 : Integration QS025 12 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 3 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.2 INTEGRATION OF TRIGONOMETRIC FUNCTIONS LEARNING OUTCOMES : At the end of the lesson, students should be able to integrate trigonometric functions in the form sinax , cosax , ax 2 sec , ax 2 sin and ax 2 cos . Integration of Trigonometric Functions ax c a ax a ax ax dx dx d ax c a ax a ax ax dx dx d ax c a ax a ax ax dx dx d tan 1 tan sec sec cos 1 cos sin sin sin 1 sin cos cos 2 2 Example 1: Find the following integrals. a) sin 4x dx d) 3sin2x cos6x dx b) dx x 4 3 2cos e) cos7x 5sec 3x dx 2 c) dx x 2 sec2 f) x dx x cos4 cos 2 3 2 Solution:

Chapter 1 : Integration QS025 13 Mathematics Unit KMK 06010 Changlun Kedah Using Double Angle Formulae sin2x 2sin xcos x x x x 2 2 cos2 cos sin 2 1 cos 2 sin2 x x and 2 1 cos 2 cos2 x x 2cos 1 2 x x 2 1 2sin Example 2: Find the following integrals. a) x dx 2 sin b) x dx 2 2cos c) dx x 2 3sin2 d) 4sin xcos x dx Solution:

Chapter 1 : Integration QS025 14 Mathematics Unit KMK 06010 Changlun Kedah Using Compound Angle Formulae sinA B sin AcosB cos AsinB cosA B cos AcosB sin AsinB Example 3: Find the following integrals. a) sin x cos 2x cos x sin 2x dx b) cos 2x cos 3x sin 2x sin 3x dx Solution:

Chapter 1 : Integration QS025 15 Mathematics Unit KMK 06010 Changlun Kedah Exercises 3 Questions : Answer: 1. sec 2x cos3x dx 2 c x x 3 sin 3 2 tan 2 2. dx 2cos ec3x 9 cos 3x c 2 3 3. dx x x 2 sin 3sec 5 1 2 2 x x sin x c 2 1 sin10 60 1 3 2 4. dx x x 3 2 cos 3 2 sin c x 3 4 cos 8 3 5. sin x sin 2x cos x cos 2x dx sin 3x c 3 1

Chapter 1 : Integration QS025 16 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 4 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.3 TECHNIQUES OF INTEGRATION LEARNING OUTCOMES : At the end of the lesson, students should be able to use substitution method to find integrals. Integration by Substitution Integration by substitution is used to compute indefinite integrals of the form f gxg xdx ……… (1) To compute (1) we make a change of variable, or substitution. We let u = g(x) ………….. (2) then g'(x) dx du or du g'(x)dx ………….(3) Making these substitutions (2) and (3) in (1), we obtain the integral f (g(x))g'(x)dx = f (u)du This method is illustrated in the following examples. Example 1 Find x 3 dx 2 3 2 1 Solution:

Chapter 1 : Integration QS025 17 Mathematics Unit KMK 06010 Changlun Kedah Example 2 Find (2x 3) x 1dx Solution:

Chapter 1 : Integration QS025 18 Mathematics Unit KMK 06010 Changlun Kedah Note: ax b C a dx ax b ln( ) 1 1 Example 3 Find dx 5 3x 1 Solution: Example 4 Find dt t e t 2 1 3 Solution:

Chapter 1 : Integration QS025 19 Mathematics Unit KMK 06010 Changlun Kedah Example 5 Find dx 2x 3 5 Solution: Example 6 Find dx x x 3 (ln 2 ) Solution :

Chapter 1 : Integration QS025 20 Mathematics Unit KMK 06010 Changlun Kedah Example 7 Prove that du u u dx x x 2 2 1 2 1 2 . Hence, find dx x x 1 2 2 . Solution: Example 8 Find x x dx 2 3 3 sin Solution:

Chapter 1 : Integration QS025 21 Mathematics Unit KMK 06010 Changlun Kedah Example 9 Find cot xdx Solution: Example 10 Find cos xsin x dx 3 Solution:

Chapter 1 : Integration QS025 22 Mathematics Unit KMK 06010 Changlun Kedah Exercise 4 Find the following integrals: dt t t 4 2 3 1 a) dx 2x ln x 1 b) x e dx x 4 3 c) 2 d) cos 2x sin 2xdx 4 Answers: c t 3 2 18 3 1 1 a) - b) ln ln x c e c x 4 2 1 c) sin 2x c 10 1 d) 5

