定积分上线3下线1 根号下x(x-2)的绝对值dx

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/04/28 03:46:02

∫[1-->3] √|x(x-2)|dx
=∫[1-->2] √(x(2-x))dx+∫[2-->3] √(x(x-2))dx
=∫[1-->2] √(2x-x^2)dx+∫[2-->3] √(x^2-2x)dx
=∫[1-->2] √(1-(x-1)^2)dx+∫[2-->3] √((x-1)^2-1)dx
前一个积分令x-1=sint,dx=costdt,√(1-(x-1)^2)=cost,t：0-->π/2
后一个积分令x-1=secu,dx=secutanudu,√((x-1)^2-1)=tanu,u：0-->π/3
=∫[0-->π/2] (cost)^2dt+∫[0-->π/3] (tanu)^2secudu
=1/2∫[0-->π/2] (1+cos2t)dt+∫[0-->π/3] (secu)^3du-∫[0-->π/3] secudu
=π/4+∫[0-->π/3] (secu)^3du-ln|secu+tanu| |[0-->π/3]
=π/4+∫[0-->π/3] (secu)^3du-ln|2+√3| （1）
下面计算
∫[0-->π/3] (secu)^3du
=∫[0-->π/3] secudtanu
=[0-->π/3] secutanu-∫[0-->π/3] (tanu)^2secudu
=2√3-∫[0-->π/3] (secu)^3du+∫[0-->π/3] secudu
=2√3-∫[0-->π/3] (secu)^3du+ln|2+√3|
将-∫[0-->π/3] (secu)^3du移到等式左边与左边合并后,除去系数得：
∫[0-->π/3] (secu)^3du=√3+1/2ln|2+√3|
将此结果代入（1）得：
原式=π/4+√3+1/2ln|2+√3|-ln|2+√3|
=π/4+√3-1/2ln|2+√3|
用数学软件验算,结果正确
>>int((abs(x*(2-x)))^(1/2),x=1..3);
3^(1/2)-1/2*ln(2+3^(1/2))+1/4*Pi

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