Answer
$-2$
Work Step by Step
Our aim is to solve the integral $\int_{-\pi}^{0} \sin x \ dx$
Therefore,
$\int_{-\pi}^{0} \sin x \ dx=[-\cos x]_{-\pi}^{0}$
Now, $[-\cos x]_{-\pi}^{0}=-[\cos 0-\cos (-\pi)]$
or, $= -[1-(-1)]$
or, $=-2$
Finite Math and Applied Calculus (6th Edition)
by Waner, Stefan; Costenoble, Steven
Published by
Brooks Cole
ISBN 10:
1133607705
ISBN 13:
978-1-13360-770-0
Chapter 16 - Section 16.3 - Integrals of Trigonometric Functions and Applications - Exercises - Page 1178: 21
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
