Question

Improper Integrals:

Evaluate the integral from 0 to 1 of x^-1 dx, or show that it is divergent?

Answer 1

\[=\int\limits_{0}^{1}dx/x=lnx\] | 0 to 1
ln(1) = 0, ln (0) -> -infinity
=> divergent

Answer 2

the integrand is raised to the negative one, in other words int from 0 to 1 of 1/ sqrt x dx

Answer 3

negative 1 is not not negative 1/2.... what is suppose to be...?
\[1/\sqrt{x}\] or 1/x ...?

Answer 4

Sorry for the confusion. The problem is int from 0 to 1 of dx/ sqrt x.

Answer 5

OK... then:
\[\int\limits_{0}^{1}dx/\sqrt{x}=2 * \sqrt{x}\] |0 to 1
=2
right?