简单解定积分,求问这个怎么解出来的?
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根据a的大小分段去绝对值然后积分得到的
追问
求过程
追答
a的范围?
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case 1: 0<a<1
∫(0->1) x|x-a| dx
=-∫(0->a) x(x-a) dx +∫(a->1) x(x-a) dx
=- [ (1/3)x^3 -(1/2)ax^2]|(0->a) + [ (1/3)x^2 -(1/2)ax^2]|(a->1)
=-[ (1/3)a^3 -(1/2)a^3] + [ (1/3 -1/2) - ( (1/3)a^3 -(1/2)a^3 ) ]
=-1/6 + (1/3)a^3
case 2: a≤0
∫(0->1) x|x-a| dx
=-∫(0->1) x(x-a) dx
= - [ (1/3)x^3 -(1/2)ax^2] |(0->1)
=(1/6)a^3
case 3: a≥1
∫(0->1) x|x-a| dx
=∫(0->1) x(x-a) dx
= [ (1/3)x^3 -(1/2)ax^2] |(0->1)
=-(1/6)a^3
∫(0->1) x|x-a| dx
=-∫(0->a) x(x-a) dx +∫(a->1) x(x-a) dx
=- [ (1/3)x^3 -(1/2)ax^2]|(0->a) + [ (1/3)x^2 -(1/2)ax^2]|(a->1)
=-[ (1/3)a^3 -(1/2)a^3] + [ (1/3 -1/2) - ( (1/3)a^3 -(1/2)a^3 ) ]
=-1/6 + (1/3)a^3
case 2: a≤0
∫(0->1) x|x-a| dx
=-∫(0->1) x(x-a) dx
= - [ (1/3)x^3 -(1/2)ax^2] |(0->1)
=(1/6)a^3
case 3: a≥1
∫(0->1) x|x-a| dx
=∫(0->1) x(x-a) dx
= [ (1/3)x^3 -(1/2)ax^2] |(0->1)
=-(1/6)a^3
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