Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.4 - The Fundamental Theorem of Calculus - Exercises 5.4 - Page 287: 35

Answer

$\frac{1}{2}(e-1)$

Work Step by Step

$\int^{1}_{0}xe^{x^2}dx $ =$\frac{1}{2}e^{x^2}|^1_0 $ =$\frac{1}{2}e^1-\frac{1}{2}e^0$ =$\frac{1}{2}(e-1)$
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