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Chapter 6: Problem 13

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges. $$ \int_{1}^{\infty} \frac{1}{x^{2}} d x $$

Short Answer

Expert verified
The improper integral converges and its value is 1.

Step by step solution

01

Turn the upper limit of integration from infinity to a variable

This is done by replacing the upper limit of integration, which is infinity, with a variable 'b'. This transforms the integral into \( \int_{1}^{b} \frac{1}{x^{2}} dx \)
02

Simplify the integral

The integral \( \int_{1}^{b} \frac{1}{x^{2}} dx \) simplifies to \( \int_{1}^{b} x^{-2} dx \) because \( \frac{1}{x^{2}} \) is equivalent to \( x^{-2} \)
03

Evaluate the integral using the power rule

A rule in calculus states that \( \int x^{n} dx = \frac{x^{n+1}}{n+1} + C \) provided \( n \neq -1 \). This rule can be applied here with \( n = -2 \) to get \( -x^{-1} \) from 1 to b or \( -(\frac{1}{b} - \frac{1}{1}) \)
04

Take the limit as b approaches infinity

The value of the integral as b approaches infinity can be computed by taking the limit as b approaches infinity of \( -(\frac{1}{b} - \frac{1}{1}) \) or \( 1 - \frac{1}{b} \). As b approaches infinity, \( \frac{1}{b} \) approaches 0, so the limit is 1.
05

Conclusion

The original integral converges, and its value is 1.

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Most popular questions from this chapter

Explain why \(\int_{-1}^{1} \frac{1}{x^{3}} d x \neq 0\)

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