Final answer:
The value of the integral from 1 to infinity of (1/(x^p))dx is 1/(p-1) if p > 1, corresponding to option B) 1/(p-1), and it is infinite if p ≤ 1, corresponding to option A) [infinity].
the value of the integral from 1 to infinity of (1/(x^p))dx is equal to option B) 1/(p-1) when p > 1, and it is infinite (option A) [infinity]) when p ≤ 1.
Step-by-step explanation:
The student is asking about the value of the improper integral of the function (1/(x^p)) from 1 to infinity. This is a problem that involves evaluating an integral to determine if it converges to a finite number or diverges to infinity. When p > 1, the integral converges, and when p ≤ 1, it diverges.
To find the value of the integral, we integrate the function from 1 to some arbitrary point L (which will later approach infinity). Thus, we have the integral ∫1^L (1/x^p)dx, which we can evaluate as follows:
If p ≠ 1, then the antiderivative of (1/x^p) is (1/(1-p))*x^(1-p).
Evaluating this from 1 to L gives (1/(1-p))*(L^(1-p) - 1).
Now we take the limit as L approaches infinity. If p > 1, the term L^(1-p) approaches 0, and thus the integral converges to 1/(p-1). However, if p ≤ 1, the term L^(1-p) does not approach 0 (for p=1, the function does not have an antiderivative in the form we used), and hence the integral diverges to infinity.
Therefore, the value of the integral from 1 to infinity of (1/(x^p))dx is equal to option B) 1/(p-1) when p > 1, and it is infinite (option A) [infinity]) when p ≤ 1.
Learn more about integral convergence here: