Chapter 5: Q. 77 (page 479)

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

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 $p \geq 1,$ then $\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges, let aand bbe any real numbers and f(x) be a function, and if fcontinuous on role="math" localid="1649068637899" $(a, b],$ but not x = a, then $\int_{a}^{b} f (x) d x = \lim_{A \to a^{+}} \int_{A}^{b} f (x) d x .$

So,

$\int_{0}^{1} \frac{1}{x^{p}} d x = \lim_{A \to 0^{+}} \int_{A}^{1} x^{- p} d x = \lim_{A \to 0^{+}} {[\frac{x^{- p + 1}}{- p + 1}]}_{A}^{1} = \lim_{A \to 0^{+}} [\frac{1}{1 - p} - \frac{1}{1 - p} A^{1 - p}] = \frac{1}{1 - p} + \frac{1}{(1 - p) 0^{+}} = \infty (p \geq 1)$

Thus, $p \geq 1$ and integral diverges on [0, 1].

If $p \geq 1,$ then $\frac{1}{x^{p}} \geq \frac{1}{x} for x \in [0, 1] and \int_{0}^{1} \frac{1}{x^{p}} d x diverges .$

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 $p \geq 1,$ then $\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges.

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.Ifp≥1,then∫011xpdxdiverges.

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

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