Comparison Test for improper integral

[sanhuy]

Homework Statement

use the comparison theorem to determine whether ∫ 0→1 (e^-x/√x) dx converges.

Homework Equations

I used ∫ 0 → 1 (1/√x) dx to compare with the integral above

The Attempt at a Solution

i found that ∫ 0 → 1 (1/√x) dx = 2 ( by substituting 0 for t and take the limit of the defenite integral as t → 0^+) thus it is convergent. and (1/√x) > (e^-x/√x) so ∫ 0→1 (e^-x/√x) dx also converges.

But in my textbook they only use the comparison theorem for the limits of integration from 0 to ∞. would it still be acceptable to use the comparison theorem for this problem for limits of integration from 0 to 1.
Thanks for reading :).

 

[Samy_A]

But in my textbook they only use the comparison theorem for the limits of integration from 0 to ∞. would it still be acceptable to use the comparison theorem for this problem for limits of integration from 0 to 1.
Thanks for reading :).

I don't see any particular reason why the comparison theorem for integrals wouldn't be valid with other limits of integration than $0$ to $\infty$.

See for example here.