Area-very tough!!!!!!

Find the area bounded by the curves x2+y2=4, y=x2+x+1,
y=[sin2(x/4)+cos(x/4)] and the x-axis,where [.] denotes greatest integer function!!

44 Answers

1
krish1092 ·

Sorry guys,
Instead of plotting y=1 graph,i plotted x=1 ,I'm sorry for the inconvenience!!!!!!! :( :( :(

1
Pavithra Ramamoorthy ·

area of d shaded portion sld be found rit????????????is dat rit???????

1
greatvishal swami ·

but ram the curve for 3rd dat is green one is wrong i guess
i tried to prove dat in my above posts

1
Pavithra Ramamoorthy ·

seems 2 b a strght line.............not a curve.........v r sittin wit d same prob fr a lon time......

1
greatvishal swami ·

which straight line !!!!!!!
which curve!!!!!!

1
Pavithra Ramamoorthy ·

leave it yaar........ i am blabberin smthn in sleep.....

62
Lokesh Verma ·

well ramkumar .. the green curve seems wrong to me :(

1
greatvishal swami ·

yup bro
& i am still getin it =1 btw x=-120 to +120

1
Pavithra Ramamoorthy ·

dis guy s stickin wit d same one.......

1
Pavithra Ramamoorthy ·

byah.... i need a 3d designer tool to design dis graph.....

1
greatvishal swami ·

:D

1
greatvishal swami ·

plzz ram giv some eg or proof too

62
Lokesh Verma ·

vishal... u need to realise that when we plot on the x and y axis.. we take length in radians .. not degrees :)

1
greatvishal swami ·

yup and 120 deg =2Î /3 eqiv to radius of given circle

1
krish1092 ·

This is the correct graph!!

1
greatvishal swami ·

ya true btw -2 to 2 acc to me cant say bout rest
will rechk tmorw

9
Celestine preetham ·

the q is simple if u conc on

y = [ sin2 X/4 + cosX/4 ] within -2,2

it is always greater than 1 as y = [ 1 - λ2 + λ] λ ≥λ2 as its cosθ

so dia reduces to parabola ,circle, y = 1 and 0

just find intersection and integrate areas

11
Anirudh Narayanan ·

Won't any GIF have a graph with horizontal stripes? ONly where the stripes come and at what intervals they come is decided by the nonsense inside. Anyway x can be in radians as well! You ppl took x as 120 deg. For all we know, it can be 2Ï€/3 !

1
Pavithra Ramamoorthy ·

i think d third curve must be lik dis...

341
Hari Shankar ·

The expression y = sin2(x/4)+ cos (x/4)

= 1-cos2(x/4) + cos (x/4)

= 1 + cos (x/4) [1-cos (x/4)]

Here we must remember that -1/2≤x/4≤1/2

So cos (x/4) will always be positive

So recalling that when 0≤t≤1, t(1-t) lies between 0 and 1/4,

the given expression lies between 1 and 5/4 (Of course t will be greater than 1/2, so 5/4 is not attained)

In any case [ 1-cos2(x/4) + cos (x/4)] = 1 for this range

So this function is just y = 1

edit: typo indicated by aragorn corrected

11
Anirudh Narayanan ·

Why -1/4≤x≤1/4 ?

1
greatvishal swami ·

thanks prophet i was shouting yesterday dat its =1
bt never thougt like this
thanks again:)

1
greatvishal swami ·

sry correctn for u bro
dat is fr x=720 & radious of cicle = 2 so no need to wry bout dat

1
skygirl ·

well thats a parabola dear.. :)
if u are joking then :P

1
Pavithra Ramamoorthy ·

BUT SKY... can u plz tell me where dat parabola cuts d axis?????

62
Lokesh Verma ·

y=x2+x+1,

This is (x+1/2)2+3/4

So this will be a parablola on x=-1/2 and 3/4 units above the x axis!! :)

y=[sin2(x/4)+cos(x/4)]

find when will sin2(x/4)+cos(x/4)=1 and sin2(x/4)+cos(x/4)=0

and also when it will be -ve!

This is the rough outline of plotting the graph

To be fair it is a tough question.

11
Anirudh Narayanan ·

Well, Nishant. that's the problem. the graph of y=[sin2(x/4)+cos(x/4)] is so tough to plot.

62
Lokesh Verma ·

yeah it is aragon

that is why i said that u need to find the points where the value of this expression is 0, 1, -1 etc

then in between we will have to unse greatest integer function :)

1
Parul Kohli ·

Really nice question
On first see i am also confused

1
greatvishal swami ·

is 3rd one a const func i am gettin it 1 for many Q
Q=angles

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