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integrate {2x-[2x]} from 0-100[x] ,where [.] is greatest integer funtion

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anupama | eNotes Newbie

Posted October 25, 2013 at 2:53 PM via web

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integrate {2x-[2x]} from 0-100[x] ,where [.] is greatest integer funtion

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 25, 2013 at 3:47 PM (Answer #1)

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You need to use the following definitions of floor and ceiling functions, such that:

`2x - [2x] => {(2x - 0, x in [0,1]),(2x - 2, x in [1,2]),(2x - 2*2, x in [2,3]),...,(2x - 2*99, x in [99,100]):}`

`int_0^100 (2x - [2x]) dx = int_0^1 2x dx + int_1^2 (2x - 2) dx +.... + int_99^100 (2x - 2*99) dx`

`int_0^100 (2x - [2x]) dx = 2(x^2/2|_0^1 + x^2/2|_1^2 + ... + x^2/2|_98^99 + x^2/2|_99^100) - 2(x|_1^2 + 2x|_2^3 + 3x|_3^4 + ... + 99x|_99^100)`

Using the fundamental theorem of calculus yields:

`int_0^100 (2x - [2x]) dx = 2(1/2 + 4/2 - 1/2 + 9/2 - 4/2 + ... + 99^2/2 - 98^2/2 + 100^2/2 - 99^2/2) - 2(2 - 1 + 2(3 - 2) + 3(4 - 3) + ... + 99(100 - 99))`

`int_0^100 (2x - [2x]) dx = 100^2 - 2(1 + 2 + ..... + 99)`

`int_0^100 (2x - [2x]) dx = 100^2 - 2*(1 + 99)*99/2`

`int_0^100 (2x - [2x]) dx = 100^2 -100*99`

Factoring out 100 yields:

`int_0^100 (2x - [2x]) dx = 100(100 - 99)`

`int_0^100 (2x - [2x]) dx = 100`

Hence, evaluating the given definite integral, using the defintion of floor and ceiling functions, yields `int_0^100 (2x - [2x]) dx = 100.`

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anupama | eNotes Newbie

Posted October 27, 2013 at 2:02 PM (Reply #1)

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question says 0 to 100[x],where [.] is g.i.f

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