Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.2 Regions Between Curves - 6.2 Exercises - Page 417: 26

Answer

$$A = \frac{{71}}{6}$$

Work Step by Step

$$\eqalign{ & {\text{The area of the region is given by}} \cr & A = \int_0^1 {\left[ {\left( {{y^3} - 4{y^2} + 3y} \right) - \left( {{y^2} - y} \right)} \right]d} y \cr & \,\,\,\,\, + \int_1^4 {\left[ {\left( {{y^2} - y} \right) - \left( {{y^3} - 4{y^2} + 3y} \right)} \right]} dy \cr & {\text{Simplifying the integrand}} \cr & A = \int_0^1 {\left( {{y^3} - 4{y^2} + 3y - {y^2} + y} \right)} dy + \int_1^4 {\left( {5{y^2} - y - {y^3} - 3y} \right)} dy \cr & A = \int_0^1 {\left( {{y^3} - 5{y^2} + 4y} \right)} dy + \int_1^4 {\left( {5{y^2} - {y^3} - 4y} \right)} dy \cr & {\text{Integrating}} \cr & A = \left[ {\frac{{{y^4}}}{4} - \frac{{5{y^3}}}{3} + 2{y^2}} \right]_0^1 + \left[ {\frac{{5{y^3}}}{3} - \frac{{{y^4}}}{4} - 2{y^2}} \right]_1^4 \cr & A = \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{5{{\left( 1 \right)}^3}}}{3} + 2{{\left( 1 \right)}^2}} \right] - 0 \cr & \,\,\,\,\,\, + \left[ {\frac{{5{{\left( 4 \right)}^3}}}{3} - \frac{{{{\left( 4 \right)}^4}}}{4} - 2{{\left( 4 \right)}^2}} \right] - \left[ {\frac{{5{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^4}}}{4} - 2{{\left( 1 \right)}^2}} \right] \cr & A = \frac{7}{{12}} - 0 + \frac{{32}}{3} + \frac{7}{{12}} \cr & A = \frac{{71}}{6} \cr} $$
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