Integrals - Mathematics, Study notes of Mathematics

Download Integrals - Mathematics and more Study notes Mathematics in PDF only on Docsity! 7.1 Overview 7.1.1 Let d dx F (x) = f (x). Then, we write ( )f dxx∫ = F (x) + C. These integrals are called indefinite integrals or general integrals, C is called a constant of integration. All these integrals differ by a constant. 7.1.2 If two functions differ by a constant, they have the same derivative. 7.1.3 Geometrically, the statement ( )f dxx∫ = F (x) + C = y (say) represents a family of curves. The different values of C correspond to different members of this family and these members can be obtained by shifting any one of the curves parallel to itself. Further, the tangents to the curves at the points of intersection of a line x = a with the curves are parallel. 7.1.4 Some properties of indefinite integrals (i) The process of differentiation and integration are inverse of each other, i.e., ( ) ( )d f dx fx x dx =∫ and ( ) ( )' Cf dx fx x= +∫ , where C is any arbitrary constant. (ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. So if f and g are two functions such that ( ) ( ) d d f dx g x dxx dx dx =∫ ∫ , then ( )f dxx∫ and ( )g dxx∫ are equivalent. (iii) The integral of the sum of two functions equals the sum of the integrals of the functions i.e., ( ) ( )( ) dxf gx x+∫ = ( )f dxx∫ + ( )g dxx∫ . Chapter 7 INTEGRALS 144 MATHEMATICS (iv) A constant factor may be written either before or after the integral sign, i.e., ( )a f dxx∫ = ( )a f dxx∫ , where ‘a’ is a constant. (v) Properties (iii) and (iv) can be generalised to a finite number of functions f 1 , f 2 , ..., f n and the real numbers, k 1 , k 2 , ..., k n giving ( ) ( ) ( )( )1 1 2 2 ... , n nk f k f k f dxx x x+ + +∫ = ( ) ( ) ( )1 1 2 2 ... n nk f dx k f dx k f dxx x x+ + +∫ ∫ ∫ 7.1.5 Methods of integration There are some methods or techniques for finding the integral where we can not directly select the antiderivative of function f by reducing them into standard forms. Some of these methods are based on 1. Integration by substitution 2. Integration using partial fractions 3. Integration by parts. 7.1.6 Definite integral The definite integral is denoted by ( ) b a f dxx∫ , where a is the lower limit of the integral and b is the upper limit of the integral. The definite integral is evaluated in the following two ways: (i) The definite integral as the limit of the sum (ii) ( ) b a f dxx∫ = F(b) – F(a), if F is an antiderivative of f (x). 7.1.7 The definite integral as the limit of the sum The definite integral ( ) b a f dxx∫ is the area bounded by the curve y = f (x), the ordi- nates x = a, x = b and the x-axis and given by ( ) b a f dxx∫ = (b – a) ( ) ( )( )1 lim ( ) ... –1 n f a f f a ha h n n→∞ + + ++  INTEGRALS 147 Example 2 Evaluate 2 2 2 3ax dx b c x+∫ Solution Let v = b2 + c2x2 , then dv = 2c2 xdx Therefore, 2 2 2 3ax dx b c x+∫ = 2 3 2 a dv c v∫ = 2 2 2 2 3 log C 2 a b c x c + + . Example 3 Verify the following using the concept of integration as an antiderivative. 