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Find the average of f(x)=x^3-x over [0,2]

1 Answer

2 votes

9514 1404 393

Answer:

1

Explanation:

The average value of the function is the integral over the interval, divided by the width of the interval.


\displaystyle\overline{y}=(1)/(2-0)\int_0^2{f(x)}\,dx=\left.(1)/(2)\left((x^4)/(4)-(x^2)/(2)\right)\right|_0^2=(1)/(2)\left((2^4)/(4)-(2^2)/(2)\right)=(4-2)/(2)\\\\\boxed{\overline{y}=1}

Find the average of f(x)=x^3-x over [0,2]-example-1
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