Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 68

Answer

$$2x{e^x}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\int_{{e^x}}^{{e^{2x}}} {\ln {t^2}dt} \cr & {\text{Property }}\int_a^b {f\left( x \right)} dx = \int_a^p {f\left( x \right)} dx + \int_p^b {f\left( x \right)} dx \cr & = \frac{d}{{dx}}\int_{{e^x}}^0 {\ln {t^2}dt} + \frac{d}{{dx}}\int_0^{{e^{2x}}} {\ln {t^2}dt} \cr & {\text{Property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr & = - \frac{d}{{dx}}\int_0^{{e^x}} {\ln {t^2}dt} + \frac{d}{{dx}}\int_0^{{e^{2x}}} {\ln {t^2}dt} \cr & {\text{by the chain rule}} \cr & = - \left( {\frac{d}{{du}}\int_0^{{u_1}} {\ln {t^2}dt} } \right)\frac{d}{{dx}}\left( {{e^x}} \right) + \left( {\frac{d}{{du}}\int_0^{{u_2}} {\ln {t^2}dt} } \right)\frac{d}{{dx}}\left( {{e^{2x}}} \right) \cr & with{\text{ }}{u_1} = {e^x}{\text{ and }}{u_2} = {e^{2x}}{\text{ respectively }} \cr & then \cr & = - \left( {\frac{d}{{du}}\int_0^{{u_1}} {\ln {t^2}dt} } \right)\left( {{e^x}} \right) + \left( {\frac{d}{{du}}\int_0^{{u_2}} {\ln {t^2}dt} } \right)\left( {2{e^{2x}}} \right) \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & = - \ln {\left( {{e^x}} \right)^2}\left( {{e^x}} \right) + \ln {\left( {{e^{2x}}} \right)^2}\left( {{e^x}} \right) \cr & {\text{simplify}} \cr & = - \ln \left( {{e^{2x}}} \right)\left( {{e^x}} \right) + \ln \left( {{e^{4x}}} \right)\left( {{e^x}} \right) \cr & = - 2x\left( {{e^x}} \right) + 4x\left( {{e^x}} \right) \cr & = 2x{e^x} \cr} $$
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