3 2 3 – – log 1 C 1 2 3 x dx x x x x x = + + + +∫ Solution 2 3 – – log 1 C 2 3 d x x x x dx   + + +   = 1 – 22 3 1– 2 3 1 x x x + + = 1 – x + x2 – 1 1x + = 3 1 x x + . Thus 2 3 3 – – log + 1 C 2 3 1 x x x x x dx x   + + =  +  ∫ Example 4 Evaluate 1 1 – x dx x + ∫ , 1.x ≠ Solution Let 1 I = 1 – x dx x + ∫ = 2 1 1– dx x ∫ + 21– x dx x ∫ = –1 1sin Ix + , 148 MATHEMATICS where I 1 = 2 1 – x dx x . Put 1 – x2 = t2 ⇒ –2x dx = 2t dt. Therefore 1I = – dt = – t + C = 2 – 1– Cx + Hence I = sin–1x 2 – 1– Cx + . Example 5 Evaluate ( )( ) , β – – dx x x >∫ α α β Solution Put x – α = t2. Then – xβ = ( )2 – tβ α+ = 2 – –tβ α = 2 – –t α β+ and dx = 2tdt. Now ( )2 2 2 I = – – t dt t t ∫ β α ( )2 2 = β – –α dt t ∫ 2 2 2 – dt k t = , where 2 –k β α= = –1 –1 – 2sin C 2sin C – t x k + = + α β α . Example 6 Evaluate 8 4 tan secx x dx∫ Solution I = 8 4 tan secx x dx∫ = ( )8 2 2 tan sec secx x x dx∫ = ( )8 2 2 tan tan 1 secx x x dx+∫ INTEGRALS 149 = 10 2 8 2tan sec tan secx x dx x x dx+∫ ∫ = 11 9tan tan C 11 9 x x+ + . Example 7 Find 3 4 23 2 x dx x x+ +∫ Solution Put x2 = t. Then 2x dx = dt. Now I = 3 4 2 2 1 23 2 3 2 x dx t dt x x t t = + + + +∫ ∫ Consider 2 A B 1 23 2 t t tt t = + + ++ + Comparing coefficient, we get A = –1, B = 2. Then I = 1 2 – 2 2 1 dt dt t t    + + ∫ ∫ = 1 2log 2 log 1 2 t t + − +  = 2 2 2 log C 1 x x + + + Example 8 Find 2 22sin 5cos dx x x+∫ Solution Dividing numerator and denominator by cos2x, we have I = 2 2 sec 2tan 5 x dx x +∫ 152 MATHEMATICS Example 11 Find 8 2 10 – 10 – x dx x x+∫ Solution We have I = 8 2 10 – 10 – x dx x x+∫ ...(1) = ( ) 8 2 10 – (10 – ) 10 – 10 – 10 – x dx x x+∫ by (P 3 ) ⇒ 8 2 I = 10 – x dx x x+∫ (2) Adding (1) and (2), we get 8 2 2I 1 8– 2 6dx= = =∫ Hence I = 3 Example 12 Find 4 0 1 sin 2x dx π +∫ Solution We have I = ( ) 4 4 2 0 0 1 sin 2 sin cosx dx x x dx π π + = +∫ ∫ = ( ) 4 0 sin cosx x dx π +∫ INTEGRALS 153 = ( ) 4 0 cos sinx x π − + I = 1. Example 13 Find 2 –1tanx x dx∫ . Solution 2 –1I = tanx x dx∫ = 3 –1 2 2 1tan – . 1 3 x x x dx dx x+∫ ∫ = 3 –1 2 1tan – 3 3 1 x x x x dx x   − +  ∫ = 3 2 –1 21tan – log 1 C 3 6 6 x x x x+ + + . Example 14 Find 210 – 4 4x x dx+∫ Solution We have 2I = 10 – 4 4x x dx+∫ ( ) ( )2 2= 2 –1 3x dx+∫ Put t = 2x – 1, then dt = 2dx. Therefore, ( )221 I = 3 2 t dt+∫ 2 21 9 9 = log 9 C 2 2 4 t t t t + + + + + ( ) ( ) ( ) ( )2 21 9= 2 –1 2 –1 9 log 2 –1 2 –1 9 C 4 4 x x x x+ + + + + . 