Chapter 1 : Integration QS025 23 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 5 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.3 TECHNIQUES OF INTEGRATION LEARNING OUTCOMES : At the end of the lesson, students should be able to perform integration by parts. Integration by Parts Suppose u and v are differentiable functions of x. By the product rule of differentiation dx du v dx dv u dx d uv ( ) …………. (1) Integrating both sides of equation (1) with respect to x, dx dx du v dx dv dx u dx d(uv) dx dx du dx v dx dv uv u Thus, dx dx du dx uv v dx dv u This technique is called integration by parts. The idea is to choose u and dx dv , so that dx dx du v is simpler than the dx dx dv u . Example 1 Find x x 1dx Solution:

Chapter 1 : Integration QS025 24 Mathematics Unit KMK 06010 Changlun Kedah Example 2 Find xe dx x Solution : Example 3 Find xe dx 3x 1 2 Solution:

Chapter 1 : Integration QS025 25 Mathematics Unit KMK 06010 Changlun Kedah Example 4 Find ln x dx Solution : Example 5 Find x ln x dx Solution

Chapter 1 : Integration QS025 26 Mathematics Unit KMK 06010 Changlun Kedah Example 6 Find x e dx 2 x Solution Example 7 Find x sin 2x dx 2 Solution :

Chapter 1 : Integration QS025 27 Mathematics Unit KMK 06010 Changlun Kedah Exercises 5 Find the following integrals: dx x x 3 ln a) xe dx 1 2x b) c) x cos3xdx t e dt 3 3t d) Answers: c x x x - 2 2 4 1 2 ln a) e c xe x x 1 2 1 2 4 1 2 b) x x cos3x c 9 1 sin 3 3 1 c) t e t e te e c t t t t 3 3 2 3 3 3 27 2 9 2 3 1 3 1 d) -

Chapter 1 : Integration QS025 28 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 6 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.3 TECHNIQUE OF INTEGRATION LEARNING OUTCOMES : At the end of the lesson, student should be able to evaluate the integral of a rational function by means of decomposition into partial fractions. Rational Functions A rational function is a fraction where both the numerator and the denominator are polynomials Example: 2 1 x x , 1 1 x x , ( 1)( 2) 3 x x , ( 2)(3 4) 2 1 x x x , 1 2 x x If the degree of the numerator is less than the degree of the denominator, the fraction is called proper fraction and if the degree of the numerator is greater or equal to that of the denominator, then the function is called improper fraction If the denominator can be factorised then the function can be expressed as the sum of partial fraction. Some functions and their partial fractions are given below: Function Partial fraction (x 1)(x 1) x 1 1 x B x A 2 ( 1) 1 x 2 1 ( 1) x B x A ( 1) 1 2 x x x ( 1) 2 x Bx C x A 2 2 ( 1) 2 x 2 2 2 1 ( 1) x Cx D x Ax B

Chapter 1 : Integration QS025 29 Mathematics Unit KMK 06010 Changlun Kedah Example 1 Find dx x(1 x) 5 Solution Example 2 Use the method of partial fraction to find dx x x x 2 3 2 1 2 Solution

Chapter 1 : Integration QS025 30 Mathematics Unit KMK 06010 Changlun Kedah Example 3 Find dx x x x x ( 2) 4 5 4 12 2 2 Solution

Chapter 1 : Integration QS025 31 Mathematics Unit KMK 06010 Changlun Kedah Exercise 6 Find the following integrals by using partial fractions dx x x 2 5 3 a) ans: c x x 5 2 ln 5 dx x x x 2 2 3 2 4 3 b) ans: x- x c 2 ln 3 2 2 1 ln 2

Chapter 1 : Integration QS025 32 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 7 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.4 DEFINITE INTEGRALS LEARNING OUTCOMES : At the end of the lesson, students should be able to: a) use the properties of definite integral b) evaluate definite integrals. The Properties of Definite Integral i. cdx c(b a) b a , where c is any constant ii. b a b a b a f (x) g(x) dx f (x)dx g(x)dx iii. b a b a cf (x)dx c f (x)dx , where c is any constant iv. a b b a f (x)dx f (x)dx v. ( ) 0 a a f x dx vi. c a b a c b f (x)dx f (x)dx f (x)dx , where a b c and if f (x) 0 for a x b , then ( ) 0 b a f x dx .

