Both integrands are undefined at x = 0, which is why the integrals are improper. Both integrands are also undefined at the other endpoint, x = 1.aaaa202 said:Homework Statement
Check if the following integrals diverge:
Int(1/(x^3-2x^2+x), from 0 to 1) and Int(1/(x^3+x^2-2x)
Homework Equations
Ratio-test(not sure if that's the name)
The Attempt at a Solution
I have solved the problem and found that both integrals diverge at x=0.
aaaa202 said:I just want to check if you guys agree with me.
An improper integral is an integral that does not have a finite value due to one or more of the following reasons: the interval of integration is infinite, the integrand is unbounded, or the integrand has a discontinuity within the interval of integration.
Divergence at x=0 means that the integral is not defined at x=0 and thus does not have a finite value. This can occur when the integrand has a singularity or discontinuity at x=0.
Yes, an improper integral can still have a finite value if it has divergence at x=0. This can happen if the integral is convergent for values of x close to 0, but diverges at x=0.
We can determine if an improper integral has divergence at x=0 by evaluating the integral using different approaches, such as using limits or breaking the integral into smaller pieces. If the integral does not converge to a finite value, it has divergence at x=0.
No, we cannot evaluate an improper integral with divergence at x=0 as it does not have a finite value. However, we can still study its behavior and properties using different approaches, such as determining its convergence or divergence and finding its limits.