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求定积分S 0到a x(a^2-x^2)^1/2dx (a>0) 要求是换元法和凑微分法
人气:256 ℃ 时间:2020-04-14 08:11:03
解答
令x=asint
则定积分等于∫(0,π/2)asint*acostdasint (换元法)
= a^3∫ (0,π/2)sint*cos²tdt
= - a^3∫ (0,π/2)cos²tdcost (凑微分法)
= - (1/3)a^3(cost)^3|∫(0,π/2)
= (1/3)a^3
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