Prove that $\int _0^4\:f_n(x)\,dx < 4^{n+1}$ where $f_n(x)= \left(4x-x^2\right)^n$

Hint:

$4x-x^2$ is a quadratic polynomial , with a negative leading coefficient. Furthermore, it vanishes for $x=0$ and $x=4$. Can you find it (absolute) maximum?

Once you've found this maximum, you can use the mean value inequality for integrals.

You cannot just blindly apply the reverse power rule. Remember the chain rule? You have a function within another function.

Consider the case when n=2

$ \int_0^4 (4x + x^2)^2 dx = \int_0^4 16x^2 + 8x^3 + x^4 dx$

Below is a suggestion for a very tedious route, but I wanted to post it anyway. This is probably not how the question is meant to be done.

Observe how by the binomial theorem we have that: $$I=\int _{x=0}^4\:f_n(x)\,dx =\int _{x=0}^4\: (4x-x^2 )^n\,dx =\int _{x=0}^4\: \sum_{k=0}^n \binom{n}{k} (4x)^{k} \cdot (-x^2)^{n-k}\,dx$$ We make this look a bit prettier: $$I=\int _{x=0}^4\: \sum_{k=0}^n \binom{n}{k} 4^{k} \cdot (-1)^{n-k}x^{2n-k}\,dx$$

We may switch the order of integration and summation since this concerns polynomials (what I mean is that they are of finite order), we compute: $$I=\sum_{k=0}^n \binom{n}{k} \frac{1}{2n-k+1} 4^{k} \cdot (-1)^{n-k}4^{2n-k+1}-0$$ We collect some terms: $$I=\sum_{k=0}^n \binom{n}{k} \frac{1}{2n-k+1} 4^{2n+1} \cdot (-1)^{n-k}$$ Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be: $$ I =\frac{\sqrt {\pi} \cdot n! \cdot 2^{2n+1}}{(n+\frac{1}{2})!}=\frac{\sqrt \pi \cdot n!}{(n+\frac{1}{2})!} \cdot 4^{n + \frac{1}{2}}$$ We then need to prove by induction that $\frac{\sqrt \pi \cdot n!}{(n+\frac{1}{2})!} > \sqrt{4}$ for positive $n$.

Yes. This is a lot of hard work. You could also just take the approach that was posted by Bernard and observe that $4x-x^2$ has a global maximum on the interval $[0,4]$ (the begin and endpoint are zeros of this polynomial), this maximum by symmetry occurs at $x=2$. The function has maximum value $8-4=4$. Observe then by the estimation lemma that $M =\max_{x \in [0,4]}((4x-x^2)^n)=4^n$, and the length of our path is $4$. We get: $$ \int _{x=0}^4 f_n dx \leq M L = 4^n \cdot 4 = 4^{n+1} $$

Also see: ML-inequality for real integrals

Or alternatively, if you don't like applying this theorem, we see that the $f_n$ is continuous as it is a polynomial, it is bounded on $[0,4]$ , thus by the mean value theorem for integrals there must exist a $\xi \in [0,4]$ such that: $$ (4-0) \cdot f_n(\xi) =\int _{x=0}^4\:f_n(x)\,dx$$ $$ \int _{x=0}^4\:f_n(x)\,dx = 4 f_n(\xi) $$ However, observe that $f_n(\xi) \leq 4^n$ for all $n \in \mathbb N$ thus: $$ \int _{x=0}^4\:f_n(x)\,dx \leq 4 \cdot 4^n = 4^{n+1} $$