Prove that $\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$ Let $f:[a,b]\to\mathbb{R}$ be continuous and increasing, show that
$$\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$$
I am thinking of using integration by parts. First let
$$F(x)=\int_a^xf(t)dt$$
Then
$$\int_a^bxf(x)dx=bF(b)-\int_a^bF(x)dx=b\int_a^bf(x)dx-\int_a^b\int_a^xf(t)dtdx$$
So far I only have these in mind. Any one have any hints?

Note that $\int_{a}^{b} \left(x - \frac{a + b}{2}\right)dx = 0$. Since $f$ is continuous on a compact set it is bounded and it achieves minimum and since it is increasing the minimum is achieved at the lowest end of an interval and maximum at the highest end of an interval.
Let $I_{1} = \left[a, \frac{a + b}{2} \right]$ and $I_{2} = \left[\frac{a + b}{2}, b\right]$, and $g(x) = \left(x - \frac{a + b}{2} \right)$.
We have  $g(x) \leq 0, f(x) \leq f(\frac{a + b}{2})$ on $I_{1}$. So that $g(x)f(x) \geq f(\frac{a + b}{2}) g(x)$.
Similarly, $g(x) \geq 0$,  $f(x) \geq f(\frac{a + b}{2})$ on $I_{2}$. So that $g(x) f(x) \geq f(\frac{a + b}{2})g(x)$ on $I_{2}$. 
Now we have:
$\int_{a}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$=  \int_{a}^{\frac{a + b}{2}}\left(x - \frac{a + b}{2} \right) f(x) dx  +  \int_{\frac{a + b}{2}}^{b}\left(x - \frac{a + b}{2} \right) f(x) dx$
$\geq f(\frac{a + b}{2})\left[\int_{I_{1}}g(x)dx + \int_{I_{2}} g(x) dx \right]$
$ = f(\frac{a + b}{2} )\int_{a}^{b} g(x)dx = 0$, which implies the desired inequality. 
