Basic question about Integration (Monotocity)

Question

Why is $$ \int_{a}^{\infty}xf(x)dx\geq \int_{a}^{\infty}af(x)dx $$ assuming $f(x)\geq 0$ and $a>0$. Is it because the integral is only defined for $x\in[a,\infty]$ and otherwise $0$ so that $x\geq a$?

Answer 1

Well, one thing you can think about is that $$\int_a^\infty xf(x) \ \text{d}x - \int_a^\infty af(x) \ \text{d}x = \int_a^\infty \left(x-a \right)f(x) \ \text{d}x $$ And over the whole region of integration $x \geq a$ so that $(x-a) \geq 0$ and we have also that $f(x) \geq 0$ so that $$ \left(x-a\right)f(x) \geq 0$$ which gives $$\int_a^\infty \left(x-a \right)f(x) \ \text{d}x \geq \int_a^\infty 0 \ \text{d}x = \lim_{b \to \infty}\left. \left[0 \right] \right|_a^b = \lim_{b \to \infty}0 = 0 $$ so that, altogether $$ \int_a^\infty xf(x) \ \text{d}x \geq \int_a^\infty af(x) \ \text{d}x$$

Answer 2

If $F(x)\leq G(x)$ on $[a,b]$, then $$ \int_a^b F(x) dx \leq \int_a^b G(x) dx; $$ this can be proven by looking at the Riemann sums. On $[a, \infty)$ we have $x\geq a$, so $xf(x)\geq af(x)$ for any $f(x)\geq 0$, and the required inequality follows.

Hope this helps!