Hot answers tagged

7

Consider the option that the "average" is not zero. Then I guess we can agree that it must be positive. However, for every positive number $\varepsilon>0$ we can choose a large enough $b$ such that the average on $[0,b]$ is less than $\varepsilon$. Therefore, the average has to be zero. The thing is, when using infinity one can often find some results ...


2

For a function $f(x)$, its average value on an interval $[a, b]$ is given as; $$f_{avg} =\frac{1}{b-a}\int_a^b f(x)dx$$ Suppose the given distance function respect to time ($t$) is given by $s(t)$. Then the velocity would be given as $v(t)=s'(t)$ Now if we were to calculate the average velocity, we would compute; $$v_{avg}=\frac{1}{b-a}\int_a^b v(t)dt$$ ...


1

The definition of average value as $\overline f =\dfrac{\int_a^b{f(x)dx}}{\int_a^b{dx}}=\frac{1}{b-a}\int_a^b{f(x)dx}$ makes a key assumption: Every value of $x$ in $[a,b]$ is equally likely. For finite intervals, this is a perfectly acceptable model. But it is not the only one possible, however, and for infinite intervals it is not an option at all ($x$ ...


1

I find it intuitive. We start with a value of $1=f(0)$ and decrease to $0$ as $n \to \infty$. If we find $f(n)$ for $n=0,1,2,3,...$ The more terms we choose, the more terms we have close to zero, and thus our average gets closer to $0$. Just think about it if you have $100$ numbers close to $0$ despite that $1$ in the begging isn't our average going to be ...



Only top voted, non community-wiki answers of a minimum length are eligible