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Showing if the improper integral is convergent or divergent
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#6
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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#7
(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
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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#8
(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
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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