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: 56

Answer

$$A = \frac{3}{4}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = {x^3} - 1{\text{ on the interval }}\left[ { - 1,2} \right] \cr & {\text{so}}{\text{, the area is}} \cr & A = \int_{ - 1}^2 {\left( {{x^3} - 1} \right)} dx \cr & {\text{use the power rule for integration}} \cr & A = \left. {\left( {\frac{{{x^4}}}{4} - x} \right)} \right|_{ - 1}^2 \cr & {\text{using the fundamental theorem}} \cr & A = \left( {\frac{{{{\left( 2 \right)}^4}}}{4} - \left( 2 \right)} \right) - \left( {\frac{{{{\left( { - 1} \right)}^4}}}{4} - \left( { - 1} \right)} \right) \cr & A = \left( {4 - 2} \right) - \left( {\frac{5}{4}} \right) \cr & A = 2 - \frac{5}{4} \cr & A = \frac{3}{4} \cr} $$
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