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: 30

Answer

\[\frac{{{\pi ^2}}}{2} - \pi {\tan ^{ - 1}}\,2\]

Work Step by Step

\[\begin{gathered} the\,\,area\,\,is\,\,bounded\,\,by\,\,\,{\left( {{x^2} + 1} \right)^{\frac{1}{2}}} \hfill \\ and\,\,y \geqslant 0\,\,and\,\,x \geqslant 2\,\,is\,\,given\,by. \hfill \\ \hfill \\ \int_2^\infty {\frac{1}{{\sqrt {{x^2} + 1} }}dx} \hfill \\ \hfill \\ which\,\,diverges\,\,to\,\,\,infinity.\, \hfill \\ But\,the\,\,volume\,\,is\,\,given\,\,by. \hfill \\ \hfill \\ \int_2^\infty {\pi \,\,\,{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}\,dx} \hfill \\ \hfill \\ \int_2^\infty {\pi \,\,\,{{\left( {\frac{1}{{\sqrt {{x^2} + 1} }}} \right)}^2}\,dx} = \pi \int_2^\infty {\frac{1}{{{x^2} + 1}}} \,dx \hfill \\ \hfill \\ {\text{integrate}}\,\, \hfill \\ = \pi \,\,\left[ {{{\tan }^{ - 1}}\,\left( x \right)} \right]_2^\infty \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \pi \,\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}\,2} \right) \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ \frac{{{\pi ^2}}}{2} - \pi {\tan ^{ - 1}}\,2 \hfill \\ \hfill \\ \end{gathered} \]
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