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12 points) Note: The formulas for the Fourier transform on half intervals are often given in the form { ). f(x)cos L dx, with2 II + 1 1 2 be = -2sin((npix)/2) dx+ (5-x)sin((npix)/2) dx 0 1 u= 5-x du = sin((npix)/2)dx v = (-cos((npix)/2)/((npi)/2) duTherefore on (0,2] 8W f(x) = 3/2 + n=1 плх COS 2 плх f(x) = Σ sin () n=1 f(x) = 3/2 + n=1 2nax cos 2 + sin 2nax 2

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