## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 7 - Integration - 7.5 The Area Between Two Curves - 7.5 Exercises - Page 405: 19

#### Answer

$$A = \frac{1}{{20}}$$

#### Work Step by Step

\eqalign{ & y = {x^4} + \ln \left( {x + 10} \right),\,\,\,\,\,\,y = {x^3} + \ln \left( {x + 10} \right) \cr & {\text{Find the points of intersection of the graphs}} \cr & {x^4} + \ln \left( {x + 10} \right) = {x^3} + \ln \left( {x + 10} \right) \cr & {x^4} = {x^3} \cr & x = 0,\,\,x = 1 \cr & {\text{Then with the critical value we have the intervals }}\left( {0,1} \right) \cr & {\text{For the interval }}\left( {0,1} \right){\text{ we have that }}{x^3} + \ln \left( {x + 10} \right) > {x^4} + \ln \left( {x + 10} \right) \cr & \cr & {\text{The area between the curves is given by}}: \cr & A = \int_0^1 {\left( {{x^3} + \ln \left( {x + 10} \right) - {x^4} + \ln \left( {x + 10} \right)} \right)} dx \cr & A = \int_0^1 {\left( {{x^3} - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {\frac{1}{4}{x^4} - \frac{1}{5}{x^5}} \right)_0^1 \cr & A = \left( {\frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{5}{{\left( 1 \right)}^5}} \right) - \left( {\frac{1}{4}{{\left( 0 \right)}^4} - \frac{1}{5}{{\left( 0 \right)}^5}} \right) \cr & A = \frac{1}{4} - \frac{1}{5} - 0 \cr & A = \frac{1}{{20}} \cr}

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