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If this is given: $$\int_0^1f(x)dx=1$$ Then is there anything else we can say about: $$\int_0^1xf(x)dx \;\;\;\;\; \text{ or } \;\;\;\;\; \int_0^1x^2f(x)dx$$

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Since $\{1, x, x^2, \cdots \}$ forms a basis of $L^1[0, 1]$, there is nothing that we can say about those integrals as we may assign arbitrary values to them and then find a suitable integrable function $f$ yielding those values. – Sangchul Lee Nov 15 '12 at 0:13
up vote 4 down vote accepted

Take $f(x)=ax^2+bx+c$. Then $$ \int_0^1f(x)dx=\frac{a}3+\frac{b}2+c, $$ $$ \int_0^1xf(x)dx=\frac{a}4+\frac{b}3+\frac{c}2, $$ $$ \int_0^1x^2f(x)dx=\frac{a}5+\frac{b}4+\frac{c}3. $$ You get a determined system of three equations on three unknowns. That means that for any values that you choose, you will be able to find coefficients $a,b,c$ such that the three integrals will take exactly those values.

In all, there is absolutely no general relation between the values of those integrals.

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If in addition $f(x) \ge 0$ for all $x \in [0,1]$, $f(x)$ is the probability density function for a random variable $X$ with values in $[0,1]$, while $E[X] = \int_0^1 x f(x)\ dx$ and $E[X^2] = \int_0^1 x^2 f(x)\ dx$. In that case $0 \le E[X] \le \sqrt{E[X^2]} \le 1$.

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