Inequality releating squared absolute value of an integral to the integral of the squared absolute values of the integrand

Question

Is this inequality $\left| \int_{0}^{x} f(t) \ dt \right|^2 \leq \int_{0}^{x} |f(t)|^2 \ dx$ true for $x\in [0,1]$. In case it is how to prove it?

If there is no square in both sides it is easy since it will be the triangle inequality $\left| \int_{0}^{x} f(t) \ dt \right| \leq \int_{0}^{x} |f(t)| \ dx$. But if I try to rise the both sides of this inequality on a power of two I can't get the desired result.

Answer 1

Use Cauchy-Schwarz inequality, then we can get \begin{align} \Bigl|\int_{0}^{x} f(t)dt\Bigr|^{2}\leq \left(\int_{0}^{x} 1^{2}dx\right)\left(\int_{0}^{x}|f(t)|^{2}dt\right)=x\left(\int_{0}^{x}|f(t)|^{2}dt\right) \end{align} So for $x\in [0,1]$, your inequality is true.