Prove that $\int_a^bxf(x)dx\geq\frac{b+a}{2}\int_a^bf(x)dx$

Solution 1:

You want to show that $$ I = \int_a^bxf(x)dx - \frac{b+a}{2}\int_a^bf(x)dx = \int_a^b \bigl(x - \frac{b+a}{2} \bigr) f(x) dx $$ is $\ge 0$. First split the integral in two parts: $$ I = \int_a^{(a+b)/2} \bigl(x - \frac{b+a}{2} \bigr) f(x) dx + \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2} \bigr) f(x) dx $$ Now substitute $x = a+b-y$ in the first integral: $$ I = \int_{(a+b)/2}^b \bigl(\frac{b+a}{2} - y\bigr) f(a+b-y) dy + \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2}\bigr) f(x) dx \\ = \int_{(a+b)/2}^b \bigl(x - \frac{b+a}{2}\bigr) \bigl( f(x) - f(a+b - x)\bigr) dx \ge 0 $$ because $x \ge a+b-x$ in $[(a+b)/2 , b]$ and $f$ is increasing.

Solution 2:

Using Chebyshev'inequality, we already have $f(x)$ and $g(x)=x$ are monotone, so $$\frac{1}{b-a}\int_a^b xf(x) \geq \left(\frac{1}{b-a}\int_a^b f(x)\right)\left(\frac{1}{b-a}\int_a^b x\right).$$

Then one has the conclusion.

Solution 3:

Since $f$ is increasing, we have $$ \frac1{b-t}\int_t^b\,f(x)\,\mathrm{d}x \ge\frac1{b-a}\int_a^b\,f(x)\,\mathrm{d}x\\ $$ for $a\le t\le b$. Therefore, $$ \begin{align} \int_a^b(x-a)\,f(x)\,\mathrm{d}x &=\int_a^b\int_a^x\,f(x)\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_a^b\int_t^b\,f(x)\,\mathrm{d}x\,\mathrm{d}t\\ &\ge\int_a^b\frac{b-t}{b-a}\int_a^b\,f(x)\,\mathrm{d}x\,\mathrm{d}t\\ &=\frac1{b-a}\int_a^b(b-t)\,\mathrm{d}t\int_a^b\,f(x)\,\mathrm{d}x\\ &=\frac{b-a}2\int_a^b\,f(x)\,\mathrm{d}x \end{align} $$ Adding $a\int_a^bf(x)\,\mathrm{d}x$ to both sides, we get $$ \int_a^bx\,f(x)\,\mathrm{d}x\ge\frac{b+a}2\int_a^b\,f(x)\,\mathrm{d}x $$

Solution 4:

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.