University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 35


$$\int^{3}_{0}(\sqrt 2+1)x^{\sqrt2}dx=3^{\sqrt2+1}$$

Work Step by Step

$$A=\int^{3}_{0}(\sqrt 2+1)x^{\sqrt2}dx$$ $$A=(\sqrt2+1)\frac{x^{\sqrt2+1}}{\sqrt2+1}\Big]^{3}_{0}$$ $$A=(x^{\sqrt2+1})\Big]^{3}_{0}$$ $$A=3^{\sqrt2+1}-0^{\sqrt2+1}$$ $$A=3^{\sqrt2+1}$$
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.