若f(x)=(x-1/2)^3,则f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=____
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/29 09:34:59
若f(x)=(x-1/2)^3,则f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=____
f(0.1)=(0.1-1/2)^3=-(0.9-1/2)^3=-f(0.9)
f(0.2)=(0.2-1/2)^3=-(0.8-1/2)^3=-f(0.8)
f(0.3)=(0.3-1/2)^3=-(0.7-1/2)^3=-f(0.7)
f(0.4)=(0.4-1/2)^3=-(0.6-1/2)^3=-f(0.6)
f(0.5)=(0.5-1/2)^3=0
所以,f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=0.
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f(0.2)=(0.2-1/2)^3=-(0.8-1/2)^3=-f(0.8)
f(0.3)=(0.3-1/2)^3=-(0.7-1/2)^3=-f(0.7)
f(0.4)=(0.4-1/2)^3=-(0.6-1/2)^3=-f(0.6)
f(0.5)=(0.5-1/2)^3=0
所以,f(0.1)+f(0.2)+f(0.3)+...+f(0.9)=0.
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