Proving an integral inequality

Question

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 $$\frac{-5}{4}< \int_{0}^{1}e^xf(x)dx< \frac{3}{2}.$$

Answer 1

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 $M=\int_{0}^{1}(e^x-x)dx$.