Indefinite Integration - MATH1010 Questions Indefinite...

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MATH1010 Questions: Indefnite Integration 1 Preliminaries Exercise 1. (Level 3) (a) Let f : R \{ 0 } −→ R be the function deFned by f ( x ) = 1 x 2 for any x R \{ 0 } . Consider the functions F 1 , F 2 , F 3 , F 4 , F 5 , whose ‘formulae of deFnition’ are given below. Each of them has domain R \{ 0 } . Which of them are primitives of f ? Which not? Why? i. F 1 ( x ) = 1 x for any x R \{ 0 } . ii. F 2 ( x ) = 1 x + 1 for any x R \{ 0 } . iii. F 3 ( x ) = b 1 /x if x < 0 1 /x + 1 if x > 0 iv. F 4 ( x ) = b 1 /x + 3 if x < 0 1 /x 2 if x > 0 v. F 5 ( x ) = 1 /x + 3 if x < 0 1 /x 2 if 0 < x < 1 1 /x + 4 if x 1 (b) Let f : R \{ 0 R be the function deFned by f ( x ) = 1 x for any x R \{ 0 } . Consider the functions F 1 , F 2 , F 3 , F 4 , F 5 , whose ‘formulae of deFnition’ are given below. Each of them has domain R \{ 0 } . Which of them are primitives of f ? Which not? Why? i. F 1 ( x ) = ln( | x | ) for any x R \{ 0 } . ii. F 2 ( x ) = ln( | x | ) + 1 for any x R \{ 0 } . iii. F 3 ( x ) = b ln( x ) if x < 0 ln( x ) + 1 if x > 0 iv. F 4 ( x ) = b ln( x ) + 3 if x < 0 ln( x ) 2 if x > 0 v. F 5 ( x ) = b ln( x ) if x < 0 ln( x ) if x > 0 1
(c) Let f : R \{− 1 , 1 } −→ R be the function deFned by f ( x ) = 1 if x < 1 0 if 1 < x < 1 1 if x > 1 . Consider the functions F 1 , F 2 , F 3 , F 4 , whose ‘formulae of deFnition’ are given below. Each of them has domain R \{− 1 , 1 } . Which of them are primitives of f ? Which not? Why? i. F 1 ( x ) = x if x < 1 0 if 1 < x < 1 x if x > 1 ii. F 2 ( x ) = x + 1 if x < 1 0 if 1 < x < 1 x 1 if x > 1 iii. F 3 ( x ) = x 1 if x < 1 0 if 1 < x < 1 x + 1 if x > 1 iv. F 4 ( x ) = x 2 if x < 1 100 if 1 < x < 1 x + 3 if x > 1 Solution. (a) F 1 , F 2 , F 3 , F 4 are primitives of f , because their respective Frst deriva- tives on R \{ 0 } are f . F 5 is not even di±erentiable at 1. Then F 5 is not a primitive of f . (b) F 1 , F 2 , F 3 , F 4 are primitives of f , because their respective Frst derivatives on R \{ 0 } are f . F 5 is di±erentiable on R \{ 0 } but F 5 ( 1) n = f ( 1). Then F 5 is not a primitive of f . (c) F 1 , F 2 , F 3 , F 4 are primitives of f , because their respective Frst derivatives on R \{− 1 , 1 } are f . Exercise 2. (Level 3) ²ind i f ( x ) dx for the following functions f ( x ). (a) f ( x ) = 4 x 1 , x < 1 3 x , x 1 (b) f ( x ) = b e 2 x , x < 0 cos 2 x, x 0 2
(c) f ( x ) = | x 3 | (d) f ( x ) = 1 1 + | x | (e) f ( x ) = 3 | x 2 4 | (f) f ( x ) = | ln x | Solution. (a) F ( x ) + C where F ( x ) = b 2 x 2 x + 5 , x < 1 6 x, x 1 (b) F ( x ) + C where F ( x ) = b e 2 x 2 , x < 0 2 x +sin 2 x +2 4 , x 0 (c) | x 3 | ( x 3) 2 + C (d) F ( x ) + C where F ( x ) = b ln(1 x ) , x < 0 ln(1 + x ) , x 0 (e) F ( x ) + C where F ( x ) = x 3 12 x, x < 2 x 3 + 12 x + 32 , 2 x < 2 x 3 12 x + 64 , x 2 F ( x ) + C where F ( x ) = b x (1 ln x ) , 0 < x < 1 x (ln x 1) + 2 , x 1 2 Integration by substitution Exercise 1. (Level 1) Evaluate the following integrals by trigonometric substitution. (a) i x 2 1 + x 2 dx (b) i dx (1 x 2 ) 3 2 (c) i r 1 + x 1 x dx (d) i dx (1 + x 2 ) 3 2 3
(e) i x 2 dx 9 x 2 (f) i dx 4 + x 2 (g) i x 2 r 16 x 2 dx (h) i dx x 2 x 2 + 4 (i) i dx (4 x 2 + 1) 3 / 2 (j) i 1 (2 x x 2 ) 3 / 2 Solution. (a) x tan 1 x + C (b) x 1 x 2 + C (c) 1 x 2 + sin 1 x + C (d) x 1 + x 2 + C (e) 9 2 sin 1 x 3 x 2 9

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