Thomas' Calculus 13th Edition

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

Chapter 15: Multiple Integrals - Practice Exercises - Page 933: 9


$$\sin (4)$$

Work Step by Step

We calculate the integral as follows: $$I=\int_{0}^{1} \int_{2y}^{2} 4 \cos x^2 \ dx \ dy \\=\int_{0}^{2} \int_{0}^{x/2} 4 \times \cos x^2 \ dy \ dx \\=\int_{0}^{2} 2x \times \cos x^2 \ dx \\=[\sin x^2]_0^2 \\=\sin (2)^2-\sin 0 \\=\sin (4)$$
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