154 MATHEMATICS Long Answer (L.A.) Example 15 Evaluate 2 4 2 2 x dx x x+ −∫ . Solution Let x2 = t. Then 2 4 2 2 A B ( 2) ( 1) 2 12 2 x t t t t t tx x t t = = = + + − + −+ − + − So t = A (t – 1) + B (t + 2) Comparing coefficients, we get 2 1 A , B 3 3 = = . So 2 4 2 2 2 2 1 1 1 3 32 2 1 x x x x x = + + − + − Therefore, 2 4 2 2 2 2 1 1 3 32 2 1 x dx dx dx x x x x = + + − + −∫ ∫ ∫ = –12 1 1 1 tan log C 3 6 12 2 x x x − + + + Example16 Evaluate x x x dx 3 + ∫ 4 9 Solution We have I = 3 x x x dx + ∫ 4 9 = 3 x x dx x dx x 4 4 9 9  ∫ + = I 1 + I 2 . INTEGRALS 157 Solution I = ( ) 1 2–1 0 tanx dxx∫ . Integrating by parts, we have I = ( ) 2 12–1 0 tan 2 x x   – 1 –1 2 2 0 1 tan .2 2 1 x x dx x+∫ = 12 2 –1 2 0 π – .tan 32 1 x x dx x+∫ = 2π 32 – I 1 , where I 1 = 1 2 –1 2 0 tan 1 x xdx x+∫ Now I 1 = 1 2 2 0 1 –1 1 x x + +∫ tan–1x dx = 1 1 –1 –1 2 0 0 1tan – tan 1 x dx x dx x+∫ ∫ = I 2 – ( )( )12–1 0 1 tan2 x = I 2 – 2π 32 Here I 2 = 1 –1 0 tan x dx∫ = ( ) 1 1–1 0 2 0 –tan 1 x dxx x x+∫ = ( )1 2 0 π 1– log 1 4 2 x+ = π 1– log 2 4 2 . Thus I 1 = 2π 1 π– log 2 4 2 32 − 158 MATHEMATICS Therefore, I = 2 2π π 1 π– log 2 32 4 2 32 + + = 2π π 1– log 2 16 4 2 + = 2π – 4π log 2 16 + . Example 19 Evaluate 2 –1 ( )f x dx∫ , where f (x) = |x + 1| + |x| + |x – 1|. Solution We can redefine f as ( ) 2 – , if –1 0 2, if 0 1 3 , if 1 2 x x f x xx x x < ≤ = + < ≤  < ≤ Therefore, ( ) ( ) ( ) 2 0 1 2 –1 –1 0 1 32 – 2f dx dx dx x dxx x x= + ++∫ ∫ ∫ ∫ (by P 2 ) = 0 1 22 2 2 –1 0 1 32 – 2 2 2 2 x x x x x      + ++           = 1 1 4 10 – 3–2 – 2 – 2 2 2 2      + ++           = 5 5 9 19 2 2 2 2 + + = . Objective Type Questions Choose the correct answer from the given four options in each of the Examples from 20 to 30. Example 20 ( )cos –sinxe dxx x∫ is equal to (A) cos Cxe x + (B) sin Cxe x + (C) – cos Cxe x + (D) – sin Cxe x + INTEGRALS 159 Solution (A) is the correct answer since ( ) ( ) ( )' Cx xe dx e ff f xx x = + + ∫ . Here f (x) = cosx, f (x) = – sin x. Example 21 2 2sin cos dx x x∫ is equal to (A) tanx + cotx + C ( B ) ( t a n x + cotx)2 + C (C) tanx – cotx + C (D) (tanx – cotx)2 + C Solution (C) is the correct answer, since I = 2 2sin cos dx x x∫ = ( )2 2 2 2 sin cos sin cos dxx x x x + ∫ = 2 2sec cosecx dx x dx+∫ ∫ = tanx – cotx + C Example 22 If – – 3 – 5 4 5 x x x x e e dx e e+∫ = ax + b log |4ex + 5e–x| + C, then (A) –1 7 , 8 8 a b= = (B) 1 7 , 8 8 a b= = (C) –1 –7 , 8 8 a b= = (D) 1 –7 , 8 8 a b= = Solution (C) is the correct answer, since differentiating both sides, we have – – 3 –5 4 5 x x x x e e e e+ = a + b ( )– – 4 –5 4 5 x x x x e e e e+ , giving 3ex – 5e–x = a (4ex + 5e–x) + b (4ex – 5e–x). Comparing coefficients on both sides, we get 3 = 4a + 4b and –5 = 5a – 5b. This verifies –1 7 , 8 8 a b= = . 162 MATHEMATICS Example 27 If 1 0 1 te dt t+∫ = a, then ( ) 1 2 0 1 te dt t+∫ is equal to (A) a – 1 + 2 e (B) a + 1 – 2 e (C) a – 1 – 2 e (D) a + 1 + 2 e Solution (B) is the correct answer, since I = 1 0 1 te dt t+∫ = ( ) 1 1 2 0 0 1 1 1 t t e dte t t + + +∫ = a (given) Therefore, ( ) 1 2 0 1 te t+∫ = a – 2 e + 1. Example 28 2 –2 cos dxx xπ∫ is equal to (A) 8 π (B) 4 π (C) 2 π (D) 1 π Solution (A) is the correct answer, since I = 2 –2 cos dxx xπ∫ = 2 0 2 cos dxx xπ∫ = 2 1 3 22 2 1 30 2 2 cos cos cosdx dx dxx x x x x x    + +π π π      ∫ ∫ ∫ = 8 π . Fill in the blanks in each of the Examples 29 to 32. Example 29 6 8 sin cos x dx x∫ = _______. INTEGRALS 163 Solution 7tan C 7 x + Example 30 ( ) – a a f dxx∫ = 0 if f is an _______ function. Solution Odd. Example 31 ( ) 2 0 a f dxx∫ = ( ) 0 2 a f dxx∫ , if f (2a – x) = _______. Solution f (x). Example 32 2 0 sin sin cos n n n x dx x x π +∫ = _______. Solution 4 π . 