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 - Page 374: 58

Answer

$$A = \frac{{37}}{{12}}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = x\left( {x + 1} \right)\left( {x - 2} \right),{\text{ }}\left[ { - 1,2} \right] \cr & {\text{From the graph we can see that the area is given by}} \cr & A = \int_{ - 1}^0 {x\left( {x + 1} \right)\left( {x - 2} \right)} dx + \int_0^2 {\left[ {0 - x\left( {x + 1} \right)\left( {x - 2} \right)} \right]} dx \cr & A = \int_{ - 1}^0 {x\left( {x + 1} \right)\left( {x - 2} \right)} dx - \int_0^2 {x\left( {x + 1} \right)\left( {x - 2} \right)} dx \cr & {\text{Simplifying the integands}} \cr & A = \int_{ - 1}^0 {\left( {{x^3} - {x^2} - 2x} \right)} dx - \int_0^2 {\left( {{x^3} - {x^2} - 2x} \right)} dx \cr & {\text{Integrate and evaluate}} \cr & A = \left[ {\frac{1}{4}{x^4} - \frac{1}{3}{x^3} - {x^2}} \right]_{ - 1}^0 - \left[ {\frac{1}{4}{x^4} - \frac{1}{3}{x^2} - {x^2}} \right]_0^2 \cr & A = - \left[ {\frac{1}{4}{{\left( { - 1} \right)}^4} - \frac{1}{3}{{\left( { - 1} \right)}^2} - {{\left( { - 1} \right)}^2}} \right] - \left[ {\frac{1}{4}{{\left( 2 \right)}^4} - \frac{1}{3}{{\left( 2 \right)}^2} - {{\left( 2 \right)}^2}} \right] \cr & A = \frac{5}{{12}} + \frac{8}{3} \cr & A = \frac{{37}}{{12}} \cr} $$
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