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可以化作定积分做:
dx = [(2n+1) - (2n-1)]/(4n) = 1/(2n), n->oo
1/(4n) = 0, (2n-1)/(4n) = 1/2, as n->oo
原积分 = ∫[0,1/2] 2cosπx dx = (2/π)sinπx|[0,1/2] = 2/π
dx = [(2n+1) - (2n-1)]/(4n) = 1/(2n), n->oo
1/(4n) = 0, (2n-1)/(4n) = 1/2, as n->oo
原积分 = ∫[0,1/2] 2cosπx dx = (2/π)sinπx|[0,1/2] = 2/π
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