Prove that $\int _0^4\:f_n(x)\,dx < 4^{n+1}$ where $f_n(x)= \left(4x-x^2\right)^n$ $ f_n(x) = \left(4x-x^2\right)^n$
I'm trying to prove that $\int _0^4\:f_n(x)\,dx < 4^{n+1}$
When I'm doing a general integral to $n > 1$, I'm getting an expression that equals to zero when $X = 4$ or $X = 0$.
This is the expression:
$$\frac{\left(4x-x^2\right)^{n+1}}{(4-2X)(n+1)} $$
What am I missing? Because it doesn't make sense that every integral with $n > 1$ is equal to zero
 A: 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.
A: 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$ 
A: 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:
$$ \int _{x=0}^4\:f_n(x)\,dx \leq  4 \cdot  4^n = 4^{n+1} $$ 
A: If you let $I_n = \int_0^4 (4x-x^2)^ndx$ for every positive integer $n$, you can even find a recurrence relation for this sequence using integration by parts:
$$I_n = \int_0^4 (4x-x^2)^n dx = \int_0^4 (4x-x^2)^n \cdot (x-2)'dx = \cdots = -2nI_n + 8n I_{n-1}.$$
Using this you can easily solve your question using induction.
A: Here is a nice solution without using the binomial expansion.
$$\int_0^4 (4x -x^2)^ndx < 4^{n+1}$$
If we let n=0, the integral resolves to 0 which is less than 4. So our base case is covered.
Now assume there exists a $k$ such that
$$\int_0^4 (4x -x^2)^kdx < 4^{k+1}$$
Holds. 
Now we need to prove this inequality holds for the $k+1$.
$$\int_0^4 (4x -x^2)^{k+1}dx = \int_0^4 (4x -x^2)^k (4x - x^2)dx $$
If we apply integration by parts and let $u= 4x -x^2$ and $dv = (4x - x^2)^k$ we get
$$(4x - x^2) \int (4x -x^2)^kdx - (4 - 2x)\int(4x -x^2)^kdx $$
We let our bounds go from 0 to 4 and we get the following
$$0 \int_0^4 (4x -x^2)^kdx + 4 \int_0^4 (4x -x^2)^k$$
We know that $\int_0^4 (4x -x^2)^kdx < 4^{k+1}$ therefore we can conclude
$$4\int_0^4 (4x -x^2)^kdx < 4^{k+2}$$
