Discussion
An integral is improper of the second kind when
Answer: A
The first kind has an infinite limit of integration; the second has a finite interval but an integrand that blows up at an endpoint or an interior point, as \( \int_{0}^{1} dx/\sqrt{x} \) does at the origin. A change of sign raises no difficulty at all, and a removable discontinuity does not affect the value
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