Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Section 6.1 - Areas Between Curves - 6.1 Exercises - Page 443: 12

Answer

$A = \frac{7}{4}$

Work Step by Step

$$\eqalign{ & {\text{Let the functions}} \cr & y = 1 + {x^3},{\text{ }}y = 2 - x,{\text{ }}x = - 1,{\text{ }}x = 0 \cr & {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr & \cr & {\text{We can find the area integrating with respect to }}x \cr & {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr & {\text{Let }} \cr & f\left( x \right) = 2 - x{\text{, }}g\left( x \right) = 1 + {x^3} \cr & x = a = - 1{\text{ and }}x = b = 0 \cr & {\text{Therefore}}{\text{,}} \cr & \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr & A = \int_{ - 1}^0 {\left[ {\left( {2 - x} \right) - \left( {1 + {x^3}} \right)} \right]} dx \cr & A = \int_{ - 1}^0 {\left( {2 - x - 1 - {x^3}} \right)} dx \cr & A = \int_{ - 1}^0 {\left( {1 - x - {x^3}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {x - \frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right]_{ - 1}^0 \cr & A = \left[ 0 \right] - \left[ {\left( { - 1} \right) - \frac{1}{2}{{\left( { - 1} \right)}^2} - \frac{1}{4}{{\left( { - 1} \right)}^4}} \right] \cr & {\text{Simplifying}} \cr & A = - \left[ { - \frac{7}{4}} \right] \cr & A = \frac{7}{4} \cr} $$
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