Otdfbt

$ f_n(x) = left(4x-x^2right)^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:

$$fracleft(4x-x^2right)^n+1(4-2X)(n+1) $$

What am I missing? Because it doesn't make sense that every integral with $n > 1$ is equal to zero

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$

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.

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:
$$I=int _x=0^4:f_n(x),dx =int _x=0^4: (4x-x^2 )^n,dx =int _x=0^4: sum_k=0^n binomnk (4x)^k cdot (-x^2)^n-k,dx$$
We make this look a bit prettier:
$$I=int _x=0^4: sum_k=0^n binomnk 4^k cdot (-1)^n-kx^2n-k,dx$$

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:
$$I=sum_k=0^n binomnk frac12n-k+1 4^k cdot (-1)^n-k4^2n-k+1-0$$
We collect some terms:
$$I=sum_k=0^n binomnk frac12n-k+1 4^2n+1 cdot (-1)^n-k$$
Using combinatorial formulae and some help from Wolfram Alpha, this can be exactly computed to be:
$$ I =fracsqrt pi cdot n! cdot 2^2n+1(n+frac12)!=fracsqrt pi cdot n!(n+frac12)! cdot 4^n + frac12$$
We then need to prove by induction that $fracsqrt pi cdot n!(n+frac12)! > sqrt4$ 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:
$$ int _x=0^4:f_n(x),dx leq 4 cdot 4^n = 4^n+1 $$

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 $k+1$.

$$int_0^4 (4x -x^2)^k+1dx = int_0^4 (4x -x^2)^k (4x - x^2)dx $$

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

$$4int_0^4 (4x -x^2)^kdx < 4^k+2$$