For example 10 11  c  9  c 1.7 (B) Basic Formulae for Integration of Exponential functions. 3. x a x 4. e x dx  e x  c For example x 2x ln 2  c 5x dx  5x ln c 1.8 (C) Basic Formulae for Integration of Trigonometric functions 5. cos xdx  sin x  c , since d dx (sin x  c)  cos x 8 6. sin xdx  cos x  c , since dxd (cos x  c)  sin x 7. sec 2 xdx  tan x  c , since d (tan x  c)  sec2 x dx 8. x tan xdx  sec x  c , since d dx (sec x  c)  sec x tan x 9. cos ecx cot xdx  cos ecx  c , since d dx (cos ecx  c)  cos ecx cot x 10.         
cos ec2 xdx  cot x  c , since d cot s  cos ec2x dx Note that we shall consider integration of tan x, cot x, sec x and cosec x afterwards. 1.9 Addition and subtraction Rule in Integration Just like d dx (u  v w)  d dx u  d dx v d dx w , we have (u  v w)dx  udx  vdx wdx , where u, v, w are functions of x. For example 2 x = x3 3 cos x ex  c 1.10 Integration of constant a function kf ( x)dx  k For example 3 3 5x4 8cos xdx  8 Example 1 Find (2x5  3ex  4x  5)dx 9 Solution 5 x 5 x = 2x6 6 x 4x2 2  5x  c = x6 3  3ex  2x2  5x  c We should put one constant c after the process of integration.      
Note that dx  x  c 5dx  5 xdx  2  c Example 2 Find (2x  3)(x 2)dx Solution we find the product and then we integrate (2x  3)(x 2)dx = (2x2  x 6)dx = 2 x2 3  x2 2 6x  c Example 3 2x3  5x2  8x  9 x Solution we divide each term by x and then we integrate 2x3  5x2  8x  9 2x3 x x  5x2 x  8x x 9 x 10 9 x = 2x3 3  5x2 2  8x  9 ln x  c Example 4 x2  5x  6 (x  2) Solution First we simplify the integrand x  2     
and then integrate. x2  5x  6 (x  3)(x  2) (x  2) dx = (x  3)dx Example 5 = x2 2  3x  c Integrate 9x3  9 3 2x 3ex x 1 Solution 3  9 x3  2x  3ex x  1)dx =   4 ln2 x x2 2  Example 6 Find 2 cos x  3sin x  4sec2 x 5cos ecx cot x dx Solution 2 = 2sin x 3cos x  4 tan x  5cos ecx  c 11 Example 7 2 2 x Solution 2 2 x = 1 x = 2 ln x  5 sec x  6 cot x  e   
4 x ln 4  8x  c Exercise 1 Using the ten basic formula for integration find 3 4 x x 2 2 4. 15  1 15  8x4 3ex  7 x 5. 2 (x  5) dx 6. x2  6x  7 7. 3x2  4x3  8x5  9x  2 x2 dx 8. (x2  3x  1)(x3  7x2  2) x2 dx 9. If (2x 9)dx  0 find the values of x 10. If (3x2  2x 12)dx  0 find the values of x x3  5x2  6x 2 12 12. Find e2 x  e5x ex e x dx You have learnt the following from this chapter i). Integration is the reverse or converse process of differentiation. ii). The meaning of “integrand” and constant of integration. iii). The Ten Basic Formula for Integration, namely n xn1 n  1      
2.5 Proof of the Ten Extension Formulae We shall prove some of he Ten Extension formulae using the substitution u = ax + b so that dx = Example 1 du a a). that (ax  b)n dx  (ax  b)n1 a(n  1)  c b). Hence find (4x  9)20 dx 16 Solution Let u = a x + b so that dx = du a a) Then (ax  b)n dx  undu = 1 a n = 1 un1 a n  1  c = (ax  b)n1 a(n  1)  c 20 x21 21  c , (4x  9)20 dx = = (4x  9)21 21.4 (4x  9)21 84  c ,  c Example 2 a). Prove that 1 ln(ax  b) a  c    
24 2  (using the Basic formula we have)  9  c = 1 +c -9u 9  Type 3 ∫eu dx = e u du 1 3  e u  c  c Consider e x 3  sin x (3x 2  cos x ) dx ∫ The integrand contains the function u = x3 + sin x and its derivative (3x2 + cos x) Let u = x3 + sin x (1) du dx = 3x2 + cos x Then du = (3x2 + cosx) dx (2) Substitution of (1) and (2) in the given integral e x 3  sin x (3x 2  cos x ) dx using the basic formula u  e u  c 25  e x
