yeahyeah<3 said:Now when I check the answer in my calc I get infinity. Thus my answer is probably wrong. Where did I go wrong?
Limit
t --> infinity integral from -1 to t of 3/x^2 dx + limit t --> infinity integral from t to 1 of 3/x^2 dx
An improper integral is an integral that does not have both limits of integration that are finite. This means that either the upper or lower limit of integration is infinite, or that the integrand is undefined at one or more points within the interval of integration.
Improper integrals are important because they allow us to extend the concept of integration to functions that would otherwise be impossible to integrate. They also have many important applications in physics, engineering, and other scientific fields.
To evaluate an improper integral, you must first determine whether it is convergent or divergent. If it is convergent, you can use various techniques such as substitution, integration by parts, or partial fractions to evaluate the integral. If it is divergent, the integral does not have a finite value.
Yes, improper integrals can have a finite value if they are convergent. This means that the integral can be evaluated and the result will be a real number. However, if the integral is divergent, it does not have a finite value.
Some common examples of improper integrals include integrals with infinite limits of integration, integrals with discontinuous integrands, and integrals with vertical asymptotes in the interval of integration. These types of integrals often arise in calculus problems involving infinite series, differential equations, and areas under curves.