Problem
Let $f(x)$ be continuous and increasing over $[a,b]$. Prove $$\displaystyle \int_a^b \left(\frac{b-x}{b-a}\right)^{2018}f(x){\rm d}x \leq \frac{1}{2019}\int_a^b f(x){\rm d}x.$$
Proof
By the second mean value theorem for integrals ，$\exists \xi \in [a,b]$ such that \begin{align*} &\int_a^b \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]f(x){\rm d}x\\ =&f(a)\int_a^\xi \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]{\rm d}x+f(b)\int_\xi^b \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]{\rm d}x\\ =&f(a)\int_a^b \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]{\rm d}x+[f(b)-f(a)]\int_\xi^b \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]{\rm d}x\\ =&f(a)\cdot 0+[f(b)-f(a)]\int_\xi^b \left[\left(\frac{b-x}{b-a}\right)^{2018}-\frac{1}{2019}\right]{\rm d}x\\ =&\frac{1}{2019}\cdot (b-\xi)\cdot [f(b)-f(a)]\cdot \left[\left(\frac{b-\xi}{{b-a}}\right)^{2018}-1\right]\\ \leq &0, \end{align*} which implies what we want.
HOPE TO SEE OTHER PROOFS. THANKS.