Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.1 Area Between Two Curves - Exercises Set 5.1 - Page 353: 25

Answer

$$A = \frac{{9152}}{{105}}$$

Work Step by Step

$$\eqalign{ & {\text{From the graph we can note that the intersection points between}} \cr & {\text{the curves are at }}x = - 3{\text{ and }}x = 1 \cr & {\text{The area is given by:}} \cr & A = \int_{ - 3}^1 {\left( {3 - 2x - {x^6} - 2{x^5} + 3{x^4} - {x^2}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {3x - {x^2} - \frac{{{x^7}}}{7} - \frac{{{x^6}}}{3} + \frac{{3{x^5}}}{5} - \frac{{{x^3}}}{3}} \right]_{ - 3}^1 \cr & {\text{Evaluate and simplify}} \cr & A = \left[ {3\left( 1 \right) - {{\left( 1 \right)}^2} - \frac{{{{\left( 1 \right)}^7}}}{7} - \frac{{{{\left( 1 \right)}^6}}}{3} + \frac{{3{{\left( 1 \right)}^5}}}{5} - \frac{{{{\left( 1 \right)}^3}}}{3}} \right] \cr & {\text{ }} - \left[ {3\left( { - 3} \right) - {{\left( { - 3} \right)}^2} - \frac{{{{\left( { - 3} \right)}^7}}}{7} - \frac{{{{\left( { - 3} \right)}^6}}}{3} + \frac{{3{{\left( { - 3} \right)}^5}}}{5} - \frac{{{{\left( { - 3} \right)}^3}}}{3}} \right] \cr & A = \frac{{188}}{{105}} + \frac{{2988}}{{35}} \cr & A = \frac{{9152}}{{105}} \cr} $$
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