已知f(x)的导函数是根号下(-X^2+2X)求原函数?
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∵f′(x)=√(-x^2+2x)=√[1-(x-1)^2],∴f(x)=∫√[1-(x-1)^2]dx.
令x-1=sinu,得:u=arcsin(x-1),dx=cosudu.
∴f(x)
=∫√[1-(sinu)^2]cosudu=∫(cosu)^2du=(1/2)∫(1+cos2u)du
=(1/2)∫du+(1/4)∫cos2ud(2u)=(1/2)u+(1/4)sin2u+C
=(1/2)arcsin(x-1)+(1/2)sinucosu+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√[1-(sinu)^2]+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√[1-(x-1)^2]+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√(2x-x^2)+C.,2,-x^3/3+x^2,2,f '(x) = 1/ [√x √(2-x)]
f(x) = ∫ 1/ [√x √(2-x)] dx
= 2 arcsin√(x/2) + C,1,
令x-1=sinu,得:u=arcsin(x-1),dx=cosudu.
∴f(x)
=∫√[1-(sinu)^2]cosudu=∫(cosu)^2du=(1/2)∫(1+cos2u)du
=(1/2)∫du+(1/4)∫cos2ud(2u)=(1/2)u+(1/4)sin2u+C
=(1/2)arcsin(x-1)+(1/2)sinucosu+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√[1-(sinu)^2]+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√[1-(x-1)^2]+C
=(1/2)arcsin(x-1)+(1/2)(x-1)√(2x-x^2)+C.,2,-x^3/3+x^2,2,f '(x) = 1/ [√x √(2-x)]
f(x) = ∫ 1/ [√x √(2-x)] dx
= 2 arcsin√(x/2) + C,1,
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