Hence e sin x cos xdx = eu + c  eu du  sin x  c Example 5 10 Let u = tan x, then du = sec2 xdx Hence ∫ (tan x )10 sec2 xdx  u10 du  u11 11  c  tan11 x 11  c 4. Adjustment of the derivative Sometimes the integrand may not have the exact derivative of the function u present. In such cases we shall adjust the derivative so that the function u and its exact derivative u1 are present in the integrand. The following examples shall illustrate this principle. Example 6 Find x 2  x  1 2x3  3x3  6x  5 dx Solution The integrand has one function u = 2x3 + 3x2 + 6x + 5 Then u1 = 6x2 + 6x + 6 = 6 (x2 + x + 10) The other function present in the integrand is (x2 + x + 1). We must multiply this second function by six and divide by six to get u1. Thus x 2  x  1 1 6x 2  6x  1 Now the integrand has u and its derivative u1. Let u = 2x3 + 3x2 + 6x + 5 28 du dx
5  sec tan x Now sec5 x tan x dx = secx tan xdx . Let u = sec x du = sec x tan xdx Then sec 5 = u   c 5   c 5 Example 9 Find sin38x cos 8xdx Let u = sin 8x, du dx = 8 cos 8x du = 8 cos 8xdx Then sin38x cos 8xdx = 1 8 3 = 1 u3du  u 4 32  c  sin 4 8x 32  c 3.5 Integration of tan x and cot x (formulae) a) tan xdx  sinx cos dx Let u = cosx, du dx = -sinx, du = -sinx dx, 30 Then tan xdx  - cos dx
 du u = - ln u + c = ln u-1 + c  ln 1  ln 1 cos x  c = ln (secx) + c we write ∫tan xdx = ln secx  c showing that ln of negative values does not b) Find ∫cot xdx cot xdx  cos x sin x dx Let u = sin x so that du = cos dx Then cot xdx  du u  ln u  c cot xdx  ln (sin x )  c Exercise 3 Find the integrals of the following: dx 2. cosx  ex dx 31 3. 2
derive the formula for integration by parts apply the formula to integrate product of two functions. 4.3 Integration by parts – Formula We shall establish the formula u dv  uv v du dx dx (1) Formula (1) is also written as ∫ udv = uv – ∫vdu (2) In formula (1) the integrand contains the product of two functions namely u and dv dx The result (1) will be helpful only when du dv Formula for integration by We know that d (uv) dx = u dv dx + v du dx (3) Then u dv dx  d (uv) dx v du dx (4) Integrate both of (4) with respect to x u dv dx dx  uv v du dx dx (5) 34 or udv  uv - vdu Note: The functions u and dv dx can be Logarithmic, Inverse trigonometric, Algebraic, Trigonometric or Exponential. We must arrange the two functions in proper order in the integrand before we proceed to apply the formula. You remember the order using
Let u = x dx =cos x, dv = cosxdx du dx =1 v = ∫cos xdx v = sin x the formula dv du We have ∫x cos xdx = x sin x – ∫sin x + dx = x sin x + cos x + c Example 3 Find ∫ x3 ln x dx Solution ∫ x3 ln xdx = ∫ ln x. x3 dx writing in the order LIATE Let u = ln x dv = x3 du dx  1 x 3 x 4 4 Substituting in the formula dv v du dx dx we have 36 3 x 4 x 4 1 3 x4 4 ln x x3 4 dx  x4 4 ln x 16  c Example 4
Solution The order of x2 e2x satisfies “LIATE” Let u = x2 dv dx  e2x du dx = 2x v = ∫e2x dx e 2x 2 the formula dv du we have 2 2x x 2 e2x 2 - e2x 2 2xdx 1 2 2x 2 Now we must integrate ∫x e2x dx let u = x dv dx = e2x du dx =1 v  e 2x dx  2 e 37 Using the formula u dv dx dx  uv v du dx dx 2x 1 2 1 2 1 1 2 4
the result (2) in (1) 2 2x 2 x e 1 2 4 e )  1 2 2x 1 2 2 1 2x 4  1 2x 4 Example 5 Find ∫ x2 cos x dx Solution (In this problem Algebraic Function x2 comes before the Trigonometric function cos x. This satisfies the condition of Let u = x2 dv dx  cos x du dx  2x v  cos xdx  sin x the formula u dv dx dx  uv v du dx dx we have x 2 cos xdx  x 2 sin x - sin x 2x dx  x 2 sin x - 2 x sin xdx (1) 38 Now we must integrate x sin xdx Using again integration by parts. Let u = x dv dx  sin x du dx  1 v  sin xdx  - cosx
