![Evaluate the integral and interpret it as the area of a region. \int_0^\frac{\pi}{2} |2 \sin(x) - 2 \cos(2x)| dx | Homework.Study.com Evaluate the integral and interpret it as the area of a region. \int_0^\frac{\pi}{2} |2 \sin(x) - 2 \cos(2x)| dx | Homework.Study.com](https://homework.study.com/cimages/multimages/16/graph3238217903753176271.jpg)
Evaluate the integral and interpret it as the area of a region. \int_0^\frac{\pi}{2} |2 \sin(x) - 2 \cos(2x)| dx | Homework.Study.com
![Consider the function f(x)={{:(-2sinx,if,xle-(pi)/(2)),(Asinx+B,if,-(pi)/(2 )ltxlt(pi)/(2)),(cosx,if,xge(pi)/(2)):} Which is continuous everywhere. The value of B is Consider the function f(x)={{:(-2sinx,if,xle-(pi)/(2)),(Asinx+B,if,-(pi)/(2 )ltxlt(pi)/(2)),(cosx,if,xge(pi)/(2)):} Which is continuous everywhere. The value of B is](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/59994322_web.png)
Consider the function f(x)={{:(-2sinx,if,xle-(pi)/(2)),(Asinx+B,if,-(pi)/(2 )ltxlt(pi)/(2)),(cosx,if,xge(pi)/(2)):} Which is continuous everywhere. The value of B is
![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](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
![The value of $\\int\\limits_\\pi ^{2\\pi } {[2\\sin x]dx} $ is equal to (where[.] is the G.I.F.)A. $ - \\pi $B. $ - 2\\pi $C. $ - \\dfrac{{5\\pi }}{3}$D. $\\dfrac{{5\\pi }}{3}$ The value of $\\int\\limits_\\pi ^{2\\pi } {[2\\sin x]dx} $ is equal to (where[.] is the G.I.F.)A. $ - \\pi $B. $ - 2\\pi $C. $ - \\dfrac{{5\\pi }}{3}$D. $\\dfrac{{5\\pi }}{3}$](https://www.vedantu.com/question-sets/81a6a607-6f3b-42d1-bdc3-6feedb76da524897162654211039244.png)
The value of $\\int\\limits_\\pi ^{2\\pi } {[2\\sin x]dx} $ is equal to (where[.] is the G.I.F.)A. $ - \\pi $B. $ - 2\\pi $C. $ - \\dfrac{{5\\pi }}{3}$D. $\\dfrac{{5\\pi }}{3}$
Evaluate the following limit : lim(x→π/2) (√(2 - sin x) - 1)/(π/2 - x)^2 - Sarthaks eConnect | Largest Online Education Community
How to plot a graph for y=2sin (x+pi/3) +1, x E (0,4pi) show full working out for all intercepts - Quora
![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](https://homework.study.com/cimages/multimages/16/fig_52c2250900925405944793.png)