Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises - Page 578: 10

Answer

$${\text{diverges}}$$

Work Step by Step

$$\eqalign{ & \int_{ - \infty }^0 {\frac{{dx}}{{\sqrt {2 - x} }}} \cr & {\text{definition of improper integral}} \cr & \int_{ - \infty }^0 {\frac{{dx}}{{\sqrt {2 - x} }}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{{dx}}{{\sqrt {2 - x} }}} \cr & = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {{{\left( {2 - x} \right)}^{ - 1/2}}} dx \cr & {\text{evaluate the integral}} \cr & = - \mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\frac{{{{\left( {2 - x} \right)}^{1/2}}}}{{1/2}}} \right)} \right|_a^0 \cr & {\text{simplify}} \cr & = - 2\mathop {\lim }\limits_{a \to - \infty } \left. {\left( {\sqrt {2 - x} } \right)} \right|_a^0 \cr & {\text{evaluate the limit}} \cr & = - 2\left( {\sqrt {2 - 0} } \right) + 2\left( {\sqrt {2 - \left( { - \infty } \right)} } \right) \cr & = - 2\sqrt 2 + \infty \cr & = \infty \cr & {\text{diverges}} \cr} $$
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