University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises: 29

Answer

$\frac{1}{2}$

Work Step by Step

Because $1\lt2$, by Equation (2): $\int_{1}^{\sqrt 2} x dx=\frac{(\sqrt 2)^2}{2}-\frac{1^2}{2}=\frac{2}{2}-\frac{1}{2}=\frac{1}{2}$
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