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Mathematics
Lyla
24 May, 09:38
Integrate [0, 1/2] xcos (pi*x).
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Vanessa Rosario
24 May, 11:31
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To answer your question, this could be the possible answer and i hope you understand and interpret it correctly:
[Integrate [0, 1/2] xcos (pi*x
let u=x so that du=dx
and v=intgral cos (xpi) dx
v = (1/pi) sin (pi*x)
integration by parts
uv-itgral[0,1/2]vdu just plug ins
(1/pi) sinpi*x] - (1/pi) itgrlsin (pi*x) dx from 0 to 1/2
(1/pi) x sinpi*x - (1/pi) [ - (1/pi) cos pi*x] from 0 to 1/2
= (1/2pi) + (1/pi^2) [cos pi*x/2-cos 0]
=1/2pi - 1/2pi^2
= (pi-2) / 2pi^2 ans
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