7.3 EXERCISE Short Answer (S.A.) Verify the following : 1. 2 –1 2 3 x dx x +∫ = x – log |(2x + 3)2| + C 2. 2 2 3 3 x dx x x + +∫ = log |x2 + 3x| + C Evaluate the following: 3. ( )2 2 1 dxx x + +∫ 4. 6log 5log 4log 3log – – x x x x e e dx e e∫ 164 MATHEMATICS 5. ( )1 cos sin x dx x x + +∫ 6. 1 cos dx x+∫ 7. 2 4tan secx x dx∫ 8. sin cos 1 sin 2 x x dx x + +∫ 9. 1 sin xdx+∫ 10. 1 x dx x +∫ (Hint : Put x = z) 11. – a x a x + ∫ 12. 1 2 3 41 x dx x+ ∫ (Hint : Put x = z4) 13. 2 4 1 x dx x + ∫ 14. 216 – 9 dx x ∫ 15. 23 – 2 dt t t ∫ 16. 2 3 –1 9 x dx x +∫ 17. 25 – 2x x dx+∫ 18. 4 –1 x dx x∫ 19. 2 41 – x dx x∫ put x2 = t 20. 22 –ax x dx∫ 21. ( ) –1 3 2 2 sin 1 – x dx x ∫ 22. ( )cos5 cos 4 1 – 2cos3 x x dx x + ∫ 23. 6 6 2 2 sin cos sin cos x x dx x x + ∫ INTEGRALS 167 49. ( ) ( )sin sin– – dx x a x b∫ is equal to (A) sin (b – a) log sin( – ) sin( – ) x b x a + C (B) cosec (b – a) log sin( – ) sin( – ) x a x b + C (C) cosec (b – a) log sin( – ) sin( – ) x b x a + C (D) sin (b – a) log sin( – ) sin( – ) x a x b + C 50. –1tan x dx∫ is equal to (A) (x + 1) –1tan – Cx x + (B) –1tan – Cx x x + (C) –1– tan Cx x x + (D) ( ) –1– tan C1x xx ++ 51. 2 2 1– 1 x x e dx x    + ∫ is equal to (A) 2 C 1 xe x + + (B) 2 – C 1 xe x + + (C) ( )22 C 1 xe x + + (D) ( )22 – C 1 xe x + + 52. ( ) 9 624 1 x x + ∫ dx is equal to (A) –5 2 1 1 C4 5x x   ++   (B) –5 2 1 1 C4 5 x   ++   (C) ( )–51 C1 4 10x ++ (D) –5 2 1 1 C4 10 x   ++   168 MATHEMATICS 53. If ( )( )22 1 dx x x+ +∫ = a log |1 + x2| + b tan–1x + 1 5 log |x + 2| + C, then (A) a = –1 10 , b = –2 5 (B) a = 1 10 , b = – 2 5 (C) a = –1 10 , b = 2 5 (D) a = 1 10 , b = 2 5 54. 3 1 x x +∫ is equal to (A) 2 3 – log C1 – 2 3 x x x x+ + + (B) 2 3 – – log C1 – 2 3 x x x x+ + (C) 2 3 – – – log C1 2 3 x x x x ++ (D) 2 3 – – log C1 2 3 x x x x+ ++ 55. sin 1 cos x x dx x + +∫ is equal to (A) log C1 cos x ++ (B) log Csinx x ++ (C) – tan C 2 x x + (D) .tan C 2 x x + 56. If 33 2 22 2 (1 ) 1 C, 1 = + + + + +∫ x dx a x b x x then (A) a = 1 3 , b = 1 (B) a = –1 3 , b = 1 (C) a = –1 3 , b = –1 (D) a = 1 3 , b = –1 INTEGRALS 169 57. π 4 – π 4 d 1 + cos2 x x∫ is equal to (A) 1 (B) 2 (C) 3 (D) 4 58. π 2 0 1 – sin 2xdx∫ is equal to (A) 2 2 (B) 2 ( )2 1+ (C) 2 (D) ( )2 2 –1 59. π 2 sin 0 cos xxe dx∫ is equal to _______. 60. ( )2 3 4 xx e dx x + +∫ = ________. Fill in the blanks in each of the following Exercise 60 to 63. 61. If 2 0 1 1 4 a dx x+∫ = π 8 , then a = ________. 62. 2 sin 3 4cos x dx x+∫ = ________. 63. The value of π −π ∫ sin3x cos2x dx is _______.