integral of modulous of cosx in interval (0,2pi)

2
by yvsravan

2014-04-29T10:46:24+05:30

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Since cos x must always be positive
it just area under cosine function
so integral

oh srrry its value is 4 only
it is ok
kkkk
look at grapg of |cos(x)|
mark as best
2014-04-29T10:46:49+05:30
The graph of y = |cos x| over the interval [0, 2pi] is exactly two "humps" of the cosine curve (see picture). So, we just need to integrate over one hump, then double the result:
INT |cos x| dx (from 0 to 2pi)
= 2 * INT cos x dx (from -pi/2 to pi/2, because this would be an interval for one hump)
= 2 * sin x, evaluated from -pi/2 to pi/2
= 2 * [sin(pi/2) - sin(-pi/2)]
= 2 * [1 - (-1)]
= 2 * 2
= 4
yes
its true only yaaar
ok cool u can mark anyone as best