Thomas' Calculus 13th Edition

Published by Pearson

Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 55

Answer

$$e - 1$$

Work Step by Step

\eqalign{ & \int_{ - 2}^{ - 1} {{e^{ - \left( {x + 1} \right)}}} dx \cr & {\text{use substitution}}{\text{: }} \cr & {\text{ }}u = - \left( {x + 1} \right),{\text{ so that }}\frac{{du}}{{dx}} = - 1,\,\,\,dx = - du \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = - 1,{\text{ }}u = - \left( { - 1 + 1} \right) = 0 \cr & \,\,\,\,\,\,{\text{If }}x = - 2,{\text{ }}u = - \left( { - 2 + 1} \right) = 1 \cr & {\text{Then}} \cr & \int_{ - 2}^{ - 1} {{e^{ - \left( {x + 1} \right)}}} dx = \int_1^0 {{e^u}} \left( { - du} \right) \cr & {\text{integrate}} \cr & = - \left( {{e^u}} \right)_1^0 \cr & {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = - \left( {{e^0} - {e^1}} \right) \cr & {\text{simplifying, we get:}} \cr & = - \left( {1 - e} \right) \cr & = e - 1 \cr}

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