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Integral Calculus
Gourav's Avatar
Gourav
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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. 


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Gourav
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20 Feb 2012 22:37:12 IST
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??  
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Vishal U. Karve's Avatar

Vishal U. Karve
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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.
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Manasi's Avatar

Manasi
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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

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Vishal U. Karve
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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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