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Chapter 7: Q. 2TF (page 617)

Improper Integrals: Determine whether the following improper integrals converge or diverge.
${鈭�}_{1}^{鈭�} \frac{1}{x^{2}} d x .$

Short Answer

Expert verified

The series is a convergent series.

Step by step solution

Step 1. Given Information.

$G i v e n a n i n t e g r a l : {鈭�}_{1}^{鈭�} \frac{1}{x^{2}} d x .$

Step 2. Solving the integral.

${鈭�}_{1}^{鈭�} \frac{1}{x^{2}} d x = {[\frac{- 1}{x}]}_{1}^{鈭�} = [\frac{- 1}{鈭�} - (\frac{- 1}{1})] = [0 + 1] = 1 . T h e r e f o r e, t h e s e r i e s i s a c o n v e r g e n t s e r i e s .$

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Improper Integrals: Determine whether the following improper integrals converge or diverge.鈭�1鈭�1x2dx.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Solving the integral.

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Improper Integrals: Determine whether the following improper integrals converge or diverge.
${鈭�}_{1}^{鈭�} \frac{1}{x^{2}} d x .$