5.4 Integration the fundamental identities Sin2 x + cos2 x = 1 (1) 1 + tan2 x = sec2 x (2) 2 Example 1 Find sin2 x 1  cos Solution Using the identity (1) sin 2 x + cos2 x = 1 sin2 x (1 cos2 x) = (1 cos x)(1  cos (1  cos x) dx = (1 cos x)dx = x – sin x + c Example 2 Find 2 xdx 43 Solution We use 1 + tan2 x = sec2 x Then 2 xdx = (sec 2 x 2 = tan x – x + c Example 3 Find cos2 x 1 sin x
1 2 1 2 = 1 2 x 1 sin 2x 2 2  c Example 6 = 1 2 1 4 Find 2 10xdx Solution the identity cos 2x = 2 cos2 x – 1 Cos 20 x = 2 cos2 10x – 1, then cos2 10x = 1 2 1 2 Then 2 1 1 2 2 = 1 2 x  1 40 sin 20x  c Example 7 Find = cos 6x Solution Using the identity cos 2x = cos2 x – sin2 x Cos 6 x = cos2 3x – sin2 3x Hence cos 6x (cos 3x sin 3x) dx = (cos 2 3x sin 2 3x) 45 =
(cos 3x  sin 3x)(cos 3x 3x)dx (cos 3x sin 3x) (cos 3x  sin 3x)dx = sin 3x 3 cos 3x 3  c 5.6 Integration using the identities, 2 sin A cos B = sin (A + B) + sin (A – B) 2 cos A cos B = cos (A + B) + cos (A – B) -2 sin A sin B = cos (A+B) – cos (A – B) These identities convert the product into sums. The following examples illustrate the use of these identities. Example 8 Find ∫sin 10x cos8x dx Solution Using the identity 2 sin A cos B = sin (A + B) + sin (A – B) We have sin 10x cos 8xdx = 1 2 sin 10x cos 8xdx = 1 (sin18x  sin 2x)dx 1 1 1 2 18 2 = 36 4 Example 9 Find cos 12 x sin 3 xdx Solution cos 12 x sin 3 xdx = sin 3 x cos 12 xdx = 1 2 3 x 12 xdx 46 = = c
= 1 46 sin 23 x  1 14 sin 7 x  c 5.7 Integration using the identity Sin 3x = 3 sin x – 4 sin3 x Cos 3x = -3 cos x + 4 cos3 x Example 12 Find 3 6 xdx Solution the identity, sin 3x = 3 sin x – 4 sin3 x, sin 18 x = 3 sin 6x – 4 sin3 6 x sin 3 6 x  3 4 sin 6 x 1 4 sin 18 x Hence 3 6 xdx = 3 1 4 sin 18 xdx = 4 6 4 18 = cos 18 x 72 3 cos 6 x 24  c Example 13 Find 3 9 xdx Solution Using the identity cos 3 x = - 3 cos x + 4 cos3 x Cos 27 x = 3 cos 9 x +4 cos3 9 x Hence cos 3 9 x  3 4 cos 9 x  1 4 cos 27 x 3 9 xdx = 3 cos 9 xdx  1 cos 27 xdx
48 = 4 9 4 27 = 1 12 sin 9x 27 x  c 5.8 Formulae for integration when the integrand contains quadratic expressions in the denominator. 1. x 2 dx  a 2  1 a tan x a  c 2. x 2 dx a 2  1 2 a ln x a x  a  c 3. 4. dx x 2  a 2 dx x 2 a 2  sinh  cosh x a x a  if x > a if –x > a a 5. dx x 2 a 2  sin x a Example 14 Prove that x 2 dx  a 2  1 a tan x a Proof
Let x = a tan θ so that θ x a 49 Then x2 + a2 = a2 tan2 θ + a2 = a2 (1 + tan2 θ) = a2 sec2 θ dx = d (a tan θ) = a sec2 θ d θ Hence 2 dx  = = a sec 2 d 1 1  a = 1 a tan x a  c Example 13 Prove dx x 2 a 2  sin x a  c Proof Let x = a sin θ so that dx = a cos θ d θ and a 2 x 2 = a 2 a 2 sin 2   = a cos  a 2 cos 2   a cos  Hence dx x 2 a 2  a cos d a cos  = = d   c = sin x a  c x = a sin θ Example 14 Find = dx 2  4
Solution dx 25 x 2  4 = dx 4 25 1 = 25 dx 2 2 5 2 = 1 25 tan 2 x 5  c Example 15 Find x 2 dx 3 x  2 Solution x2 – 3x + 2 = x2 – 3x + a2 –a2 – 2 = (x – a)2 – a2 + 2 = x2 – 2ax + a2 – a2 + 2 Equating coefficient of x on both sides 2a = 3 or a = 3 2 2 2 2 2 –  2 2 2 1 4 Then 2 dx = x dx 2 2 2 2
51 = 1 2 ln 3 2 3 2 1 2 1 2  c Hence 2 dx  ln dx 2 x 2 x 1  c 2  1 2 a ln x a x  a Note: since x2 – 3x + 2 = (x – 2) (x – 1), we can resolve dx x 2 3 x  2 into partial fraction and then integrate to get the same answer. Try this method yourself. Example 16 Find dx 2 x 2 5 x  2 Solution 5 Let 2x2 + 5x + 1 = 2(x2 + = 2[(x + a)2 – a2 + 1] = 2[x2 + 2ax + a2 – a2 + 1] Comparing the coefficient of x on both 4a = 5 or a = 5 4 . 2 2 x  2 2  4 4 = 2 x  2 4 9 16
52 = 2 x  2 2 4 4 2 dx = 1 2 dx 2 4 4 2 = = 1 5 4 1 4 x  5 2 3 3 4 = 1 2 cosh 4 x 5 3  c if -x > 3 4 or x < - 3 4 .
