7.8 Integration Techniques
Question #2 (Easy): Improper Integral With Infinite Discontinuity
Whether the improper integral converges or diverges, they are still considered improper integrals and
grouped under Type 1 (infinite interval) and Type 2 (finite interval, but infinite discontinuity due to
vertical asymptotes caught within the finite interval).
Explain why the given integral is improper.
Notice that the interval is fixed over real numbe