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?
 A: You have already correctly identified the area of the rectangle but you are vague in stating the integral to compute the area between the parabola and the x-axis. Your approach is correct but there are some simplifications we can use to make the computation easier.
We first translate the parabola to the origin and let $b-a=2w$. This gives the area of the rectangle as $A_{\text{rec}}=2wc$ and makes the algebra easier.
Since the parabola whose vertex is the origin is of the form $y=kx^2$ we immediately get that $$kw^2=c\implies k=\frac{c}{w^2}\implies y=\frac{c}{w^2}x^2$$
Although we want the area bounded by the parabola and the line $y=c$, we can take the area under the parabola and subtract that from the area of the rectangle to prove the result.
We exploit the symmetry of the parabola to get $$A_{\text{par}}=2\int_{0}^{w}\frac{c}{w^2}x^2 dx=\frac{2}{3}wc=\frac{1}{3}2wc=\frac{1}{3}A_{\text{rec}}$$
Subtracting the result from the area of the rectangle, we get that the area bounded by the parabola and $y=c$ is $A_{\text{rec}}-\frac{1}{3}A_{\text{rec}}=\frac{2}{3}A_{\text{rec}}$
A: 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.
A: I prefer to show that the area $A$ of the region below the parabola inside the rectangle, $$A=\frac{1}{3}\textrm{  of the rectangle }= \frac{c}{3}(b-a).$$
Since $(\frac{b+a}{2}, -c)$ is the vertex of the parable, we have $$
y=k\left(x-\frac{b+a}{2}\right)^2-c \tag*{(1)} 
$$
for some constant $k$.
Putting $(b,0)$ into $(1)$ yields $$
k=\frac{4 c}{(b-a)^2}
$$
Then
$$
\begin{aligned}
A & =\int_a^b[y-(-c)] d x \\
& =\frac{4 c}{(b-a)^2} \int_a^b\left(x-\frac{b+a}{2}\right)^2 d x \\
& =\frac{4 c}{(b-a)^2}\left[\frac{\left(x-\frac{b+a}{2}\right)^3}{3}\right]_a^b\\&= \frac{c}{3}(b-a)
\end{aligned}
$$
