Final answer:
The integral from 0 to 1 of the function f(x) = 1/x^p exists for values of p < 1, and its value is 1/(1-p) for those values of p.
Step-by-step explanation:
The student has asked when the integral from 0 to 1 of the function f(x) = 1/xp exists and its value. To determine this, we can set up the integral \( \int_{0}^{1} \frac{1}{x^{p}} dx \) and analyze it for convergence. For integration within the range of 0 to 1, we must consider the behavior of the function at the lower bound, where \( x = 0 \).
If \( p < 1 \), the function becomes less steep as it approaches 0 and the integral is likely to converge. Specifically, for \( p = 0 \), the integral is simply \( \int_{0}^{1} dx \), which equals 1. However, if \( p \geq 1 \), the function approaches infinity as \( x \) approaches 0, which would make the integral diverge. Therefore, the integral \( \int_{0}^{1} \frac{1}{x^{p}} dx \) exists for values of \( p < 1 \).
To evaluate the integral for the case where it converges (\( p < 1 \)), we employ the power rule for integration, resulting in:
\( \int_{0}^{1} \frac{1}{x^{p}} dx = \int_{0}^{1} x^{-p} dx = \left.\frac{x^{1-p}}{1-p}\right|_{0}^{1} = \frac{1^{1-p}}{1-p} - \frac{0^{1-p}}{1-p} = \frac{1}{1-p} \), provided that \( p \\eq 1 \).
In conclusion, the integral exists and its value is \( \frac{1}{1-p} \) for all real numbers \( p < 1 \), excluding \( p = 1 \).