53 Exercise 5 Find the integrals of the following 18. cos3 5x dx 19. cos3 8x dx 1. 2. 3. 4. 5. 6. sin 2 x 2 cos 2 x 2 2 cos 8 x (cos 4 x sin 4 x dx 20. 21. 22. 23. 24. 25. dx x 2  16 dx 2 dx x 2  25 dx x 2 25 dx 4 x 2 dx 16 x 2 7. 8. 9. cos 2 ax cos 10x sin 4x dx cos sin 9x dx 26. 27. 28. dx dx dx 9 x 2 6 x  1 10. 11. 12. 13. 14. 15. 16. dx 2 sin 5x cos 7x dx cos 8x cos 5x dx cos 3x cos 11x dx sin 4x sin dx sin sin dx sin3 sin sin3 dx 29. 30. dx 9 x 2  6 x  1 dx 25 x 2 54
Summary You have learnt the following from this chapter: i). The technique of integration of trigonometric expressions using the following trigonometric identities. 1. 2. sin 2 x  cos 2 x  1 2 1 2  1 2 1 2 cos 2 x cos 2 x 3. tan 2 x  sec 2 x 1 4. cot2 x = cosec2 x - 1 5. 2 sin A cos B = sin (A + B) + sin (A – B) 6. 2 cos A cos B = cos (A + B) + cos (A – B) 7. -2 sin A sin B = cos (A+B) – cos (A – B) 8. sin 3x = 3 sin x – 4 sin3 x 9. cos 3x = -3 cos x + 4 cos3 x ii). The technique of integration using the following formulae: 1. dx x 2  a 2  1 a tan x a  c 2. 1 2  1 2 a ln x a x  a  c 3. dx x 2  a 2  sinh x a 4. dx x 2  a 2  cosh x a if x > a  x a 55 if x < -a
2 x  3 dx 1 x 4 dx We the integrand into two partial fractions each can be integrated using the extension of the formulae for (ax + b) Then (x 11) = 2 ln (x + 3) – ln (x – 4) + A = ln (x +3)2 – ln (x – 4) + A = ln (x  3)2 x 4  A 6.4 Integration of expression with repeated linear factors in the denominator. Consider 1 (x  2)(x 1)2 dx (1) The integrand is a proper fraction and with repeated linear factors It is written as 1 (x  2)(x 1)(x 1)  A x  2  B x C (x 1)2 = A(x 1)2  B(x  2)(x 1)  C(x  2) (x  2)(x 1)(x 1) Then 1 = A (x-1)2 + B (x +2) (x – 1) + C (x + 2) (2) Let us put x = 1 on both sides of (1) so that the terms containing A and B will become zero. Then 1 = A (0) + B (0) + C (1+2), Hence C = 1 3 Similarly put x = -2 on both sides of (1) so that the terms containing B and C become zero. Then 1 = A (-2 – 1)2 + B (0) + C (0), Hence A = 1 9 . To get the value of B either we can give one more value other than 1 and –2 for x or else we can compare the coefficients of x2 on both sides of (2) 0=A+B 0= 1 9 + B then B = - 1 9 . Hence the integral (1) becomes 58 1 2 dx  1 1 1
is an improper fraction we shall divide f (x) by F (x) and the remainder is resolved into partial fractions. This principle is illustrated from the following example: Example 3 Find 3x2 2x 7 x2 x 2 dx Solution The integrand is an improper algebraic fraction. using long division we find that the quotient is 3 and the remainder is (x – 1) Then 3x2 2x 7 2  3  (x 1) (x 2)(x  1) Note: sometimes instead of long division it is easier to proceed as follows: 3x2 2x 7 x2 x 2  3(x2 x 2)  x 1 (x2 x 2) = 3  = 3  x 1 2 (x 1) (x 2)(x  1) Now we can resolve x 1 (x 2)(x  1) into partial fractions. Let x 1 (x 2)(x  1)  A x 2  B x  1 = A(x  1)  B(x 2) (x 2)(x  1) (x – 1) = A (x + 1) + B (x – 2) (2) Putting x = 2 on both sides of (2) A = 1 3 . Putting x = -1 on both sides of (2), we have – 2 = B (-3) or 2 3 =B x 1 (x 2)(x   ) e h
1 3(x 2)  2 3(x  1) the integral (1), is written as 3x2 2x 7 (x 1) 1 3(x 2)  2 1 2 3 3 Exercise 6 Find 1. x 11 2. x 3. 3x2 21x  24 (x 1)(x 2)(x 3) dx 4. 5x2 10x  11 5. x 5 2 dx 6. 5x  4 2 dx 7. 8. 3x  7 2x3 x 1 61
