Calculus / Calculus: Early Transcendentals 9 / Chapter 7.8 / Problem 66

Calculus: Early Transcendentals | 9th Edition | ISBN: 9781337613927 | Authors: Daniel K. Clegg, Saleem Watson, James Stewart

?Improper Integrals that Are Both Type 1 and Type 2 The integral is improper because the

Chapter 7, Problem 66

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Improper Integrals that Are Both Type 1 and Type 2 The integral ${∫}_{a}^{\infty{}} f(x)dx$ is improper because the interval $[a,∞)$ is infinite. If $f$ has an infinite discontinuity at $a$ , then the integral is improper for a second reason. In this case we evaluate the integral by expressing it as a sum of improper integrals of Type 2 and Type 1 as follows:

$\int\limits_{a}^{\infty{}} f(x)dx=\int\limits_{a}^{c} f(x)dx+\int\limits_{c}^{\infty{}} f(x)dx\ c>a$

Evaluate the given integral if it is convergent.

${∫}_{0}^{\infty{}} \frac{1}{\sqrt{x}}dx$

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