Justify the following inequality $ \left|\int^b_a u\right| \leq \sqrt{b-a}\sqrt{\int^b_a u^2}$

Question

Can someone please help me justify the inequality in the question?

Thanks.

Answer 1

Cauchy-Schwarz inequality asserts that if $u,v:[a,b]\to\mathbb{R}$ are piecewise continuous, then :

$$\left(\int_a^bf(t)g(t)\,dt\right)^2\le\left(\int_a^bf(t)^2\,dt\right)\left(\int_a^bg(t)^2\,dt\right)$$

If we choose $g:t\mapsto 1$, we get :

$$\left(\int_a^bf(t)\,dt\right)^2\le(b-a)\int_a^bf(t)^2\,dt$$

Answer 2

By applying the Cauchy-Schwarz inequality, $$ \left|\int_{[a,b]} f(u) g(u)\,du\right|^2 \leq \int_{[a,b]} |f(u)|^2\,du \cdot \int_{[a,b]} |g(u)|^2 \,du $$ to $f(u)=1$ and $g(u)=u$, one obtains the given result.