Chapter 5: Q. 9 (page 477)

Why does it make sense that $\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges when $p \geq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.

Short Answer

Expert verified

For $p \geq 1$ , $\frac{1}{x^{p}}$ is greater than $\frac{1}{x}$ in the interval $[0, 1]$ , whose improper integral on $[0, 1]$ is known to diverge.

Step by step solution

Step 1. Given information.

We are given,

$\int_{0}^{1} \frac{1}{x^{p}} d x$

Step 2. Comparing the integral

When the integral $\int_{0}^{1} \frac{1}{x^{p}} d x$ is solved in the interval $p \geq 1$ , it gives an result of $\infty$ . Consider $\frac{1}{x}$ , since $p = 1$ .

Then,

\int_{0}^{1} \frac{1}{x} d x = [\ln x]_{0}^{1} = \ln 1 - \ln 0 = \infty

So, the integral is diverges.

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Why does it make sense that $\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges when $p \geq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Comparing the integral

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Most popular questions from this chapter

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Why does it make sense that $\int_{0}^{1} \frac{1}{x^{p}} d x$ diverges when $p \geq 1$ ? Consider how $\frac{1}{x^{p}}$ compares with $\frac{1}{x}$ in this case.