SOLUTION: graph the function y=sin x on the interval -pi/2 <= x <= pi/ 2
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![SOLVED:The graph of y=sinx over [0,2 π] is the of y=sinx.](https://cdn.numerade.com/previews/148a0a7c-9c49-4b88-9d80-bf34c531003a.gif "SOLVED:The graph of y=sinx over [0,2 π] is the of y=sinx.")
SOLVED:The graph of y=sinx over [0,2 π] is the of y=sinx.
The area of the region bounded by the curve y = "sin" x between the ordinates x=0, x=pi/2 and the X-"axis" is
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![Draw the graph of [y] = sin x, x in [0,2pi] where [*] denotes the grea](https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_GRA_C03_S01_031_S01.png "Draw the graph of [y] = sin x, x in [0,2pi] where [*] denotes the grea")
Draw the graph of [y] = sin x, x in [0,2pi] where [*] denotes the grea
Find the area of the region bounded by the curve y = sinx , the lines x = - ( pi ) /(2), x = ( pi ) /(2) and X- axis.
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![Sketch the graph of y = - 2 sin (x - pi / 2) + 3 over the interval [0, 2 pi] by comparing it to the graph of y = sin x. | Homework.Study.com](https://homework.study.com/cimages/multimages/16/fig_52a4944619864405640501.png "Sketch the graph of y = - 2 sin (x - pi / 2) + 3 over the interval [0, 2 pi] by comparing it to the graph of y = sin x. | Homework.Study.com")
Sketch the graph of y = - 2 sin (x - pi / 2) + 3 over the interval [0, 2 pi] by comparing it to the graph of y = sin x. | Homework.Study.com
