2个回答
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(x-1)的平方的原函数为F(x)=∫(x-1)²dx
令u=x-1,则du=dx,
∴∫(x-1)²dx=∫u²du=(1/3)u³ + c (c为任意常数)
将u=x-1代回,得:
F(x)=(1/3)(x-1)³ + c (c为任意常数)
也可令u=1-x,则du=-dx,
∴∫(x-1)²dx=∫(-u)²(-du) = -∫u²du = (-1/3)u³ + c (c为任意常数)
将u=1-x代回,得:
F(x)=(-1/3)(1-x)³ + c =(1/3)(x-1)³ + c(c为任意常数)
令u=x-1,则du=dx,
∴∫(x-1)²dx=∫u²du=(1/3)u³ + c (c为任意常数)
将u=x-1代回,得:
F(x)=(1/3)(x-1)³ + c (c为任意常数)
也可令u=1-x,则du=-dx,
∴∫(x-1)²dx=∫(-u)²(-du) = -∫u²du = (-1/3)u³ + c (c为任意常数)
将u=1-x代回,得:
F(x)=(-1/3)(1-x)³ + c =(1/3)(x-1)³ + c(c为任意常数)
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