Assignment 3

# Assignment 3 - MAT 1332 Winter 2009 Assignment#3 Solutions...

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MAT 1332, Winter 2009, Assignment #3, Solutions Total marks=29. 1. Using the substitution u =2+5 x , then du =5 dx , and then we fnd that dx = du 5 , and the limits oF integration [2] are From u =2to u = . ThereFore, ! 0 1 (2+5 x ) 4 dx = ! 2 1 5 u 4 dx = 1 15 u 3 | 2 = 1 120 . 2. This is the type II improper integral. Using the substitution u = x 1, then du = dx , and the limits oF [2] integration are From u =0to u = 2. ThereFore, ! 3 1 5 x 1 dx = ! 2 0 5 u du = lim ± 0 + ! 2 ± 5 u du = lim ± 0 + 5 · 2 u 1 2 | 2 ± = lim ± 0 + (10 2 10 ± 1 2 )=10 2 . 3. ±or x> 1, 0, then x + e x >e x ,so [3] ! 1 1 x + e x dx < ! 1 1 e x dx = lim T →∞ ! T 1 1 e x dx = lim T →∞ ( e x | T 1 ) = lim T →∞ ( e T + e 1 )= e 1 . [2] So this integral is convergent. [1] An upper bound is e 1 = 1 e . 4. Let b ( t ) denote the population oF bacteria at time t . Then b (0) = 1000. [4] (a) [1] The pure-time di²erential equation is db dt = 1000 (2+3 t ) 1 . 5 . (b) [2] Approach 1: The population starting with b (0) = 1000, the total change between t = 0 and t will be given by the defnite integral " t 0 1000 (2+3 s ) 1 . 5 ds , so we can write down the solution For this equation as Follows, b ( t ) = 1000 + ! t 0 1000 s ) 1 . 5 ds (using the substitution u =2+3 s, then ds = du 3 , the limits oF integration are From u u t. ) = 1000 + ! 2+3 t 2 1000 3 · 1 u 1 . 5 du = 1000 + 1000 3 ! 2+3 t 2 · 1 u 1 . 5 du = 1000 + 1000 3 · ( 1 0 . 5 u 0 . 5 | 2+3 t 2 ) = 1000 + 2000 3 # u 0 . 5 | 2+3 t 2 \$ = 1000 + 2000 3 [ t ) 0 . 5 +2 0 . 5 ] = 1000 + 2000 3 · 2 0 . 5 2000 3 t ) 0 . 5 1471 667 · t ) 0 . 5 . Since the population would approach a limiting size oF 1471, the rule oF this sort oF growth can be maintained indefnitely. 1

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Approach 2: To answer this question, we can just need to test the convergence of the improper integral !
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Assignment 3 - MAT 1332 Winter 2009 Assignment#3 Solutions...

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