Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 681: 36

Answer

$\frac{1}{3}$

Work Step by Step

We have $g(t)=\sin t\cos t$ and $f(t)=\sin t $ for $0\leq t\leq \pi/2$. For $t=0$, $x=\sin 0=0$ For $t=\pi/2$, $x=\sin\pi/2=1$ Then, area enclosed the curve and the x-axis is given by the integral: $A=\int_0^{\pi/2}g(t)f'(t)dt$ Evaluate $A$: $A=\int_0{\pi/2}\sin t\cos t\cdot \cos tdt$ $A=\int_0^{\pi/2}\sin t\cos^2 tdt$ $A=[-\frac{\cos^3t}{3}]_0^{\pi/2}$ $A=(-\frac{\cos^3(\pi/2)}{3})-(-\frac{\cos^30}{3})$ $A=(-0^3/3)-(-1^3/3)$ $A=0+\frac{1}{3}$ $A=\frac{1}{3}$ Thus, the area is $\frac{1}{3}$.
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