x3  2x2 10x 9 (x2 9) dx 10. (x 2) 3 dx Hint: a3 + b3 = (a + b) (a2 – a b + b2) 11. dx 3 a3 - b3 = (a - b) (a2 + a b + b2) 12. 3 dx Summary You have learnt the following from this chapter: i). Splitting a fraction into partial fractions. Type I: (x a)(x b) (x a) (x b) Type II: ax  b (x a)(x b)(x b)  A x a  B x b  C (x b)2 Type III: ax  b 2  A x a  Bx  C ( px2  qx  r) Type IV: For improper algebraic fraction, we must divide and consider only the remainder for partial fraction. ii). After partial fractions, integrand is transformed into the form of basic formula which can be integrated. Further Reading 1. Additional pure mathematics By L. Harwood Clarke. Heinemann Educational Books Ltd London. 62 CHAPTER 7
b a x b  Second definition of the definite Integral If f(x) is continuous in the interval a x b we definite the area enclosed by the y = f(x), the ordinates x = a, x = b, and the x axis as the definite integral of b f ( x)dx a 7.4 Some important properties of definite Integral: If f(x) is continuous in 1. a a f ( x ) dx  0 b a 2. f x )dx  f ( x a b 3. If f is integrable in a x b  and c is a point between a and b b c b f ( x )dx  f ( x ) dx  f ( x )dx a a c 4. If f(x) is integrable in a x b  and c is any constant then b b  c f ( x ) dx  f ( x c )dx a a  c b  a 5. If f(x) is integrable in a x b  and m is any constant then 64 b mb f ( x )dx  ma f m dx b b m a a m
7.5 Definition of an Indefinite Integral [or primitive of f(x)] Suppose that f(x) is integrable in the interval x d f (t )dt  f ( x ) at each point x in the interval. dx a a x b  then A function F(x) with the property that F1(x) = f(x) at point x of the interval a x b  is called an indefinite integral or primitive of f(x) Thus for a continuous function f(x) we have F ( x )  x f (t )dt a b Consider if a f ( x ) dx  G ( x ) , we say in G(x) is the indefinite integral of b f ( x ) dx a 7.6 The fundamental Theorem of the integral calculus If f(x) is continuous in f(x) then b f ( x ) dx  G (b ) G (a ) a This is called the fundamental theorem of the integral calculus Proof Let F be given by 65 F ( x )  x f (t dt a Then Since F ( x )  G ( x )  c a F (a )  (t ) dt  0 we have a F ( a )  0  G ( a )  c c  ( a )
F ( x )  G ( x )  G (a ) or F (b )  b f ( x )dx  G (b ) G (a ) where G is the primitive of f(x). a This is the Fundamental theorem of the Integral calculus. b 7.7 Evaluation of definite integrals Using the fundamental theorem of integral calculus we can evaluate definite integrals. The following examples will illustrate the evaluation of definite integrals. Example 1 Evaluate 3 2 2  2 x dx Solution 3 2 2 3 3 8 3 4 1 = 18 – 20 3 66 Example 2 = 34 3 Evaluate 2 0 2 xdx Solution 2 2 2 0 2  1 2 1 = x 
1 4 2 0  4 Example 3 2 1  sin x dx 4 Solution Cos x is the derivative of 1 + sin x Put u = I + sin x, du dx  cos x du = cos x dx 2 0 cos x 1  sin x dx  2 0 du u = ln u (For definite integral we need not put the constant of integration)  ln 1  ln 1 67  ln 2  (0 ) = ln 2 Example 4 Evaluate 3 e 0 2 x dx Solution 3 2 x 3 0 6 2 2 1 2 l t
Example 5 Evaluate 2 x sin xdx 0 Solution Using integration by parts 0 2 2 0 2  cos xdx 0 = (0 + 1) – (0 + 0) =1 Example 6 Evaluate sin 8 x sin 3 xdx 0 68 Solution sin 8 x sin 3 xdx  2 2 sin 8 x sin 3 xdx  1 2 0  2 11 0 2 5 0  1 2 0 0  1 0 0  2 =0
