Find integral, what's wrong with these 2 solution?

Question

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, $f'(0)=f'(2)$

$$$$

Answer 1

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.

Answer 2

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 $\int_{2}^{4}f(x)dx = -42e^{-3}+42e^{-5}+48$

Answer 3

$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\\ $