Let $f:(0,\infty) \rightarrow \mathbb{R}$ be continuous. I need to show that
$\left(\int_1^ef(x)dx \right)^2 \leq \int_1^e xf(x)^2dx$ I have been trying to use C-S to prove this but with no luck.
Let $f:(0,\infty) \rightarrow \mathbb{R}$ be continuous. I need to show that
$\left(\int_1^ef(x)dx \right)^2 \leq \int_1^e xf(x)^2dx$ I have been trying to use C-S to prove this but with no luck.
$\left(\int_1^ef(x)dx \right)^2 = \left(\int_1^e{1\over\sqrt{x}}\cdot \sqrt{x}f(x)dx \right)^2 \leq \int_1^e{1\over x}\,dx\cdot\int_1^e xf(x)^2dx=\int_1^e xf(x)^2dx$