# Calculus 2

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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!

• Calculus 2 -

∫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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