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Evenly Periodic (Posted on 2008-06-16) Difficulty: 1 of 5
The function H(s) is integrable with a period of 2. It is also known that H(s) is an even function.

The function M(y) is such that:

y
∫ H(s) ds = M(y), and:
0

M(1) = B, where B is a nonzero real constant.

Determine the value of M(7) – M(4) in terms of B.

See The Solution Submitted by K Sengupta    
Rating: 2.0000 (1 votes)

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Solution I can do this! (spoiler) | Comment 1 of 2
If H even, then H(x) = H(-x), so the integral from -1 to 0
    = the integral from 0 to 1 = M(1) = B.

And if H has period 2, the the integral from 1 to 2
  = the integral from -1 to 0 = B.

And the integral from 2 to 3
   = the integral from 0 to 1 = B

And so on, and so on.

Therefore,
M(7) = 7B
M(4) = 4B

M(7) - M(4) = 3B

  Posted by Steve Herman on 2008-06-16 17:18:57
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