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: 66

Answer

$$ - \frac{1}{{{x^2} + 1}}$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\int_x^0 {\frac{{dp}}{{{p^2} + 1}}} \cr & {\text{Property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr & = - \frac{d}{{dx}}\int_0^x {\frac{{dp}}{{{p^2} + 1}}} \cr & {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( p \right)dp{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( p \right)dp} = f\left( x \right) \cr & {\text{In this exercise}} \cr & f\left( p \right) = \frac{1}{{{p^2} + 1}},{\text{ and }}a = 0 \cr & then \cr & - \frac{d}{{dx}}\int_0^x {\frac{{dp}}{{{p^2} + 1}}} = - \frac{1}{{{{\left( x \right)}^2} + 1}} \cr & = - \frac{1}{{{x^2} + 1}} \cr} $$
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