Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Answer 1

Answer:

$\int_{1}^{\infty} \frac{1}{x} dx = \infty -0$

And with that we see that our integral diverges.

Step-by-step explanation:

For this case we assume the following integral:

$\int_{1}^{\infty} \frac{1}{x} dx$

Since 1.000 = 1 are equivalent. We can solve the integral and since we know that $\int \frac{1}{x} dx = ln|x| +c$ we got this:

$\int_{1}^{\infty} \frac{1}{x} dx = ln |x| \Big|_1^{\infty}$

And when we evaluate the limits using the fundamental theorem of calculus we got:

$\int_{1}^{\infty} \frac{1}{x} dx = \lim_{x\to\infty} ln|x| - ln |1|$

By properties we know that $ln |1|=0$. But the $\lim_{x\to\infty} ln|x|$ is not defined since the natural log is a continuous increasing function so then:

$\lim_{x\to\infty} ln|x| = \infty$

And if we repplace this into our integral we got:

$\int_{1}^{\infty} \frac{1}{x} dx = \infty -0$

And with that we see that our integral diverges.