Calculate the 5-unit moving average of the function. f(x)=x^3-x

Question

anonymous · Answer

Is the interval not specified? If it isn't the answer cannot be a value but instead must be a function of the start of the interval.

anonymous · Answer

yeah it has to be in the form of a function.

anonymous · Answer

And you know you must integrate over some interval (I assume of length 5) and the divide by the interval length to find the average?

anonymous · Answer

What i got so far is F(x)=[1/5(x^4/4-x^2/2)]-[1/5(x-5)^4/4-(x-5)^2/2】

Its not right though...

anonymous · Answer

$f(a)=\int_a^{a + 5}x^3 - x$

$f(a)=(\int_{0}^{a + 5}x^3 - x) - (\int_{0}^{a}x^3 - x)$

f(a)=$\left(1/4(a + 5)^2 (-2 + (a + 5)^2)\right) - \left(1/4a^2 (-2 + a^2)\right)$

Something like that I suppose

anonymous · Answer

I get $f(x)=5/4 (115 + 96 x + 30 x^2 + 4 x^3)$

anonymous · Answer

Sorry, you still need to divide that by 5.

anonymous · Answer

$\displaystyle f(x)=x^3+\frac{15 x^2}{2}+24 x+\frac{115}{4}$