Chapter 1 : Integration QS025 33 Mathematics Unit KMK 06010 Changlun Kedah Example 1: Given ( ) 5 3 1 f x dx and 8 3 f (x) 10 . Find (a) 3 3 f (x)dx (d) 8 1 f (x)dx (b) 1 3 f (x)dx (e) 8 3 ( f (x) 5)dx (c) f x dx 3 1 4 ( ) Solution: Example 2: Given ( ) 5, ( ) 10 8 5 5 2 f x dx f x dx and 5 2 g(x) 6 . Evaluate (a) 5 2 3 f (x)dx (c) f x g x dx 5 2 2 ( ) 3 ( ) (b) 2 5 g(x)dx (d) 8 2 f (x)dx Solution:

Chapter 1 : Integration QS025 34 Mathematics Unit KMK 06010 Changlun Kedah Evaluate Definite Integrals Example 1: Find 2 0 4 2 e dx x Solution: Example 2: Evaluate dx x x 1 0 4 3 1 2 Solution:

Chapter 1 : Integration QS025 35 Mathematics Unit KMK 06010 Changlun Kedah Example 3: Find 4 0 tan3 xdx Solution: Example 4: Find e xdx 1 ln 2 Solution:

Chapter 1 : Integration QS025 36 Mathematics Unit KMK 06010 Changlun Kedah Example 5: Evaluate 3 0 2 x 3x 4 dx Solution: Example 6: Evaluate x x dx 2 0 (cos 4 sin ) Solution:

Chapter 1 : Integration QS025 37 Mathematics Unit KMK 06010 Changlun Kedah Exercise 7 1. Given 7 2 f (x)dx 5 , find the value of (a) 7 2 3 f (x)dx (c) 7 2 (x 2 f (x))dx (b) 7 2 ( f (x) 2)dx (d) 2 7 2 ( f (x) x )dx 2. Evaluate (a) 3 1 1 0 (x 3)dx e dx x (b) 3 3 sin x cos xdx 3. Evaluate the following (a) dx e e e x x x 2 1 (c) 1 2 (ln x)xdx (b) dx x x 3 2 1 3 (d) x x x dx 4 0 2 3 (6 1)(3 ) Answers: 1. a) 15 b) 15 c) 2 25 d) 3 320 2. a) 9 + e b) 0 3. a) 1.059 b) 0.3863 c) -0.6363 d) 1,827,904

Chapter 1 : Integration QS025 38 Mathematics Unit KMK 06010 Changlun Kedah a b x f(x) LECTURE 8 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.4 DEFINITE INTEGRALS LEARNING OUTCOMES : At the end of the lesson, students should be able to determine the area of a region. (a) Area of a Region Bounded By The Curve And The x – Axis Area under a curve can be computed using definite integration. Hence the formal definition: Notes : Area is always positive area = b a f (x)dx area = b a f (x)dx a b x f(x) If f is continuous throughout [a, b] , then the area of the region between the curve y = f( x ) and the x – axis from x = a to x = b is given by area = b a f (x) dx

Chapter 1 : Integration QS025 39 Mathematics Unit KMK 06010 Changlun Kedah Example 1: Find the area of the region bounded by the line y 2x , x-axis and the line x 4 . Solution: Example 2: Find the area of region bounded by f (x) x 2x 2 , x-axis, the lines x 0 and x 3. Solution:

Chapter 1 : Integration QS025 40 Mathematics Unit KMK 06010 Changlun Kedah d x f(x) c d c x f(x) (b) Area of a Region Bounded By The Curve And The y – Axis area = d c g( y)dy area = d c g( y)dy Example 3: Find the area enclosed by y 4x 2 , y-axis, the lines y 2 and y 2 . Solution: If g is continuous throughout [c , d ] , then the area of the region between the curves x = g( y ) and the y – axis from y = c to y = d is given by area = d c g( y) dy

Chapter 1 : Integration QS025 41 Mathematics Unit KMK 06010 Changlun Kedah Example 4: Find the area of region bounded by x y 1 , 2 y 4 , 4 y 2 and y axis. Solution:

