Asked by Sara
The question is:
Evaluate the improper integral for a>0.
The integral is:
the integral from 0 to infinity, of e^(-y/a)dy
Can anyone help me solve this? When I try I get 'a', which apparently is incorrect. Thank you!
Evaluate the improper integral for a>0.
The integral is:
the integral from 0 to infinity, of e^(-y/a)dy
Can anyone help me solve this? When I try I get 'a', which apparently is incorrect. Thank you!
Answers
Answered by
Reiny
∫e^(-y/a) dy
= -a(e^(-y/a) from 0 to ∞
= -a/e^(y/a) from 0 to ∞
as y -->∞
a/e^(y/a) ---> 0
so the integral from above
= 0 - (a/e^0) )
= 0 - a/1
= -a
= -a(e^(-y/a) from 0 to ∞
= -a/e^(y/a) from 0 to ∞
as y -->∞
a/e^(y/a) ---> 0
so the integral from above
= 0 - (a/e^0) )
= 0 - a/1
= -a
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