Question

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

Answer 1

Answer:

Divergent.

Step-by-step explanation:

We have been given an integral $\int\limits^\infty _1 {\frac{1}{x^{0.999}} \, dx$. We are asked to determine whether the given integral diverges or converges.

$\int _1^{\infty }\:\:\:\frac{1}{x^{0.999}}\:\:\:dx$

$\int _1^{\infty }\:\:\:\frac{1}{x^{0.999}}\:\:\:dx=\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx$

$\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=\frac{x^{-0.999+1}}{-0.999+1}$

$\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=\frac{x^{0.001}}{0.001}$

$\int _1^{\infty }\:\:\:x^{-0.999}\:\:\:dx=1000x^{0.001}$

Let us compute the boundaries.

$1000(\infty)^{0.001}=\infty$

$1000(1)^{0.001}=1000$

Since $\infty-1000$ is not a finite number, therefore, the given integral diverges.