Chapter 1 : Integration QS025 42 Mathematics Unit KMK 06010 Changlun Kedah (c) Area of a Region Bounded By Two Curves Example 5: Find the area of the region bounded by the curve y x 4x 2 and the line 2 x y . Solution: x a b f(x) g(x) Definition If f and g are continuous with f(x) ≥ g(x) throughout [ a , b ] , then the area of the region between the curves y = f( x ) and y = g ( x ) from x = a to x = b is the integral of [f – g] from a to b. area = b a [ f (x) g(x)]dx y

Chapter 1 : Integration QS025 43 Mathematics Unit KMK 06010 Changlun Kedah Example 6: Calculate the area of the region bounded by the curve y x 2 and y x 2 . Solution: Example 7: Calculate the area of the region bounded by the curve 3 2 y x , 3y 2x 14 and x-axis. Solution:

Chapter 1 : Integration QS025 44 Mathematics Unit KMK 06010 Changlun Kedah Exercise 8 1. Find the area enclosed by the curve y ln x the line y 1 and x and y axes. 2. Show that the area enclosed by the line y 4 2x and curve 2 y 4 x is 2 3 4 unit 3. Find the area of the region bounded by the graphs y 8 x 2 and y x 2 . 4. Find the area of the region bounded by the curves y 2 x 1 and y x between x 0 and x 4 . Answer: 1. 2 e 1 unit 2. 2 3 4 unit 3. 2 3 64 unit 4. 2 3 28 unit

Chapter 1 : Integration QS025 45 Mathematics Unit KMK 06010 Changlun Kedah LECTURE 9 OF 9 TOPIC : 1.0 INTEGRATION SUBTOPIC : 1.4 DEFINITE INTEGRALS LEARNING OUTCOMES : At the end of the lesson, students should be able to find the volume of a solid of revolution. Volume of Solids Generated By Revolving The Region Bounded By The Curve And The x or y Axis In this section we use integration to calculate the volumes of solids. Consider the solids of revolution in the following figures, which are formed by revolving certain regions about an axis. cylinder cone sphere

Chapter 1 : Integration QS025 46 Mathematics Unit KMK 06010 Changlun Kedah y = f (x) (a) Revolve about the x–axis The volume of the solid generated by revolving the region about the x-axis between the graph of a continuous function y = f (x) and the x-axis from x = a to x = b is Volume, V y dx b a 2 V π f(x) dx b a 2 x = a x = b x (b) Revolve about the y–axis The volume of the solid generated by revolving the region about the y axis between the graph of a continuous function x = g (y ) and the y-axis from y = c to y = d is Volume, V x dy d c 2 y V g( y) dy d c 2 y = d x = g ( y) y = c x y

Chapter 1 : Integration QS025 47 Mathematics Unit KMK 06010 Changlun Kedah Example 1: Find the volume of the solid formed when the area bounded by the curves x y 1 , y-axis, y 2 and y 5 is rotated about y-axis. Solution:

Chapter 1 : Integration QS025 48 Mathematics Unit KMK 06010 Changlun Kedah Example 2: Let R be the region bounded by the curve 1 2 y x . Calculate the volume of the solid formed when it is rotated about x axis Solution: Example 3: Let R be the region bounded by curve x y e , and 2 x 5. Calculate the volume of the solid formed when R is rotated about the x-axis. Solution:

Chapter 1 : Integration QS025 49 Mathematics Unit KMK 06010 Changlun Kedah Example 4: The area R bounded by the curves y x 2 , and between 0 y 2 is revolved completely about y axis . Calculate the volume of the solid formed. Solution:

Chapter 1 : Integration QS025 50 Mathematics Unit KMK 06010 Changlun Kedah y x a b y1=f(x) y2=g(x) R Volume of Solids Generated By Revolving The Region Bounded By Two Curves (a) Revolve About x Axis Suppose that y 1 = f (x ) and y 2 = g (x ) are nonnegative continuous function such that , f (x ) ≥ g (x ) for a ≤ x ≤ b and let R be the region enclosed between the graphs of these functions and the lines x = a and x = b. When this region is revolved about the x-axis, it generates a solid . The volume of the solid generated by revolving R about x-axis is V ( y y ) dx 2 2 2 1 b a or V f(x) g(x) dx b a 2 2 (b) Revolve About y-Axis Suppose that x 1 = f (y ) and x 2 = g ( y) are nonnegative continuous function such that , f (y ) ≥ g ( y) for c ≤ y ≤ d and let R be the region enclosed between the graphs of these functions and the lines y = c and y = d. When this region is revolved about the y- axis, it generates a solid.