Answer
$2$
Work Step by Step
Given: $f(x)$$=$$\sin x$ on $[0,\pi]$
$\int f(x)dx=$$\int_{0}^{\pi}\sin xdx$
$=>[-\cos x]_{0}^{\pi}$
Now;we apply the limits:
$=>-\cos \pi-(-\cos 0)$
$=>-\cos\pi+\cos 0$
$=>1+1=2$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 278: 95
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