69 Exercise 7 Evaluate the following 2 2 1. x 0 1 2. ( x  3)( x 2) dx 0 12. 13. 14. 2 0 3 2 cos 2 x dx 3. 3 2 5  2 x 2 dx 15. 2 x 1 x dx 4. 2 0 7 dx 16. 2 0 dx ( x 1)( x 2) dx 5. cos 7 xdx 0 17. 2 ( x  1)( x 2  1) 6. 2 tan( 2 x  3) dx 0
x 3  2 x 2  sin x 0 dx 18. 19. 20. 3 ( x 1) 2 ( x 2) 1 2 1 ( x  3) dx 8. 2 x cos xdx 0 9. 1 e 0 5 x  3 dx 10. 2 7 x  3 dx 11. 2 0 2 3 xdx 70 Summary You have learnt the following from this chapter: a). The fundamental Theorem of the Integral calculus If then b f ( x ) dx  G (b ) G ( a ) a b b). can be defined as the area enclosed by the four boundaries y = f(x), a the x- axis and the ordinates x = a and x = b. Further Reading i).Additional Pure mathematics Harwood Clarke Heinemann Educational Books Ltd London
75 x4 3 x1 128 2 3 3 =42 sq units Example 3 1. Sketch the curve y  x3 3x2  2x from x = 0 to x = 3. 2. Find the area included between the curve y = x3 - 3x2 + 2x and the axis of x from x = 0 to x = 2. 3. Evaluate 2 0 3 3x2  2x dx . Interpret your result. Solution a) The following table shows some points on the curve. X 0 1 2 0.5 1.5 2 Y 0 0 0 0.375 -0.375 0 The graph is shown in the figure 0 1 2 Figure 7 b) Part of the area is positive and the other part is negative. Hence we should find each area separately. Area from x =0 to x = 1 is given by A  1 x0 3 3x2  2xdx 76 x4 4
that the improper integral 1 1 x dx is convergent or divergent. b). Show that 1 1 x dx is divergent. Solution a). If lim b b 1 1 x dx converges (or exists) we say that 1 1 x dx converges otherwise we say that it diverges. b). lim b l 1 1 x dx = b ln x1b 84 = lim b ln b – ln 1 lim b = ln = ln b since ln 1 = 0 Since b 1 1 x dx dose not exist, it diverges. Example 2 Show that the improper integral 1 1 x 2 dx is convergent Solution 1 1 x 2 b b 1 x 2 dx exists or does not. b l
b 1 1 x 2 dx = b = = lim b lim b b x 1 1 1 b = lim b 1 1 b = 1 (Limit exists and finite) Hence 1 1 x 2 dx is convergent. 85 9.5 Definition of absolute convergence and conditional convergence of an improper integral i).The improper integral, a f ( x ) dx is said to be absolutely convergent if a f ( x ) dx is convergent. ii).The improper integral a f ( x ) dx is said to be conditionally convergent if b a f ( x ) dx is convergent but a f ( x ) dx is not convergent. 9.6 Cauchy’s inequality for definite integrals Cauchy’s inequality states that b a f ( x ) dx b a f ( x ) dx This inequality is very useful to determine the absolutely convergent and conditionally convergent improper integrals.
0 xe x dx is convergent if b b 0 xe x dx is convergent. Using integration by parts lim b l 0 xe x x b 0 + b e 0 x dx = lim b  1 b b 87 Hence 0 xe x dx is convergent. 9.7 Theorem: if b a f ( x )dx is an absolutely convergent improper integral it is also convergent, where b can be finite or infinite. By Cauchy’s inequality, b a f ( x ) dx b a f ( x ) dx If b a f ( x ) dx is convergent (or absolutely convergent) then it has an upper bound M Hence Then b a b a f ( x )dx f ( x )dx M M and hence it is convergent. e
In other words b a f ( x )dx is convergent if b a f ( x ) dx is convergent. Example 5 Show that e 0 x cos 2x dx is absolutely convergent. Solution We must show that 0 e - x cos 2x dx is convergent. 88 0 e - x cos 2x dx = lim b b 0 e - x cos 2x dx b lim b 0 l 0 lim b -1 1 e b 1 lim b b 0 e - x cos 2x dx is convergent since the limit exists e 0 - x cos 2x dx is convergent. Hence e 0 x cos 2x is absolutely convergent. 9.8 Convergence and divergence of the improper integrals of the second kind. Consider the integral b a f ( x) dx suppose that f(x) has a singularity at an interior c of a
