Prove integral inequality including n-order derivative Let $f$ be $n$ times continuously differentiable on $[0,1]$, with $f(\frac{1}{2})=0$ and $f^{(i)}(\frac{1}{2})=0$ when $i$ is even and less than $n$. Prove
$$
\left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( 2n+1 \right) 4^n\left( n! \right) ^2}\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}.
$$

$f\left( x \right) =f\left( \frac{1}{2} \right) +f'\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) +\frac{1}{2!}f''\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^2+\cdots +\frac{1}{n!}f^{\left( n-1 \right)}\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^{n-1}+\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n=\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n$，$\theta\in(0,1)$.


so $\int_0^1{f\left( x \right) \mathrm{d}x}=\frac{1}{n!}\int_0^1{f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n\mathrm{d}x}$

How to deal with $\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}$ to prove this inequality?

Your solution does not work because $\theta$ (from the mean-value form of the remainder) depends on $x$.
As @leoli1 already said, one has to use Taylor's theorem with the integral remainder, this gives
$$
\int_0^1f(x) \, dx=\int_0^1\int_{1/2}^x  \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \, .
$$
Now we split the integral in two parts so that we can interchange the order of integration. First,
$$
\int_{1/2}^1\int_{1/2}^x  \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \\
=
\int_{1/2}^1\int_t^1  \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t) \, dx\, dt
=  \frac{1}{n!} \int_{1/2}^1  (1-t)^n f^{(n)}(t)\, dt \, .
$$
Second,
$$
\int_0^{1/2}\int_{1/2}^x  \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)\, dt \, dx \\
=
-\int_0^{1/2}\int_0^t \frac{(x-t)^{n-1}}{(n-1)!} f^{(n)}(t)  \, dx\, dt
=  \frac{(-1)^n}{n!} \int_0^{1/2} t^nf^{(n)}(t)  \, dt \, .
$$
Combining these equations we get
$$
\int_0^1f(x) \, dx = \frac{1}{n!} \left( (-1)^n \int_0^{1/2} t^nf^{(n)}(t)  \, dt + \int_{1/2}^1  (1-t)^n f^{(n)}(t)\, dt\right) \\
\le \frac{1}{n!}  \int_0^1 \min(t, 1-t)^n |f^{(n)}(t)| \, dt \, .
$$
Now we can apply the Cauchy-Schwarz inequality:
$$
\left( \int_0^1f(x) \, dx \right) \le \frac{1}{n!^2} \int_0^1 \min(t, 1-t)^{2n} \, dt \int_0^1 |f^{(n)}(t)|^2 \, dt \, .
$$
This gives the desired estimate because
$$
\int_0^1 \min(t, 1-t)^{2n} \, dt = 2 \int_0^{1/2} t^{2n} \, dt
= \frac{1}{(2n+1)4^n} \, .
$$
