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: 103

Answer

$$\frac{9}{t}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} - \int_{{t^2}}^1 {\frac{3}{x}dx} } \right) \cr & {\text{use the property of the derivatives}} \cr & = \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} } \right) - \frac{d}{{dt}}\left( {\int_{{t^2}}^1 {\frac{3}{x}dx} } \right) \cr & {\text{property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr & = \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} } \right) + \frac{d}{{dt}}\left( {\int_1^{{t^2}} {\frac{3}{x}dx} } \right) \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^t {f\left( x \right)dx{\text{ }} \to {\text{ }}} A'\left( t \right) = \frac{d}{{dt}}\int_a^t {f\left( x \right)dx} = f\left( t \right) \cr & {\text{use the chain rule for the second derivative}} \cr & = \frac{3}{t} + \left( {\frac{3}{{{t^2}}}} \right)\frac{d}{{dt}}\left( {{t^2}} \right) \cr & = \frac{3}{t} + \left( {\frac{3}{{{t^2}}}} \right)\left( {2t} \right) \cr & {\text{simplify}} \cr & = \frac{3}{t} + \frac{{6t}}{{{t^2}}} \cr & = \frac{3}{t} + \frac{6}{t} \cr & = \frac{9}{t} \cr} $$
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