Find integral, what's wrong with these 2 solution? I am meeting an interesting problem and solved two different method.
Could you see following 2 solution, and find errors?

QUESTION.
  $$    f(x)=2x^3 -2x^2 - \int_{0}^{x+1}f(t)dt  $$
where $f(x)$ is symmetric about the point $(1,0)$ 
and differentiable at all point.
Find $$    \int_{2}^{4}f(x)dx  $$ .

$$ $$
Sol.1
By the SYMMETRIC about $(1,0)$, we get $f(1)=0$ $  \    \  $$    $     $\cdots(a)$
From $    f(x)=2x^3 -2x^2 - \int_{0}^{x+1}f(t)dt  $
$\Longrightarrow$  $f'(x)=6x^2 - 4x -f(x+1)$
$\Longrightarrow$ $f'(0)=-f(1)=0 $ $  \    \  $ (by (a).)
$ $ 
By the SYMMETRIC about $(1,0)$,  $f'(0)=f'(2)$.
$\Longrightarrow$ $0=f'(0)=f'(2)=6  \cdot2^2 -4\cdot2-f(3)=24-8-f(3)=16-f(3)  $
Thus $f(3)=16$.
$ $ 
From $   f(x)=2x^3 -2x^2 - \int_{0}^{x+1}f(t)dt  $
$$\begin{align}
&  \Longrightarrow  16=f(3)=2\cdot 3^3-2\cdot3^2-\int_{0}^{4}f(t)dt =36-\int_{0}^{4}f(t)dt \\
&  \Longrightarrow  20=\int_{0}^{4}f(t)dt\\
\end{align}$$
$$$$
$0=f(1)=2\cdot1^3-2\cdot1^2  -   \int_{0}^{2}f(t)dt $
$ \Longrightarrow 0= \int_{0}^{2}f(t)dt $
$$$$
SO, 
$$ \int_{2}^{4}f(x)dx = \int_{0}^{4}f(x)dx  -\int_{0}^{2}f(x)dx  = 20-0=20 $$
$$$$$$$$


NOW, 2nd solution


$$$$
 Sol.2
$ \int_{0}^{2}f(t)dt = 0 $.
$ f(-1)=-4$.
By the SYMMETRIC about $(1,0)$, $  \   $   $f(3)=-f(-1)$.
So,
$$\begin{align}
 & 4 = -f(-1)=f(3)=2\cdot3^3-2\cdot3^2-\int_{0}^{4}f(t)dt\\
&  \Longrightarrow  32=\int_{0}^{4}f(t)dt\\
\end{align}$$
$$$$
Thus, $$ \int_{2}^{4}f(x)dx = \int_{0}^{4}f(x)dx  -\int_{0}^{2}f(x)dx  = 32-0=32 $$
$$$$ $$$$

Sol.1 and Sol.2 say different things.
What mistake led me this result?

$$$$
Thanks in advance.
$$$$
$\color{red}{\text{----------- **EDIT** ----- **Please read below, thank you** ---------------------------------}}$
$$$$
P.S.1 
 $$$$  Let me 'darely' DEFINE symmetric about point $(a,b)$
$$$$
If $f$ is symmetric about point $(a,b)$
if and only if
$f(x)+f(2a-x)=2b$
if and only if
$f(a-x)+f(a+x)=2b$ 
$$$$
P.S.2
$$$$
By P.S.1 
$f'(x)=f'(2-x)$
So, $f'(0)=f'(2)$
$$$$
 A: Edit 2: Ok,first I apologize for what I wrote.. It seems that I understimated the problem.. Here is what I've done: 
The simetry with respect to $(1,0)$ traduces in $$f(x+1)=-f(-x+1) $$(#). Just replacing in the equation that defines $f(x)$(call it equation (1)) we get $f(-1)=-4$. By the simmetry, $f(3)=4$, too. Replacing in (1), we get $\int_{0}^{4} f(t) dt =32$. Now, using that $f$ is differentiable in all of R, we can use the chain rule in (#), giving $$f'(x+1)=f'(-x+1)$$. Deriving (1), we get $$f'(x)=6x^2-4x-f(x+1)$$. Evaluating in $2$, we get $f'(2)=12=f'(0)=-f(1)=\int_{0}^{2} f(t) dt$. 
Finally,$$\int_{2}^{4} f(t) dt=\int_{0}^{4} f(t) dt-\int_{0}^{2} f(t) dt=32-12=20$$.
Concerning your solutions: About the first, I don't know where do you get from that $f(1)=0$. In the second I don't get $\int_{0}^{2} f(t) dt=0$. I think you are asuming that $(1,0)$ belongs to the graph of $f$, but it's not necessary.
A: $f(x)=2x^3 -2x^2 - \int_{0}^{x+1}f(t)dt \text{ solving this for } \int_{0}^{x+1} f(t) dt \\ \int_0^{x+1} f(t) dt=2x^3-2x^2-f(x)  \\ \text{ Evaluating } \int_2^4 f(t) dt= \int_0^4 f(t) dt- \int_0^2 f(t)dt \\ =(2(3)^3-2(3)^2-f(3))-(2(1)^3-2(1)^2-f(1)) \\ =(36-f(3))-(0-f(1))\\ =36-f(3)+f(1) \\ =36-f(3)+0 \text{(since } f(1)=0 \text{ is given } \\ \\ \text{ time to play with } f(3) \\ \int_0^{0} f(t) dt=2(-1)^3-2(-1)^2-f(-1) \implies 0=-2-2-f(-1) \implies f(-1)=-4 \\ \text{ and we know } f(3)=-f(-1) \text{ since } f \text{ is point symmetric about } (1,0)   \\ \text{ so we were almost done earlier and now we can finish } \\ \int_2^4 f(t) dt=36-f(3)+0=36-f(3)=36-4=32\\  $
A: As far as I can see, there are 2 problems with your solutions.  First $f(x) = 2x^2 - 2x^3 - \int_{0}^{x+1} f(t)dt$ isn't really an algebraic equation, so you cannot expect to perform various algebraic manipulations on it and get a consistent answer.  Second it isn't clear what you mean by "$f(x)$ is symmetric about the point $(1,0)$."
If you differentiate the formula for $f(x)$ you get the differential equation $f'(x) = 6x^2-4x-f(x).$  Its solution is $f(x) = ce^{-x}+6x^2-16x+16$.  Using $f(-1) = ce+38 =-4$ we have $c = -42/e$.  Now we can integrate directly to find that $\int_{2}^{4}f(x)dx = -42e^{-3}+42e^{-5}+48$ 
