sinx= sin2 (a) Region #1: Bounded by y sin r and y sin 2.r on the interval [0. 7] CoSx =} sinx=2sinxcOsx 6-2sincosx-sina Per= 2 O=sintrzcosk-I de (sinx- sinz) dx y = sin x (sin2x-sinx)dx y = sin 21 (snzx-sin) dx + "(snx-sin2x)dx Area Integral: (set-up only)
sinx= sin2 (a) Region #1: Bounded by y sin r and y sin 2.r on the interval [0. 7] CoSx =} sinx=2sinxcOsx 6-2sincosx-sina Per= 2 O=sintrzcosk-I de (sinx- sinz) dx y = sin x (sin2x-sinx)dx y = sin 21 (snzx-sin) dx + "(snx-sin2x)dx Area Integral: (set-up only)
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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solve for x to show how you got pi/3
sint=sin2e
sinx=2sinxcosx
6=2sinčosx-siìna
O =sintrzcosk-I
(a) Region #1: Bounded by y = sin r and y = sin 2.r on thc interval [0. 7]
%3D
COSX =
per= 24
YA
is
dk
(sinx- sinzx)de
y = sin x
%3D
conzi-sing)de+Cs
= sin 2x
(sinZx-sinx) dx
+ (
(sinx-sin2x)dx
Area Integral:
(set-up only)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b2d8d84-896e-4f9e-963b-da7844b9b0f5%2F1b089b8b-987a-4e25-adf9-47ebfa7b7da0%2Fgwgoho_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2
solve for x to show how you got pi/3
sint=sin2e
sinx=2sinxcosx
6=2sinčosx-siìna
O =sintrzcosk-I
(a) Region #1: Bounded by y = sin r and y = sin 2.r on thc interval [0. 7]
%3D
COSX =
per= 24
YA
is
dk
(sinx- sinzx)de
y = sin x
%3D
conzi-sing)de+Cs
= sin 2x
(sinZx-sinx) dx
+ (
(sinx-sin2x)dx
Area Integral:
(set-up only)
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