Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.5 The Area Between Two Curves - 7.5 Exercises - Page 405: 1

Answer

$$A = 21$$

Work Step by Step

$$\eqalign{ & x = - 2,\,\,\,\,\,x = 1,\,\,\,\,\,y = 2{x^2} + 5,\,\,\,\,\,\,y = 0 \cr & {\text{Find the area using the formula for the area between two curves }}\left( {{\text{page 399}}} \right) \cr & {\text{If }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right] \cr & A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr & {\text{then }} \cr & f\left( x \right) = 2{x^2} + 5,\,\,\,\,g\left( x \right) = 0,\,\,\,\,a = - 1,\,\,\,\,b = 1 \cr & A = \int_{ - 2}^1 {\left( {2{x^2} + 5 - 0} \right)} dx \cr & A = \int_{ - 2}^1 {\left( {2{x^2} + 5} \right)} dx \cr & {\text{integrating by the power rule}} \cr & A = \left[ {\frac{{2{x^3}}}{3} + 5x} \right]_{ - 2}^1 \cr & A = \left( {\frac{{2{{\left( 1 \right)}^3}}}{3} + 5\left( 1 \right)} \right) - \left( {\frac{{2{{\left( { - 2} \right)}^3}}}{3} + 5\left( { - 2} \right)} \right) \cr & A = \frac{2}{3} + 5 + \frac{{16}}{3} + 10 \cr & A = 21 \cr} $$
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