SUMMARY
The integral \(\int_0^1 \frac{\sin(x)}{x} \, dx\) is convergent due to the removable discontinuity at the lower limit of zero. The limit \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) confirms that the function is bounded on the interval [0, 1]. Therefore, the integral is integrable over this range, establishing its convergence definitively.
PREREQUISITES
- Understanding of improper integrals
- Knowledge of limits and continuity
- Familiarity with the sine function and its properties
- Basic calculus concepts, particularly integration techniques
NEXT STEPS
- Study the concept of removable discontinuities in calculus
- Learn about convergence tests for improper integrals
- Explore the properties of the sine function and its limits
- Investigate other examples of improper integrals and their convergence
USEFUL FOR
Students studying calculus, particularly those focusing on improper integrals and convergence, as well as educators looking for examples to illustrate these concepts.