Showing if the improper integral is convergent or divergent

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InfinityMist
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I’ll send pic of question and working
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InfinityMist
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InfinityMist
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(Original post by InfinityMist)
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CloutMaster
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Give me a sec boss
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DFranklin
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(Original post by InfinityMist)
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There are 3 lines there. Which don't you get?
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CloutMaster
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Ok basically, if a is less than one, it's power will be positive therefore as t tends towards infinity it also tends towards infinity and therefore diverges. If a is greater than an 1, it tends towards 0 so you can get rid of that part of the equation after saying a is greater than 1. Therefore you're left with - (1/-a+1) which gets you the answer in the book, only when a is greater than one, which is the point it converges
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CloutMaster
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(Original post by CloutMaster)
Ok basically, if a is less than one, it's power will be positive therefore as t tends towards infinity it also tends towards infinity and therefore diverges. If a is greater than an 1, it tends towards 0 so you can get rid of that part of the equation after saying a is greater than 1. Therefore you're left with - (1/-a+1) which gets you the answer in the book, only when a is greater than one, which is the point it converges
I forgot to mention, when a is greater than 1 it tends to 0 because t would be raised to a negative power. eg, 1/infinity tends towards 0. That's why you can get rid of that bit of the integral after stating the range of a
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davros
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(Original post by CloutMaster)
I forgot to mention, when a is greater than 1 it tends to 0 because t would be raised to a negative power. eg, 1/infinity tends towards 0. That's why you can get rid of that bit of the integral after stating the range of a
Please don't say "1/infinity tends towards 0". That makes my head hurt
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KaziMahathir
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(Original post by InfinityMist)
I’ll send pic of question and working
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InfinityMist
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(Original post by KaziMahathir)
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2 digits. 6.2
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KaziMahathir
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(Original post by InfinityMist)
2 digits. 6.2
BRO! You didn't have to do it!
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