$Convergence integrals$

Question

$Convergence integrals$

$\int_{15}^{\infty} \frac{-6x^4-7x^2+1}{7x^4-x^3+x}dx $

Calculus Improper Integrals Convergence

Phia

19

Report

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

Note that
\[\lim_{x \rightarrow \infty} \frac{-6x^4-7x^2+1}{7x^4-x^3+x}=-\frac{6}{7}\]
\[=\lim_{x \rightarrow \infty} \frac{x^4(-6-\frac{7}{x^2}+\frac{1}{x^4})}{x^4(7-\frac{1}{x}+\frac{1}{x^3})}\]
\[\lim_{x \rightarrow \infty} \frac{-6-\frac{7}{x^2}+\frac{1}{x^4}}{7-\frac{1}{x}+\frac{1}{x^3}}=\frac{-6}{7}\neq0.\]
Hence the integral is divergent.

Erdos

4.7K

Erdos

0

Leave a comment if you need any clarifications.
Erdos

0

@Phia: What details do you need?
Erdos

0

I added some more detail. Let me know if you need any clarifications, but the solution is pretty much what I have written.
Phia

0

Can you explain why the integral converges if the lim is 0?
Erdos

0

I am not saying that he integral converges if the lim is 0. If the limit is not zero then the integral is divergent. This is a well known fact about improper integrals which can be easily found in any calculus textbook.

$Convergence integrals$

Answer

Related Questions

Search