x l  Now b a f ( x) dx = c a f ( x) dx b + f ( x) c dx (1) The improper integral b a f ( x) dx is said to be convergent if both the improper integrals on the right side of (1) exist. The sum of these is defined as the value of the improper integral from a to b in (1) 89 b 9.9 The Cauchy’s principal value of f ( x) dx a is a singular point of f(x). If b a f ( x) dx = c a f ( x) dx + b c f ( x) dx does not exist but lim  0 a c   b f ( x) dx does exist. We call this limit the Cauchy’s principal value of b a f ( x) dx Thus the Cauchy’s principal value of b a f ( x) dx  lim  0 a c f (x)dx  b Example 6 a) Show that 2 1 1 x dx is improper. b) Find the Cauchy’s principle value of 2 1
1 x Solution a) The integrand 1 x has a singularity at x = 0 in the interval  ] Hence 2 1 1 x is an improper integral of the second kind. b) 2 1 1 x dx  0 1 1 x dx  2 0 1 x dx Clearly 0 1 1 x and 2 0 1 x dx both diverge; 90 0 1 1 x dx  lnx 1  ln 0 ln 1   ln 0 does not 2 0 1 x dx  lnx 02  ln 2 ln( 0 ) ln(0) does not exist. Hence 2 1 1 x does not exist. However -  1 1 x dx  2 1 x dx exist when ε 0 Since  1 1 x dx  lnx 1  ln   ln 1   ln  and 2 1 x dx  lnx 2  ln 2 ln  Hence -  1 1 x dx  2 1 x dx  ln   ln 2 ln 
Thus the Cauchy’s principal value of 2 1 1 x dx  ln 2 Example 7 Show that 3 2 dx x 4 i). does not converge ii). does not have a principal value Solution The integral 3 2 dx x 4 is improper x = 0 is a singularity on the interval [-2 ≤ x ≤ 3] Now 3 2 dx x 4  0 - 2 dx x 4  3 0 dx x 4 91 1 3 x 3 0 - 2 1 3 x 3 3 0  1 24 1 81 does not converge The Cauchy’s principal value of 3 2 dx x 4   - 2 dx x 4  3 dx x 4 where  0 1 3 x 3 -  - 2 1 3 x 3 3   1 3  0 = 3 1 24 1 81 1 3  3 where
Exercise 9 1. Classify the following into Improper integrals of the first kind and Improper integrals of the second kind. a). 0 x 2  1 x 1  dx b). x sin x x dx c). 3 2 dx x 4 d). 5 2 x 3 ( x 3 ) dx 2. When will you say that the improper integral a f ( x ) dx converges or diverges. 3. When will you say that the improper integral 2 2 dx x 3 converges or diverges. 92 4. Define the Cauchy’s principal value of the improper integral b a f ( x ) dx if f (x) has a singularity at c a < c < b 5. Define absolute convergence and conditional convergence of an improper integral b a f ( x ) dx 6. Show that 7. Show that 1 0 1 dx x 5 dx x 5 is divergent . is divergent. 8. Show that e 0 x cos3xdx is absolutely convergent. x 2 1 dx is absolutely convergent or not.
10. Find Cauchy’s principal value of 4 1 3 x dx 11. Evaluate 2 ln(sin 0 x ) dx 12. that cos 0 2 xdx converges (Hint: Make the substitution u = x2) a 13. Define f(x)dx - - 14. Test the following integrals for convergence a). 0 sin x b). 3 1 x lnx dx c). 0 x 2 (1  x 2 ) 2 dx d). 0 x sinx dx 93 15. Test whether the principal value for the integral 5 2 dx x 4 exists. 16. Show that 0 sin x cos 4 x x dx converges. Summary You have learnt the following from the chapter: i. Definition of improper integrals of first and second kind. ii. Convergence and divergence of improper integrals. iii. Absolute convergence and conditional convergence of improper integrals. iv. Cauchy’s inequality for absolute convergence.
