# Integral inequality (Cauchy-Schwarz)

Let $u\in \mathcal{C}^1[a,b]$ be such that $u(a)=u(b)=0$. Show that

$$\int_a^b u^2(x)dx\leq (b-a)^2\int_a^b (u')^2(x)dx$$

using the Schwarz's inequality.

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This is a possible duplicate. But, I don't seem to find the same question which was asked earlier. –  user17762 Mar 18 '12 at 20:12
We have $|u(x)|\leq \int_a^x|u'(t)|dt\leq \sqrt{x-a}\sqrt{\int_a^b|u'(t)|^2}dt$ so $$u(x)^2\leq (x-a)\int_a^bu'(t)^2dt$$ and integrating $$\int_a^bu(x)^2dx\leq \frac{(b-a)^2}2\int_a^bu'(t)^2dt.$$