Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises - Page 555: 44

Answer

$$A = \ln \left( {\frac{{2 + \sqrt 5 }}{{\sqrt 2 - 1}}} \right)$$

Work Step by Step

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{1}{{\sqrt {{x^2} - 2x + 2} }},\,\,\,{\text{on the interval }}\left[ {0,3} \right] \cr & {\text{The area under the curve is given by}} \cr & A = \int_0^3 {\frac{1}{{\sqrt {{x^2} - 2x + 2} }}} dx \cr & {\text{Completing the square}} \cr & A = \int_0^3 {\frac{1}{{\sqrt {{x^2} - 2x + 1 + 1} }}} dx \cr & A = \int_0^3 {\frac{1}{{\sqrt {{{\left( {x - 1} \right)}^2} + 1} }}} dx \cr & \cr & {\text{Let }}u = x - 1,{\text{ }}du = dx \cr & {\text{The new limits of integration are:}} \cr & x = 3 \to u = 2 \cr & x = 0 \to u = - 1 \cr & \cr & {\text{Substituting}} \cr & A = \int_0^3 {\frac{1}{{\sqrt {{{\left( {x - 1} \right)}^2} + 1} }}} dx = \int_{ - 1}^2 {\frac{1}{{\sqrt {{u^2} + 1} }}} du \cr & \cr & {\text{Integrate by tables using the formula 77}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr & A = \left[ {\ln \left( {u + \sqrt {{u^2} + 1} } \right)} \right]_{ - 1}^2 \cr & A = \left[ {\ln \left( {2 + \sqrt {{2^2} + 1} } \right) - \ln \left( { - 1 + \sqrt {{{\left( { - 1} \right)}^2} + 1} } \right)} \right] \cr & {\text{Simplify}} \cr & A = \left[ {\ln \left( {2 + \sqrt 5 } \right) - \ln \left( {\sqrt 2 - 1} \right)} \right] \cr & A = \ln \left( {\frac{{2 + \sqrt 5 }}{{\sqrt 2 - 1}}} \right) \approx 2.325 \cr} $$
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