Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 15 - Review - Exercises - Page 1062: 6



Work Step by Step

Given: $\int_{0}^{1}\int_{x}^{e^{x}}3xy^{2}dydx$ Let us solve first $\int_{x}^{e^{x}}3xy^{2}dy=xe^{3x}-x^{4}$ Now, $\int_{0}^{1}xe^{3x}-x^{4}dx=(\frac{1}{9}(3x-1)e^{3x})|_{0}^{1}-\frac{x^{5}}{5}|_{0}^{1}$ $=\frac{2}{9}e^{3}+\frac{1}{9}-\frac{1}{5}$ $=\frac{2}{9}e^{3}-\frac{4}{45}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.