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: 61

Answer

$${x^2} + x + 1$$

Work Step by Step

$$\eqalign{ & \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} \cr & {\text{Using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr & A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr & {\text{In this exercise}} \cr & f\left( t \right) = {t^2} + t + 1,{\text{ and }}a = 3 \cr & then \cr & \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} = {\left( x \right)^2} + \left( x \right) + 1 \cr & \frac{d}{{dx}}\int_3^x {\left( {{t^2} + t + 1} \right)dt} = {x^2} + x + 1 \cr} $$
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