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求广义积分∫（1/2~3/2)(√（x-x²）的绝对值)dx
题目详情
求广义积分∫（1/2~3/2)(√（x-x²）的绝对值)dx
▼优质解答
答案和解析
∫[1/2,3/2]√|(x-x^2)|dx
=∫[1/2,3/2]√|(x-1/2)^2-1/4|dx
=∫[1/2,1] √[1/4-(x-1/2)^2]dx +∫[1,3/2]√[(x-1/2)^2-1/4]dx
=(1/2)x√(x-x^2)-(1/2)√(x-x^2)+(1/4)arcsin(2x-1) |[1/2,1]
+(1/2)x√(x^2-x)-(1/2)√(x^2-x)-(1/4)ln|√(x^2-x)+(x-1/2)| |[1,3/2]
∫√(x-x^2)dx
=x√(x-x^2)+∫x(x-1/2)dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx+∫(x/2)dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx+(-1/2)∫d(x-x^2)/√(x-x^2)+(1/2)∫dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx-√(x-x^2)+(1/2)∫d(2x-1)/√(1-(2x-1)^2)
=x√(x-x^2)-√(x-x^2)+(1/2)arcsin(2x-1)-∫√(x-x^2)dx
∫√(x-x^2)dx=(1/2)x√(x-x^2)-(1/2)√(x-x^2)+(1/4)arcsin(2x-1)
∫√(x^2-x)dx
=x√(x^2-x)-∫√(x^2-x)dx-(1/2)∫xdx/√(x^2-x)
=x√(x^2-x)-∫√(x^2-x)dx-√(x^2-x)-(1/2)∫dx/√(x^2-x)
=x√(x^2-x)-∫√(x^2-x)dx-√(x^2-x)-(1/2)∫d(2x-1)/√[(2x-1)^2-1]
2x-1=secu ∫d(2x-1)/√[(2x-1)^2-1]=∫secudu=ln|secu+tanu|=ln|√(x^2-x)+(x-1/2)|
∫√(x^2-x)dx=(1/2)x√(x^2-x)-(1/2)√(x^2-x)-(1/4)ln|√(x^2-x)+(x-1/2)|
=∫[1/2,3/2]√|(x-1/2)^2-1/4|dx
=∫[1/2,1] √[1/4-(x-1/2)^2]dx +∫[1,3/2]√[(x-1/2)^2-1/4]dx
=(1/2)x√(x-x^2)-(1/2)√(x-x^2)+(1/4)arcsin(2x-1) |[1/2,1]
+(1/2)x√(x^2-x)-(1/2)√(x^2-x)-(1/4)ln|√(x^2-x)+(x-1/2)| |[1,3/2]
∫√(x-x^2)dx
=x√(x-x^2)+∫x(x-1/2)dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx+∫(x/2)dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx+(-1/2)∫d(x-x^2)/√(x-x^2)+(1/2)∫dx/√(x-x^2)
=x√(x-x^2)-∫√(x-x^2)dx-√(x-x^2)+(1/2)∫d(2x-1)/√(1-(2x-1)^2)
=x√(x-x^2)-√(x-x^2)+(1/2)arcsin(2x-1)-∫√(x-x^2)dx
∫√(x-x^2)dx=(1/2)x√(x-x^2)-(1/2)√(x-x^2)+(1/4)arcsin(2x-1)
∫√(x^2-x)dx
=x√(x^2-x)-∫√(x^2-x)dx-(1/2)∫xdx/√(x^2-x)
=x√(x^2-x)-∫√(x^2-x)dx-√(x^2-x)-(1/2)∫dx/√(x^2-x)
=x√(x^2-x)-∫√(x^2-x)dx-√(x^2-x)-(1/2)∫d(2x-1)/√[(2x-1)^2-1]
2x-1=secu ∫d(2x-1)/√[(2x-1)^2-1]=∫secudu=ln|secu+tanu|=ln|√(x^2-x)+(x-1/2)|
∫√(x^2-x)dx=(1/2)x√(x^2-x)-(1/2)√(x^2-x)-(1/4)ln|√(x^2-x)+(x-1/2)|
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