Chapter 5: Q. 76 (page 479)

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.
If 0 < p < 1, then $\int_{0}^{1} \frac{1}{x^{p}} d x$ converges to $\frac{1}{1 - p} .$

Short Answer

Expert verified

The given statement is proved.

Step by step solution

Step 1. Given Information.

The given integral is $\int_{0}^{1} \frac{1}{x^{p}} d x .$

Step 2. Prove.

To prove if 0 < p < 1, then $\int_{0}^{1} \frac{1}{x^{p}} d x$ converges to $\frac{1}{1 - p},$ let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649067893443" $(a, b]$ but not x = a, then role="math" localid="1649067215041" $\int_{a}^{b} f (x) d x = \lim_{A \to a^{+}} \int_{A}^{b} f (x) d x .$

So,

$\int_{0}^{1} \frac{1}{x^{p}} d x = \lim_{A \to 0^{+}} \int_{A}^{1} x^{- p} d x = \lim_{A \to 0^{+}} {[\frac{x^{- p + 1}}{- p + 1}]}_{A}^{1} = \lim_{A \to 0^{+}} [\frac{1}{1 - p} - \frac{1}{1 - p} A^{1 - p}] As 0 < p < 1, = \frac{1}{1 - p}$

Now, if 0 < p < 1, the pvalue is less than 1and the integral converges on [0, 1].

So, if 0 < p< 1, then $\frac{1}{x^{p}} < \frac{1}{x} for x \in [0, 1] and \int_{0}^{1} \frac{1}{x^{p}} d x converges to \frac{1}{1 - p} .$

Hence, the integral is proved.

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Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.
If 0 < p < 1, then $\int_{0}^{1} \frac{1}{x^{p}} d x$ converges to $\frac{1}{1 - p} .$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Prove.

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Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.If 0 < p < 1, then∫011xpdxconverges to11-p.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Prove.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

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Study anywhere. Anytime. Across all devices.

Prove each statement in Exercises 74–77, using limits of definite integrals for general values of p.
If 0 < p < 1, then $\int_{0}^{1} \frac{1}{x^{p}} d x$ converges to $\frac{1}{1 - p} .$