Calculus 2
posted 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!

∫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
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