20/07/29 23:57:47.73 DoToA506.net
前>>796訂正。
>>528
∫[π/2→3π/2]π(3π/2-t)^2sintdt
=[π(3π/2-t)^2(-cost)](π/2→3π/2)-∫[π/2→3π/2]π(2t-3π)(-cost)dt
=0-π^3×0+∫[π/2→3π/2]π(2t-3π)(-cost)dt
=[π(2t-3π)(-sint)](π/2→3π/2)-∫[π/2→3π/2]2π(-sint)dt
=0-π(-2π)(-1)+2π∫[π/2→3π/2]sintdt
=-2π+2π[-cost](π/2→3π/2)
=-2π-2π×0
=-2π＜0
∴飲めない。