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: 63

Answer

$$\frac{3}{{{x^4}}}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\int_2^{{x^3}} {\frac{{dp}}{{{p^2}}}} \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr & {\text{the upper limit is }}{x^3},{\text{ so is needed to apply the chain rule}} \cr & \frac{d}{{dx}}\int_2^{{x^3}} {\frac{{dp}}{{{p^2}}}} = \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}} \cr & {\text{with }}u = {x^3} \cr & = \left( {\frac{d}{{du}}\int_0^u {\frac{{dp}}{{{p^2}}}} } \right)\frac{d}{{dx}}\left( {{x^3}} \right) \cr & = \left( {\frac{d}{{du}}\int_0^u {\frac{{dp}}{{{p^2}}}} } \right)\left( {3{x^2}} \right) \cr & {\text{by the fundamental theorem }}\left( {{\text{part 1}}} \right) \cr & = \left( {\frac{1}{{{u^2}}}} \right)\left( {3{x^2}} \right) \cr & = \left( {\frac{1}{{{{\left( {{x^3}} \right)}^2}}}} \right)\left( {3{x^2}} \right) \cr & = \frac{3}{{{x^4}}} \cr} $$
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