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Integral Calculus

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22 Feb 2007 00:49:12 IST

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integral

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_[0]^[x] [cost]dt and _[0]^[2(n-1)pie] [sinx] where [.] denotest the greatest integer function.

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Bipin Dubey

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Joined: 23 Jan 2007

Posts: 7942

23 Feb 2007 01:10:43 IST

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[sinx] = 0 0<=x<=

= -1

<x<2

Also the period of this function is 2

.

₀

^2*pie [sinx].dx = ₀

^pie 0.dx + _pie

^2*pie (-1).dx = -

Hence I = _[0]

^[2(n-1)pie] [sinx].dx
= (n-1).₀

^2*pie [sinx].dx (since it repeats itself after intervals of 2

)
= (n-1).(-

)

=> I = (1-n)

are instructed to answer one query at a time. So please post your other query on a new page.Hope you understand.

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Amar Gupta

Forum Expert

Joined: 31 Jan 2007

Posts: 363

23 Feb 2007 18:26:07 IST

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Dear,

I agree with Mr.Bipin.procedure is right but i just want to make one correction though it does not make any change in the ans. as sinx=1 at x=pie/2

[sinx] = 0 0<=x<

= 1 x=

/2
= 0

/2<x<

= -1

<x<2

now if you plot the graph, you find that only region bounded between the function and x-axis is rectangle whose sides are 1 and ( 2

below x axis in the region of 0 to 2

so the integration is nothing but the area u can easily get -

now use the periodic properity and u can get the ans.

and similarly you can solve the other one two.

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