Calculate $\int\limits_{0}^{1}{f\left( x \right)dx}$ when $f$ satisfies three given conditions. 
Let $f$ be a function with $f'$ is continuous on $[0,1]$ such that $$f\left( 1 \right)=0,\quad\int\limits_{0}^{1}{{{\left[ f'(x) \right]}^{2}}dx=7},\quad\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx=\frac{1}{3}.}$$
  Calculate $\displaystyle\int\limits_{0}^{1}{f\left( x \right)dx}$.


Hint. Note that by Cauchy-Schwarz inequality
$$
\begin{align*}\frac{1}{3}=\int_0^1 x^2 f(x)dx&=\left[\frac{x^3}{3} f(x)\right]_0^1-\int_0^1\frac{x^3}{3}\, f'(x)dx=\frac{1}{3}\int_0^1 (-x^3) f'(x)dx\\
&\leq  \frac{1}{3}\left(\int_{0}^{1}x^6 dx\right)^{1/2} \left(\int_{0}^{1} (f'(x))^2dx\right)^{1/2}=\frac{1}{3}\cdot\frac{1}{\sqrt{7}}\cdot \sqrt{7}= \frac{1}{3}.\end{align*}$$
Now recall that in the Cauchy-Schwarz inequality, the equality holds if and only if one function is a scalar multiple of the other.
P.S. "Without" the Cauchy-Schwarz inequality: from the first line $\int_0^1 x^3 f'(x)dx=-1$ and therefore
$$\int_0^1(7x^3+ f'(x))^2 dx=7^2\int_0^1 x^6 dx+2\cdot 7\int_0^1 x^3 f'(x) dx +\int_0^1(f'(x))^2 dx=0$$
which implies that the non-negative continuous function $(7x^3+ f'(x))^2$ is zero, that is $f'(x)=-7x^3$. Hence
$$f(x)=f(1)+\int_1^x f'(t)dt=0-\frac{7}{4}[t^4]_1^x=\frac{7(1-x^4)}{4}.$$
