Show that area of parabola is $\frac{2}{3}$ of rectangle's area My math book has following problem on chapter of basic integration:


Show that area between parabola and x axis is 2/3 of rectangle's area.

The book has no examples for situation like this and I could not find anything helpful from internet. (Maybe I don't know correct English terms.)
I thought I could calculate the two areas and compare them.
$$ A_{\text{rectangle}} = (b-a)*c $$
If parabola is drawn by function $ f(x) $ then I can get its area by integration.
$$ A_{\text{parabola}} = \int_a^b f(x) dx $$
The parabole goes through points $ (a,0) (b,0) $ and $ (\frac{a+b}{2}, -c) $
Using those points and the quadric equation I could find equation for the parabola and integrate it. But can't figure out how I could compare area resulting from this with the area of rectangle.
What would be the correct approach to this problem?

By inspection, the equation of the parabola is $y=\frac{4c}{(b-a)^2}(x-\frac{b+a}{2})^2-c$.
The difficult part of the computation is the area. By a careful inspection, the area between the $x$-axis and the parabola is:
$$A_p=\frac{4c}{(b-a)^2}\frac{2(b-a)^3}{3\times8}-c(b-a)=-\frac{2}{3}c(b-a)=\frac{2}{3}A_r$$
where $A_r=(b-a)(-c)=-c(b-a)$ is the area of the rectangle.
