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: 35

Answer

\[ - \frac{{32}}{3}\]

Work Step by Step

\[\begin{gathered} \int_{ - 2}^2 {\left( {{x^2} - 4} \right)} dx \hfill \\ \hfill \\ {\text{Use }}\frac{1}{{{x^{1/2}}}} = {x^{ - 1/2}} \hfill \\ \hfill \\ \int_{ - 2}^2 {\left( {{x^2} - 4} \right)} dx \hfill \\ \hfill \\ {\text{Use }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }}} \hfill \\ \hfill \\ \left[ {\frac{{{x^3}}}{3} - 4x} \right]_{ - 2}^2 \hfill \\ \hfill \\ {\text{Fundamental Theorem of calculus}} \hfill \\ \hfill \\ \left[ {\frac{{{{\left( 2 \right)}^3}}}{3} - 4\left( 2 \right)} \right] - \left[ {\frac{{{{\left( { - 2} \right)}^3}}}{3} - 4\left( { - 2} \right)} \right] \hfill \\ \hfill \\ {\text{Simplify and subtract}} \hfill \\ \hfill \\ \left( {\frac{8}{3} - 8} \right) - \left( { - \frac{8}{3} + 8} \right) \hfill \\ \hfill \\ - \frac{{16}}{3} - \frac{{16}}{3} \hfill \\ \hfill \\ - \frac{{32}}{3} \hfill \\ \end{gathered} \]
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