微积分,高等数学 定积分计算 图中的划线地方是在怎么算出来的?
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(x^2/2)arcsinx|(0~1)
=(1^2/2)arcsin1-(0^2/2)arcsin0
=(1/2)(π/2)-0*0
=π/4
∫(0~1)[(x^2/2)(1/√(1-x^2))dx
=(1/2)∫(0~1)[x^2/√(1-x^2)]dx
=(1/2)∫(0~1)[1-x^2-1/√(1-x^2)]dx
=(1/2)∫(0~1)[√(1-x^2)]dx-(1/2)∫(0~1)[1/√(1-x^2)]dx
=(1/2)(∫(0~1)[√(1-x^2)]dx-∫(0~1)[1/√(1-x^2)]dx)
其中∫(0~1)[1/√(1-x^2)]dx=arcsinx|(0~1)=π/2-0=π/2
所以π/4+(1/2)(∫(0~1)[√(1-x^2)]dx-∫(0~1)[1/√(1-x^2)]dx)
=π/4+(1/2)∫(0~1)[√(1-x^2)]dx-π/4
=(1/2)∫(0~1)[√(1-x^2)]dx
满意请采纳,有问题的话我得今晚18点左右才得上线^ ^
=(1^2/2)arcsin1-(0^2/2)arcsin0
=(1/2)(π/2)-0*0
=π/4
∫(0~1)[(x^2/2)(1/√(1-x^2))dx
=(1/2)∫(0~1)[x^2/√(1-x^2)]dx
=(1/2)∫(0~1)[1-x^2-1/√(1-x^2)]dx
=(1/2)∫(0~1)[√(1-x^2)]dx-(1/2)∫(0~1)[1/√(1-x^2)]dx
=(1/2)(∫(0~1)[√(1-x^2)]dx-∫(0~1)[1/√(1-x^2)]dx)
其中∫(0~1)[1/√(1-x^2)]dx=arcsinx|(0~1)=π/2-0=π/2
所以π/4+(1/2)(∫(0~1)[√(1-x^2)]dx-∫(0~1)[1/√(1-x^2)]dx)
=π/4+(1/2)∫(0~1)[√(1-x^2)]dx-π/4
=(1/2)∫(0~1)[√(1-x^2)]dx
满意请采纳,有问题的话我得今晚18点左右才得上线^ ^
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