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Integral Calculus

Cool goIITian

 Joined: 3 Jul 2011 Post: 38
20 Feb 2012 22:31:09 IST
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Area bounded by
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Integral Calculus

The area bounded by 2[y]=3[x] is .  -

[.] is greatest integer function.

then the value of  is?? It is an integer type question!! please explain the steps.

Cool goIITian

Joined: 3 Jul 2011
Posts: 38
20 Feb 2012 22:37:12 IST
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??

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
21 Feb 2012 11:52:22 IST
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[y]=3[x]/2

So [x] is even.

Take cases

For -2<=x<-1 [x]=-2 implies [y]=-3, -3<=y<-2

For 0<=x<1 [x]=0  implies [y]=0, 0<=y<1

For 2<=x<3 [x]=2  implies [y]=3, 3<=y<4

For 4<=x<5 [x]=4  implies [y]=6, 6<=y<7

Draw a graph and shade the area given by the above inequalities.

Therefore area will be 4 sq. units.

Forum Expert
Joined: 1 Dec 2006
Posts: 2108
25 Feb 2012 09:32:06 IST
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[y]=3[x]/2

for [y] is an integer=> [x] is even.

For 0<=x<1 [x]=0

=>  [y]=0, 0<=y<1
For 2<=x<3 [x]=2
=>  [y]=3, 3<=y<4
For 4<=x<5 [x]=4
=>  [y]=6, 6<=y<7
For 6<=x<7 => [x]=6
=>  [y] = 9 => 9<=y<10

Similarly for negative values . Thus the function is valid for all x such that 2n<=x<2n+1 , whr n is any integer
When the graph is plotted with these values, there will be infinite number of bounded regions. Hence the area bounded = infinity

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
25 Feb 2012 21:22:51 IST
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Arre Manasi ma'am, click on view post...the function is bounded. So area will be finite.

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