dami said:Homework Statement
give a general rule for when Integral (1/x^n) from (x, 0, a), a>0 converges or diverges.
Homework Equations
The Attempt at a Solution
I have checked the textbooks for this answer but i can't seem to find it. The closest i got was Integral (1/x^n) from (x, 1, infinity), converges or diverges.
A convergent integral is one in which the area under the curve approaches a finite value as the limits of integration approach infinity. In contrast, a divergent integral is one in which the area under the curve approaches infinity as the limits of integration approach infinity.
There are several methods for determining the convergence or divergence of an integral. One approach is to use the comparison test, where the given integral is compared to a known convergent or divergent integral. Another method is to use the integral test, where the convergence or divergence of a given integral is determined by evaluating the corresponding series. Additionally, the ratio test and root test can also be used to determine convergence or divergence.
The exponent (n) in the integral (1/x^n) represents the power to which x is raised. This exponent is important because it affects the convergence or divergence of the integral. For example, if n > 1, the integral will converge, but if n <= 1, the integral will diverge.
No, an integral (1/x^n) cannot be both convergent and divergent. The convergence or divergence of an integral is determined by the value of n. If n > 1, the integral will converge, and if n <= 1, the integral will diverge.
Yes, there are some exceptions to the convergence/divergence rules for integral (1/x^n). For example, the integral (1/x) is an exception, as it is both divergent and convergent depending on the limits of integration. Additionally, some integrals may require more advanced techniques, such as the use of improper integrals, to determine convergence or divergence.