Geometric proof that $\int_a^c (x-a)(x-b)(x-c)\ dx=0$ if and only if $b$ is the midpoint of $a$ and $c$. Let $a<b<c$ be three real numbers, and $f(x)=(x-a)(x-b)(x-c)$. We want to show that $\int_a^c f(x)\ dx=0$ if and only if $b=\frac{a+c}{2}$.
The first move is to horizontally shift $f$ by $b$ units, that is introduce $g(x)=f(x+b)=(x-r_1)x(x+r_2)$ with $r_1<0<r_2$. Then the statement becomes $\int_{r_1}^{r_2} g(x)\ dx=0$ if and only if $r_1=-r_2$.
If $r_1=-r_2$, the result is true, since the integrand is odd. For the converse, a not difficult but boring computing shows that the integral is $\frac{(r_1-r_2)^3(r_1+r_2)}{12}$, which means that $r_1=-r_2$ (since $r_1-r_2<0$).
This works, but computations does not satisfy me and I have the strong intuition that there could be a more geometric way.
My question: is there a computationless way to show that result?
 A: Try to rewrite it in this way:
$\int_a^c f(x)=\int_a^c  (x-a)(x-m)(x-c)dx + (m-b)\int_a^c  (x-a)(x-c)dx$
The function $(x-a)(x-m)(x-c)$ is odd function if we place the grid origin in $(m, 0)$. So the integral is zero. The second integral is zero if and only if $m = b$.
Does this proof satisfy you?
A: Note that $f(x) = (x-a)(x-c) < 0$ for all $x \in (a,c)$.
It is not too hard to see (by 'symmetry', since $f(x) = - f({a+c \over 2} -(x-{a+c \over 2})$) that
$\int_a^c f(x) (x-{a+c \over 2}) dx = 0$.
Note that $\int_a^c f(x) (x-b) dx = \int_a^c f(x) ((x-b) - (x-{a+c \over 2})) dx = \int_a^c f(x) (({a+c \over 2} -b)) dx$, from which it follows
that the integral is zero iff $b={a+c \over 2}$.
Aside: To see why $\int_a^c f(x) (x-{a+c \over 2}) dx = 0$, choose a 
change of variables $t=a+c-x$. Then we note that $f(t) = f(x)$ and
$\int_a^c f(x) (x-{a+c \over 2}) dx =\int_c^a f(t) ({a+c \over 2}-t) (-1) dt = - \int_a^c f(t) (t-{a+c \over 2}) dt$.
A: Did you notice, every cubic curve is anti-symmetric with respect to its points of inflection, which can be calculated in your case by double differentiation:
$$ x_I =\frac{a+b+c}{3},y_I=f(\frac{a+b+c}{3})?  $$
Shifting the origin of coordinate system to $ x_I,y_I $ brings it into a form 
$ y_1 = A  x_1( x_1^2 - B^2) $ where $ B = (a_1+c_1)/2 $
which is an odd function, the integral or area under cubic curve vanishes between the new $ x_1=a_1, x_2=c_1. $
If not sufficiently clear, shall explain again.
EDIT2:
Sorry about delay. Recasting the cubic using $ h,k $ displacement symbols.
When 3  roots are real,
The cubic equation of third degree polynomial is taken wlog for discussion of roots as:  
$$ y = - (x-a) ( x-b) (x-c) \tag{1} $$ has an inflection point at 
$$ x = h = (a+b+c)/3 ; \, \, y = k = - (a + b- 2 c) ( b + c -2 a) ( c + a - 2 b)/27 \tag{2} $$
with a spread $\sigma$ on either side at inflection point level $ y= k: $
$$ \sigma = \pm \sqrt{ (a^2 + b^2 + c^2 - a b - b c - c a)  /3}\tag{3} $$
i.e., roots when taken as  
$$ x = ( h -\sigma, h, h + \sigma ),\,\, y = k \tag{4} $$ 
bring the cubic equation to another algebraic form equivalent to (1):
$$ ( y-k) = - ( x -h -\sigma)( x- h)( x-  h + \sigma) \tag{5} $$ 
When the roots are in arithmetic progression, letting $ ( a + c) = 2\, b, X = x -h \tag{6} $
it assumes a much simpler form:
$$ y = - X ( X^2 - \sigma ^2) \tag{7} $$ 
which is an odd function, anti-symmetric  with respect to displaced coordinates $X=0 $ . 
The integral vanishes when evaluated between $ X = \pm \sigma $ limits. It is same as saying that between three equi-spaced points  $ (a, (a+c)/2, c) $ the integral should also vanish in the un-shifted situation of wavy cubic.
The equivalence of cubic forms (1), (5) is valid when there are one real and two complex conjugate roots.
A: To expand on Narasimham's answer we need to also show that
$$x_I = (a+b)/2$$if$$c = (a+b)/2$$
Substituting and simplifying gives:
$$x_I = \frac{a+b + \frac{a+b}{2}}{3} = \frac{2(a+b) + (a+b)}{6} = \frac{a+b}{2}$$
