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?

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 $A_{\text{rec}}-\frac{1}{3}A_{\text{rec}}=\frac{2}{3}A_{\text{rec}}$
