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Chapter 7: Problem 74

Consider the family of functions \(f(x)=1 / x^{p},\) where \(p\) is a real number. For what values of \(p\) does the integral \(\int_{0}^{1} f(x) d x\) exist? What is its value?

Short Answer

Expert verified
The integral exists and converges for values of p > 1. The value of the integral for these values of p is given by: 鈭�(1/x^p)dx from 0 to 1 = 1/(1-p).

Step by step solution

01

Find the antiderivative of the function

To evaluate the integral, we first need to find the antiderivative of the function \(f(x) = \frac{1}{x^p}\). Using the power rule for integration, we can rewrite the function as \(x^{-p}\) and find its antiderivative as follows: $$ F(x) = \int x^{-p} dx = \frac{x^{(-p + 1)}}{(-p + 1)} + C, $$ where C is the constant of integration.
02

Evaluate the antiderivative from 0 to 1

To find the value of the integral, we need to evaluate the antiderivative \(F(x)\) at the limits of integration 1 and 0: $$ \int_{0}^{1} \frac{1}{x^p} dx = F(1) - F(0) = \frac{1^{(-p + 1)}}{(-p + 1)} - \frac{0^{(-p + 1)}}{(-p + 1)}. $$ However, we need to be careful when evaluating the antiderivative at the lower limit 0 since the function is undefined when \(x=0\). The integral converges when \(p>1\), within the following limits: $$ \lim_{x \to 0^+} \frac{x^{1 - p}}{1 - p} = \lim_{x \to 0^+} \frac{1}{(1 - p)x^{p - 1}}. $$
03

Analyze the conditions for convergence

For the integral to converge, the expression above should have a finite value as \(x\) approaches 0. We can see that the limit is finite when \(p - 1 > 0\), or \(p > 1\). So, the integral exists for values of \(p > 1\). When the condition \(p > 1\) holds, we can evaluate the integral as follows: $$ \int_{0}^{1} \frac{1}{x^p} dx = F(1) - \lim_{x \to 0^+} F(x) = \frac{1}{1 - p} - 0 = \frac{1}{1 - p}. $$
04

State the result

The integral \(\int_{0}^{1} \frac{1}{x^p} dx\) exists and converges for values of \(p > 1\). Its value is given by: $$ \int_{0}^{1} \frac{1}{x^p} dx = \frac{1}{1 - p}. $$

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