- #1
ozgunozgur
- 27
- 0
Can Improper Integrals Help Solve This Inequality?
- Context: MHB
- Thread starter ozgunozgur
- Start date
Click For Summary
SUMMARY
The discussion focuses on solving the inequality $\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$ using improper integrals. It is established that the integral $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ diverges, which is a critical point in analyzing the behavior of the inequality. Participants are encouraged to explore methods for continuing the solution process, particularly through the application of improper integrals in evaluating limits and convergence.
PREREQUISITES- Understanding of improper integrals and their convergence
- Familiarity with inequalities in calculus
- Knowledge of limits and asymptotic behavior
- Basic algebraic manipulation of rational expressions
- Study the properties of improper integrals and their applications in inequalities
- Learn techniques for evaluating limits involving rational functions
- Investigate the relationship between divergence of integrals and behavior of functions
- Explore advanced topics in calculus, such as L'Hôpital's Rule and series convergence
Mathematics students, educators, and anyone interested in advanced calculus techniques, particularly those focusing on inequalities and improper integrals.
- #2
skeeter
- 1,103
- 1
note ...
$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$
and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.
$\dfrac{x\sqrt{x}}{x^2-1} > \dfrac{1}{\sqrt{x}}$
and $\displaystyle \int_2^\infty \dfrac{dx}{\sqrt{x}}$ is divergent.
Similar threads
- · Replies 3 ·
- · Replies 2 ·
- · Replies 4 ·
- · Replies 10 ·
- · Replies 1 ·
- · Replies 5 ·
- · Replies 3 ·
- · Replies 3 ·
- · Replies 10 ·
- · Replies 2 ·