Between parabolas

It is well known that a parabola fills $\frac{2}{3}$ of the rectangle that contains it. WLOG let each of the three rectangles have area $3$ . The entire shaded area is then $2+3+2=7$

Intersection area is $1$ .

The proportion is $\frac{1}{7}$

Let the parabolas be $x^2$ and $2x-x^2$ , so that the vertices are $(0, 0)$ and $(1, 1)$ . Since only the relative areas matter, the exact choice is irrelevant and this is the easiest.

The area of overlap is then $\int_0^1((2x-x^2)-x^2)dx=\frac13$

While the entire area is $\int_{-1}^0(1-x^2)dx+1+\int_1^2(2x-x^2)dx=\frac73$

The ratio is therefore $\boxed{\frac17}$

Let the intersection points be $(0,0)$ and $(1,1)$ . Note that the equation of the downward-pointing parabola can be written as $y=x^2$ and the equation of the upward-pointing parabola can be written as $y=2x-x^2$ . The area of intersection is:

$\int_{0}^{1} \;(2x-x^2-x^2) \;dx =\int_{0}^{1} \;(2x-2x^2) \; dx=\left[x^2-\dfrac{2x^3}{3}\right]_{0}^{1}=1-\dfrac{2}{3}=\dfrac{1}{3}$

Note that the area bounded by the y-axis, the line $y=1$ and the graph $y=x^2$ is equal to $\int_{0}^{1} \; \sqrt{y} \; dy$ . Considering that this area is equal to that bounded by the x-axis, the line $x=1$ and the graph $y=2x-x^2$ , the total area is:

$1+2\int_{0}^{1} \; \sqrt{y} \; dy=1+2\left[\dfrac{2\sqrt{y}^3}{3} \right]_{0}^{1}=1+2\left(\dfrac{2}{3}\right)=1+\dfrac{2\times 2}{3}=1+\dfrac{4}{3}=\dfrac{3+4}{3}=\dfrac{7}{3}$

Finally the desired fraction is $\boxed{\dfrac{1}{7}}$ .

One of the parabolas is not necessarily $y=x^{2}$ . I'm going to use more general equations. Let the intersection points be (0,0) and (h,k).

$\int\limits_0^1 {{x^2}} dx = \frac{1}{3}$

Drawing it out:

So adding up the areas:

(1/6+1/6)/(2/3+1/3+1/6+1/6+1/3+2/3) = 1/7

Let the left parabola be $y=kx^2$ which $k>0$ , and it intersects another parabola at points $(0,0)$ and $(a, ka^2)$ , the another parabola can be expressed as $y=-kx^2+2akx$ . Let the area at the middle be $B$ , and the area enclosed by a parabola and horizontal line be $A$ , so the full figure is $2A-B$ . $\begin{aligned}B&=\int_0^a(-kx^2+2akx)-(kx^2)dx\\&=\frac13a^3k\end{aligned}$

$\begin{aligned}A&=\int_0^{2a}-kx^2+2akxdx\\&=\frac{4}{3}a^3k\end{aligned}$ From the result we see that $A=4B$ , so the ratio is $\begin{aligned}\frac{B}{2A-B}&=\frac{B}{8B-B}\\&=\boxed{\dfrac17}\end{aligned}$

I solved this by mentally visualizing different types of parabolas, stretched and narrow and finding out via intuitive cluster-data analysis what the approximate fraction would be.

1) Divide the area. It's seen that the whole area is 2 triangles plus the area in the middle.

2) The middle area $A_{middle} = A_{parabola} - A_{triangle}$ Figure out the area between the red curve and y-axis. An inverse function is needed: $y = cx^2$ becomes $x = \sqrt{y}/c$ (when $x$ is positive)

Integrate and multiply by 2 to get both sides. $A_{parabola} = 2* \int_{0}^{h} (\sqrt{y}/c) dy\ = \frac{4}{3c} * h^\frac{3}{2}$

Area of the triangle $A_{triangle} = ah$ . But also $a = \sqrt{h}/c$ , so $A_{triangle} = \frac{\sqrt{h}}{c} * h = \frac{h^\frac{3}{2}}{c}$

The middle area is $A_{middle} = A_{parabola} - A_{triangle}\ = \frac{4}{3c} * h^\frac{3}{2} - \frac{h^\frac{3}{2}}{c} = \frac{1}{3} * \frac{h^\frac{3}{2}}{c}$

$\frac{A_{middle}}{2*A_{triangle}+A_{middle}} = \frac {\frac{1}{3}} {\frac{7}{3}} = \frac{1}{7}$

It can be solved with simple calculus. Calculate the C area between the curves a(x) = x^2 and a inverted down replica of it intersecting its vertice. It might be something like b(x) = -(x-2)^2 + 4. The B area (above this function "b") added to the area between y=4 and the initial "a" parabola minus the area between both the curves is going to be the total area. Dividing that C area by the Total area will give you the ratio between them.

I cheated a bit by measuring the width and height of the parabolas. It turns out they have a width of 3 and a height of 9. The left parabola is the graph of the function: f(x)=x^2 . The right one has the corresponding function: g(x)=-x^2+6 x . For the intervall {0;3} : g(x)≥f(x) . Therefore our d(x) is equal to: g(x)-f(x) . d(x)=-2 x^2+6 x By taking the integral of d(x) in the intervall {0;3} we find out the overlapping area is equal to 9. In order to find out the area of the entire figure I first changed f(x) into f2(x)=x^2-9 This makes it easier to get the area within one parabola. The area of one is 36. Hence both parabolas are congruent to each other, the are of the entire figure is: A(whole)=A(parabola) 2-9 . We have to subtract 9 in order to make sure we only take the overlapping area once. In the end you divide the overlapping area by the entire are and 9/63 equals 1/7 . I apologize for bad use of english technical terms and technical spelling. I am not a native speaker. Thank you!

Use parabola y=4x^2 and parabola y=1-4(x-1/2)^2 to find the overlap area of two parabolas as 7/6. The intersection area of two parabolas is equal to 1/6, which gives the desired ratio as equal to 1/7.