Q: Proving an integral inequality 
The function $f$ is continuous on $[0, 1]$, $\int_{0}^{1}xf(x)dx=0$ and $\max_{x\in[0,1]}|f(x)|=1$. Prove that $$\frac{-5}{4}< \int_{0}^{1}e^xf(x)dx< \frac{3}{2}.$$


A: Hint. Note that $$\int_{0}^{1}e^xf(x)dx=\int_{0}^{1}(e^x-x)f(x)dx.$$
Hence, since $e^x>x$ in $[0,1]$, it follows that
$$-\int_{0}^{1}(e^x-x)|f(x)|dx\leq \int_{0}^{1}e^xf(x)dx\leq\int_{0}^{1}(e^x-x)|f(x)|dx.$$
Since $\max_{x\in[0,1]}|f(x)|=1$, we have that
$$-M\leq \int_{0}^{1}e^xf(x)dx\leq M$$
where $M=\int_{0}^{1}(e^x-x)dx$.
