$f(x)$ is non-negative and that $ \int_{-a}^b xf(x) dx = 0 $. Show $ \int_{-a}^bx^2f(x)dx \leq ab\int_{-a}^b f(x)dx $

Question

$a, b > 0$. $f(x)$ is non-negative and integrable on $[-a, b]$ and that $ \int_{-a}^b xf(x) dx = 0 $. Prove that $$ \int_{-a}^bx^2f(x)dx \leq ab\int_{-a}^b f(x) dx $$ Thanks!

score 9 · Accepted Answer · answered Mar 15 '15 at 08:21

9

This follows from $x^2\leq (b-a)x+ ab$ on $[-a,b]$.

answered Mar 15 '15 at 08:21

WimC

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$f(x)$ is non-negative and that $ \int_{-a}^b xf(x) dx = 0 $. Show $ \int_{-a}^bx^2f(x)dx \leq ab\int_{-a}^b f(x)dx $

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