Question
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
âĞe^(-y/a) dy
= -a(e^(-y/a) from 0 to â
= -a/e^(y/a) from 0 to â
as y -->â
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 â
= -a/e^(y/a) from 0 to â
as y -->â
a/e^(y/a) ---> 0
so the integral from above
= 0 - (a/e^0) )
= 0 - a/1
= -a
Related Questions
integral -oo, oo [(2x)/(x^2+1)^2] dx
(a) state why the integral is improper or involves improper...
1. integral -oo, oo [(2x)/(x^2+1)^2] dx
2. integral 0, pi/2 cot(theta) d(theta)
(a) state why...
Evaluate the integral: 16csc(x) dx from pi/2 to pi (and determine if it is convergent or divergent)....
Evaluate the integral: 16csc(x) dx from pi/2 to pi (and determine if it is convergent or divergent)....