Prove $\int_{a}^{b}xf(x)dx=a\int_a^cf(x)dx + b\int_c^bf(x)dx$

Question

$f:[a,b]\to\mathbb{R}$ continous. Show that existe c in $[a,b]$ such that $$\int_{a}^{b}xf(x)dx=a\int_a^cf(x)dx + b\int_c^bf(x)dx.$$

I already tried integration by parts, and mean value theorem for integrals... I need a light

Answer 1

One can proceed as in Prove for continous function $f$, $\int_0^1 xf (x) dx = \int_c^1 f (x) dx $, where the case $[a,b]=[0, 1]$ is handled.

Define $F(x) = \int_a^x f(t) \, dt$ and integrate by parts: $$ \int_a^b xf(x) \,dx = xF(x) \bigr]_a^b - \int_a^b F(x) \, dx = bF(b) - \int_a^b F(x) \, dx\, . $$ From the mean value theorem for integrals we have $\int_a^b F(x) \, dx = (b-a)F(c)$ for some $c \in [a, b]$. It follows that $$ \int_a^b xf(x) \,dx = bF(b) - (b-a)F(c) = a F(c) + b(F(b)-F(c)) \\ = a \int_a^c f(x) \, dx + b\int_c^b f(x) \, dx \, . $$