Prove that $\int_a^c f(t)dt - \int_c^b f(t)dt = f(c)(a+b-2c) $, for some $c\in(a,b)$ Let $f$ be a continuous on $[a,b]$ then prove that there exist some $c$ that lies in $(a,b)$ such that
$$\int_a^cf(t)\,dt  - \int_c^b  f(t)\,dt  = f(c)(a+b-2c) $$
and hence prove that 
$\int_a^c f(t)\,dt  - \int_c^b  f(t)\,dt  = n  f(c)(a+b-2c) $ where $n \ge0$
I have been able to prove the first part. but not for general $n$.
I used rolles theorem to prove the first part.
I dont know in which tag to put this question into.
 A: When $n=0$, it is the Intermediate Value Theorem applied to 
$$
g:x\to \int_a^{x} f -\int_{x}^b f .
$$ 
Suppose $n>0$.
Let $$d=\frac{1}{2}a+\frac{1}{2}b.$$
If $g (d)=0 $ the inequality is proved for all $n$. 
Suppose this is not the case : there is $c_0\in(a,b)\setminus\{d\}$ where $g(c_0)=0$. 
Replacing $f(x)$ by $f(a+b-x)$ if necessary, we may assume $c_0<d$. 
Consider the function 
$$
H(x):x\to|a+b-2x|^{1/n}g(x).
$$
Note that $H$ is continuous on both $[a,d]$ and differentiable on $[a,d)$.
As $H(c_0)=H(d)=0$, applying Rolle's Theorem on $[c_0,d]$  we obtain that for some $c\in(c_0,d)$, $H^\prime(c)=0$, that is,
$$
0=-\frac{2}{n}(a+b-2x)^{\frac{1}{n}-1}g(c)+(a+b-2x)^{1/n}\left(2f(c)\right),
$$
in other words,
$$
\int_a^cf -\int_c^b f = nf(c)(a+b-2c).
$$
A: For the case where $n \geq 0$, I am going to reformulate your problem as follows:
Show
$$
\int_a^cf(t) + nf(c)dt = \int_c^bf(t)+nf(c)dt
$$
This is just rearranging some terms, and factoring some things so that everything is inside an integral.  You want to split the brackets on the right side so that you have a $(a-c)$ term and a $(b-c)$ term.  The rest I'll leave as an exercise.  Let me know if you get stuck there, and I can expand a bit further in the comments.  
And, believe it or not, that was the hard part.
Now, take each side of this equation separately.  From the assumptions, we know that
$$\int_a^cf(t) + nf(c)dt$$
is a continuous, bounded function of $c$.  Let's call it 
$$
G(c) = \int_a^cf(t) + nf(c)dt.
$$
Now, $G(a) = 0$, and $$G(b) = \int_a^bf(t) + nf(c)dt$$.
Let's call the right hand side 
$$H(c) = \int_c^bf(t)+nf(c)dt.$$
This is also a continuous, bounded function of $c$.  We also know that 
$$H(a) = \int_a^bf(t)+nf(c)dt$$
and $H(b) = 0$.
Using the intermediate value theorem, and the knowledge we've gained about $G(c)$ and $H(c)$, we can conclude there must be some point in the middle where these two functions cross.  That is, there is some point $C$ that makes $G(C) = H(C)$.  
Thus, the conclusion holds, that there exists a point $c\in(a,b)$ such that
$$
\int_a^cf(t)dt -\int_c^bf(t)dt = nf(c)(a+b-2c)
$$
Sidenote:  Using Rolle's theorem to solve the first part of this question is pretty slick. However, I don't think there's an equivalent way to do that for the $n > 0$ case.  The IVT looks like the intended solution for both cases.
A: Define:
$$
F(x) = \int_a^x f(t) \, \mathrm{d} t
$$
Then your expression is just:
$$
F(c) - (F(b) - F(c))
  = 2 F(c) - F(b)
$$
If you substitute in your identity, by the fundamental theorem of calculus you get the differential equation:
$$
F'(c)(a + b - 2 c) - 2 F(c) = - F(b)
$$
Its solution is:
$$
F(c) = \frac{F(b) (c - a)}{2 c - a - b}
$$
Differentiate $F$, and you get the only function satifying your identity.
