what is the average value of f(x)=(x-3)^2 on the intergral [2,5]

f(x)=(x-3)^2, [2,5]
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1 Answer

 

Average value = (1/(5-2))∫[2,5](x²-6x+9)dx=

⅓[x³/3-3x²+9x]⁵₂=⅓(35/3-26/3)=1.

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