Question

Determine whether the improper integral is convergent or divergent without solving the integral. Use an appropriate inequality to support your conclusion. (a) The integral is:

A) Convergent, as 1/x is less than or equal to 1 for x greater than or equal to 1
B) Divergent, as 1/x is greater than or equal to 1 for x greater than or equal to 1
C) Convergent, as 1/x is less than or equal to 1/x² for x greater than or equal to 1
D) Divergent, as 1/x is greater than or equal to 1/x² for x greater than or equal to 1

Answer 1

The improper integral is convergent.

Step-by-step explanation:

To determine whether the improper integral is convergent or divergent, we can use an appropriate inequality. In this case, we can compare the given function, 1/x, to another function. Let's consider option D: Convergent, as 1/x is greater than or equal to 1/x² for x greater than or equal to 1.

We can see that for x greater than or equal to 1, 1/x is always greater than 1/x². And if the integral of 1/x² is convergent, then the integral of 1/x must also be convergent. Therefore, the improper integral is convergent.