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: 62



Work Step by Step

Since, the function $\sin x$ is a periodical function having period $\dfrac{2\pi}{k}$. Thus, we have $A=\int_0^{\pi/k} \sin kx dx$ $\implies [(\dfrac{-1}{k}) \cos kx]_0^{\pi/k}=\dfrac{-1}{k}\cos k(\dfrac{1}{k})-(\dfrac{-1}{k})(\cos 0)$ $\implies A=\dfrac{2}{k}$
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.