Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 33

Answer

8

Work Step by Step

\[\begin{gathered} \int_1^9 {\frac{2}{{\sqrt x }}} dx \hfill \\ \hfill \\ {\text{Use }}\sqrt x = {x^{1/2}} \hfill \\ \hfill \\ \int_1^9 {\frac{2}{{{x^{1/2}}}}} dx \hfill \\ \hfill \\ {\text{Use }}\frac{1}{{{x^{1/2}}}} = {x^{ - 1/2}} \hfill \\ \hfill \\ \int_1^9 {\frac{2}{{{x^{1/2}}}}} dx = \int_1^9 {2{x^{ - 1/2}}} dx \hfill \\ \hfill \\ {\text{Use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }}} \hfill \\ \hfill \\ \int_1^9 {2{x^{ - 1/2}}} dx = 2\left[ {\frac{{{x^{ - 1/2 + 1}}}}{{ - 1/2 + 1}}} \right]_1^9 \hfill \\ \hfill \\ = 2\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)_1^9 = 4\left( {{x^{1/2}}} \right)_1^9 \hfill \\ \hfill \\ {\text{Fundamental Theorem of calculus}} \hfill \\ \hfill \\ 4\left( {{{\left( 9 \right)}^{1/2}} - {{\left( 1 \right)}^{1/2}}} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ 4\left( 2 \right) \hfill \\ \hfill \\ 8 \hfill \\ \end{gathered} \]
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