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 445: 64

Answer

$A = \frac{1}{{12}}$

Work Step by Step

$$\eqalign{ & {\text{Let the functions }}y = {x^2}{\text{, }}y = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the point }}\underbrace {\left( {1,1} \right)}_{\left( {{x_1},{y_1}} \right)} \cr & {\text{Differentiating }}y = {x^2} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2}} \right] \cr & \frac{{dy}}{{dx}} = 2x \cr & {\text{Find the slope of the tangent line at the point }}\left( {1,1} \right) \cr & m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,1} \right)}} = 2\left( 1 \right) \cr & m = 2 \cr & {\text{The equation of the tangent line to }}y = {x^2}{\text{ at }}\left( {1,1} \right){\text{ is}} \cr & y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 1 = 2\left( {x - 1} \right) \cr & y - 1 = 2x - 2 \cr & y = 2x - 1 \cr & \cr & {\text{The graph of the functions }}y = {x^2},{\text{ }}y = 0,{\text{ }}y = 2x - 1{\text{ is}} \cr & {\text{shown below}} \cr & {\text{The enclosed area is given by}} \cr & A = \int_0^{1/2} {\left( {{x^2} - 0} \right)} dx + \int_{1/2}^1 {\left[ {{x^2} - \left( {2x - 1} \right)} \right]} dx \cr & A = \int_0^{1/2} {{x^2}} dx + \int_{1/2}^1 {\left( {{x^2} - 2x + 1} \right)} dx \cr & A = \int_0^{1/2} {{x^2}} dx + \int_{1/2}^1 {{{\left( {x - 1} \right)}^2}} dx \cr & {\text{Integrating}} \cr & A = \left[ {\frac{1}{3}{x^3}} \right]_0^{1/2} + \left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_{1/2}^1 \cr & A = \frac{1}{3}{\left( {\frac{1}{2}} \right)^3} + \left[ {\frac{{{{\left( {1 - 1} \right)}^3}}}{3} - \frac{{{{\left( {1/2 - 1} \right)}^3}}}{3}} \right] \cr & A = \frac{1}{{24}} + \left[ {\frac{0}{3} - \frac{{{{\left( { - 1/2} \right)}^3}}}{3}} \right] \cr & A = \frac{1}{{24}} + \frac{1}{{24}} \cr & A = \frac{1}{{12}} \cr} $$
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