ImproperIntegrals

ImproperIntegrals - Improper Integrals Definitions I +∞...

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Unformatted text preview: Improper Integrals Definitions I +∞ 1. t ∫ f ( x)dx = lim ∫ f ( x)dx a b 2. t →+ ∞ a b ∫ f ( x)dx = lim ∫ f ( x)dx −∞ t → −∞ t Pr ovided these limits exist; in which case the integral is said to converge to the value of the limits. Otherwise the integral is said to diverge. +∞ +∞ −∞ 2. c −∞ c ∫ f ( x)dx = ∫ f ( x)dx + ∫ f ( x)dx c +∞ −∞ c provided both of the improper integrals ∫ f ( x)dx and c ∈R ∫ f ( x)dx exist , where Examples I Example (1) +∞ t 1 1 dx = lim ∫ dx = lim (arctan x) ∫ 1+ x2 t →+ ∞ 1 + x 2 t →+ ∞ 0 0 = lim [arctan t − arctan 0] t →+ ∞ = lim [arctan t − 0] t →+ ∞ = lim (arctan t ) t →+ ∞ = ππ = 22 t 0 Example (2) 0 0 1 1 dx = lim ∫ dx = lim (arctan x) ∫∞1 + x 2 t → −∞ 1 + x 2 t → −∞ − t = lim [arctan 0 − arctan t ] t → −∞ = lim [0 − arctan t ] t → −∞ = − lim (arctan t ) t → −∞ π π = 0 − (− ) = 2 2 0 t Example (3) +∞ 1 ∫∞1 + x 2 dx − 0 +∞ 1 1 =∫ dx + ∫ dx 2 2 1+ x 1+ x −∞ 0 ππ =+ 22 =π Example (4) +∞ t dx dx I=∫ = lim ∫ = lim (ln x) x t →+ ∞1 x t →+ ∞ 1 = lim [ln t − ln 1] t →+ ∞ = lim [ln t − 0] t →+ ∞ = lim ln t t →+ ∞ =+∞ (Why ?) Thus I diverges t 1 Example (5) dx x −4 I = ∫ 4 = lim ∫ x dx = lim ( t →+ ∞ t →+ ∞ x 1 1 +∞ t −3 −3 1 1 = lim [( ) − ( )] 3 t → + ∞ − 3t −3 1 1 = lim ( ) − lim ( ) 3 t → + ∞ − 3t t →+ ∞ − 3 1 1 = 0 − (− ) = 3 3 t ) 1 Example (6) Evaluate the following integral if p is greater than 1 − p +1 x dx I=∫ = lim ∫ x dx = lim ( ) x − p +1 t +∞ t p 1 t →+ ∞ 1 −p t →+ ∞ 1 1 = lim [( )−( )] p −1 t → + ∞ ( − p + 1)t − p +1 1 1 = lim ( ) − lim ( ) p −1 t → + ∞ ( − p + 1)t t →+ ∞ − p + 1 1 1 = 0−( )= − p +1 p −1 1 Example (7) Evaluate the following integral if p is less than 1 x dx −p I = ∫ p = lim ∫ x dx = lim ( t →+ ∞ t →+ ∞ 1x 1 +∞ t − p +1 ) − p +1 t 1− p 1 = lim [( )−( )] t → + ∞ ( − p + 1) − p +1 =+∞ (Why ?) Because, t 1− p lim [( ) = + ∞ , while, t → + ∞ ( − p + 1) t 1− p 1 lim [( )= ∈R t → + ∞ ( − p + 1) − p +1 t 1 Consequence +∞ dx I=∫ p x 1 diverges { ; 1 converges to ; p −1 p ≤1 p is greater than 1 Example (8) 0 0 I = ∫ e dx = lim ∫ e dx = lim (e x ) x x t → −∞ −∞ = lim [e 0 − e t ] t → −∞ = lim [1 − e t ] t → −∞ = lim 1 − lim e t t → −∞ t → −∞ = 1− 0 = 1 t t → −∞ 0 t Example (9) +∞ t I = ∫ e dx = lim ∫ e dx = lim (− e −x t →+ ∞ 0 0 = − lim[e −t − e 0 ] t →∞ = − lim[e −t − 1] t →∞ −t = −[lim e − lim1] t →∞ = −[0 − 1] = 1 t →∞ −x t →+ ∞ −x ) t 0 Definitions II 1. ∫ b a b f ( x) dx = lim ∫ f ( x) dx + t →a t where f is contoninuous on (a, b] and x = a is a vertical asymptote for f 2. ∫ b a t f ( x) dx = lim ∫ f ( x) dx − t →b a where f is contoninuous on [a, b) and x = b is a vertical asymptote for f 3. ∫ b a c b a c f ( x) dx = ∫ f ( x) dx + ∫ f ( x) dx Pr ovided that both integrals exist where f is contoninuous on [a, c) and (c, b] and x = c is a vertical asymptote for f , c ∈ (a, b) Examples II dx 1. ∫ 0x 1 dx = lim ∫ dx +t t →0 x 1 1 = lim[ln x] + t →0 t = lim[ln 1 − ln t ] + t →0 = lim[0 − ln t ] + t →0 = lim[− ln t ] + t →0 =∞ (Why ?) Example (1) Example (2) 1. dx 1 ∫ 0 2 ( x − 1) 3 t dx = lim ∫ − t →1 = lim[ − t →1 0 2 ( x − 1) 3 ( x − 1) 1 3 1 3 t 0 1 3 = lim[3(t − 1) − (−3)] − t →1 1 3 = lim[3(t − 1) ] − lim(−3) − − t →1 = 0+3=3 t →1 Example (3) 1. ∫ dx 2 2 3 1 = lim ∫ + t →1 = lim[ + t →1 t ( x − 1) 2 dx ( x − 1) ( x − 1) 1 3 1 3 2 3 2 t 1 3 1 3 = lim[3(2 − 1) − 3(t − 1) ] + t →1 1 3 = lim 3 − lim[3(t − 1) ] + + t →1 t →1 = 3−0 = 3 Example (4) 1. ∫ 2 0 dx 2 ( x − 1) 3 1 2 dx dx =∫ 2 +∫ 2 0 1 ( x − 1) 3 ( x − 1) 3 = 3+3 =6 ...
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This note was uploaded on 10/12/2011 for the course MATH 201 taught by Professor Foad during the Spring '11 term at Qatar University.

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