v. The cauchy’s principal of the improper integrals b a f ( x ) dx when f(x) has a singularity at c where a ≤ c ≤ b Further reading 1. Advanced Calculus By Watson Fulks John and sons New York. Brisbane Toronto. 94 CHAPTER 10 Length of an Arc of a Curve 10.1 Introduction If you ask a secondary school student to measure the length of an arc of a curve, he will use a string or a flexible material like a string placed upon the arc of the curve and he measures the length of the string placing against a ruler. Also an approximation of length could be found by marking a number of points on the arc and finding the sum of the lengths of all the chords that are small. Any mathematician could do the same method unless he knows the equation of the curve either in Cartesian form or Parametric or any other form. In this chapter we shall learn the method of finding the length of an arc of a curve. 10.2 Objectives of the chapter By the end of this chapter you should be able to.  derive an expression or formula for S, the arc length of a curve if the Cartesian or parametric equation of the is known.  calculate accurately the length of the using the formula. 10.3 Length of an arc S of a curve in Cartesian equation is b S  a 1  dx dx 2 Pr ∆y Pn Po P1
1 using integration by parts. Example 2 a). Draw the curve 6 y2 = x (x – 2)2 b). Find the point of intersection of the curve with x-axis. c). Calculate the length of the loop of the curve y2 = 0 1 2 Figure 3 b) Let 6 y2 = x (x – 2)2 Differentiating with to x 1 6 x (x – 2)2 12 y dy dx = (x – 2)2 + 2 x (x – 2) dy dx = (x – 2) (3x – 2) = 12 y 1   1  2 (x 2)2 (3x 2)2 24x(x 2)2 = (x 2)2 (3x 2)2 24x(x 2)2 = 1  (3x 2)2 24x Length of the loop = a 2 x0 1  dx dx 2 = 2 0 (3x  2) 24x dx
98 = 2 0 2 6  2 2 6 x = 3 2 2  3x 2 2 6 3 2 0 = = 1 6 8 3 3 3 2 Example 3 Show that the length of the circumference of a circle of radius a is 2 π a. Solution (o, a) 0 B x2 + y2 = a2 P A (a, o) Figure 4 The equation of the circle of radius a is x2 + y2 = a2. Differentiating with respect to x, dy dy 2x x dx dx 2 y y 99 of arc APB = t h
Figure 5 Let o Pr –1 = r and o Pr = r + ∆ r x o Pr – 1 = θ and angle Pr – 1 o pr = ∆ θ Let N Pr – 1 be perpendicular to o Pr. Now N Pr – 1 = r ∆ θ and N Pr = ∆ r Hence (Pr – 1 Pr)2 = (N Pr)2 + (N Pr – 1)2 = (∆ r)2 + (r ∆ θ)2 2 2 2 2 2  S   2 d 2 where α and β are the values of θ corresponding to at the end points of the curve. Example 6 Find the length of the equiangular spiral r = a ek θ from θ = 0 to θ = 2π. Solution The equation of the is given in polar form. The length of arc of a curve in polar form is given by 103  S   2 d 2 = 0 k 2 dae k d 2 = 0 a 2 2k   d =
Example 1 Find the volume of a cone of height h and base radius r. Solution A r 0 h (h, 0) Figure 2 A cone is formed by the revolution of the area under the line OA about the x-axis. 107 The gradient of the line OA is r h and the equation of the line OA is y = r h x. If V is the volume of the cone V  xh x0 2 r 2 2 h h  r 2 h3 h2 3 = 1 3 r 2h units Example 2 Find the volume of a hemisphere of radius r units. Hence show that the volume of a sphere of radius r is given by 4 3 Solution
= 4rxo 111 = 4r 2 Example 5 axis. Solution 2 t3 3 the The curve cuts the x- axis at y=0 or t t3 3  0 , Hence t=0 or t  3 or x=0 or x=3. Eliminating t from x=t2 and y  t t3 3 we have the Cartesian equation of the curve, 2 2 2 2   since (d s)2 = (d x)2 + (d y)2 dt dt dt 2 = 2 2 ds dt  1  t 2 The required surface area = 2 yds 3 t3 3 3 0 2t 3 3 = 2 6 =  3 3 = 3 Sqr units
a) Calculate 2 x dx using trapezoidal rule with 5 intervals (Assume 1 is not known) b) Suppose that you know 1 = ln x + c Calculate the area under the curve. c) Calculate the percentage Solution y0 =1.0000 y0 = 1.0000 y1 = 0.8333 2y1 = 1.6666 y2 = 0.7143 2y2 = 1.4286 y3 = 0.6250 2y3 = 1.2500 y4 = 0.5556 2y4 = 1.1112 y5 =0.5000 y5 = 0.5000 Total Sum = 6.9564 = 2 1 5  0.2 120 Then 2 xdx = h  2 y1  2 y2  2 y3  2 y4  2 y5  = 0.2 2 6.9564 = 1 10 x6.9564 = 0.69564 = 0.6956 to four decimals. b) Now, 2 1
Example 4 Find 6 0 ydx , given the following values, using i) Simpsons Rule for the interval (0 ≤ x≤ 6) ii) Trapezoidal Rule for the interval (0 ≤ x ≤6) X 0 1 2 3 4 5 6 Y 8 12 14 11 9 3 1 Solution i) Area under curve using Simpsons Rule Y0 = 8 Y0 = 8 125 Y1 = 12 4Y1 = 48 Y2 = 14 2Y2 = 28 Y3 = 11 4Y3 = 44 Y4 = 9 2Y4 = 18 Y5 = 3 4Y5 = 12 Y6 = 1 Y6 = 1 Total Sum 159 h = 6 0 6 =1 Approximate area in therefore. A = 1 h(y0 + 4y1 + 2y2 + 4y3 + 2y4 + 4y5 + y6) 3 = 1 (1)(159) 3 = 53 sq ii) Area under curve using Trapezoidal Rule (≤ 6) Y0 = 8 Y0 = 8