Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 6 - Applications of the Integral - 6.1 Area Between Two Curves - Exercises - Page 287: 44

Answer

$$\frac{1}{8}$$

Work Step by Step

Given $$y=\sin (2 x), \quad y=\sin (4 x), \quad x=0, \quad x=\frac{\pi}{6}$$ Since \begin{align*} \sin (4 x)&=\sin (2 x)\\ 2\sin ( 2x)\cos(2x)-\sin (2x)&=0\\ \sin(2x)(2\cos (2x)-1)&=0 \end{align*} then $x=0,\ \ x=\pi/6$ and $\sin4x\geq \sin2x $ for $ 0\leq x\leq \pi/6$ Hence, \begin{aligned} A &=\int_{0}^{\pi / 6}(\sin (4 x)-\sin (2 x)) d x \\ &=\left.\left(\frac{-\cos (4 x)}{4}+\frac{\cos (2 x)}{2}\right)\right|_{0} ^{\pi / 6} \\ &=\frac{1}{8} \end{aligned}
Small 1585484594
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.