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判断反常积分的收敛性 ∫(0,1)x^-1/2 和 ∫(1,+∞)x^-1/2
人气:351 ℃ 时间:2020-08-30 10:14:21
解答
∫(0,1)x^-1/2 dx
=lim(a->0+)∫(a,1)x^-1/2 dx
=lim(a->0+)∫(a,1)dx^1/2
=lim(a->0+)x^1/2|(a,1)
=lim(a->0+)1^1/2-a^1/2
=1(反常积分收敛)
∫(1,+∞)x^-1/2
=lim(a->+∞)∫(1,a)x^-1/2 dx
=lim(a->+∞)∫(1,a)dx^-1/2
=lim(a->+∞)x^-1/2|(1,a)
=lim(a->+∞)a^-1/2-1^-1/2
=+∞(反常积分发散)
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