Answer
$\int_{0}^{\pi/3} sec^2~x~dx = \sqrt{3}$
Work Step by Step
We can find the area under the graph:
$\int_{0}^{\pi/3} sec^2~x~dx$
$= tan~x~\vert_{0}^{\pi/3}$
$=tan(\frac{\pi}{3})-tan(0)$
$=\sqrt{3}-0$
$= \sqrt{3}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises - Page 400: 52
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