Animated Solution for Mathematics - Definite Integration: Let $A$ be the area of the region $\{(x,y):y\ge x^2,y\ge (1-x{)}^2,y\le 2x(1-x)\}$. Then $540A$ is equal to ______.

Welcome, fellow traveler of the JEE landscape. Today, we are exploring a beautiful, symmetric arena defined by three parabolas: $y=x^2$, $y=(1-x{)}^2$, and $y=2x(1-x)$.

The first two are standard upward-opening parabolas, while the third is a downward-opening one. Together, they carve out a small, elegant region. Our goal is to find the area $A$ of this region and then calculate $540A$.

Before we can calculate the area, we must identify the intersection points of these curves.

By setting $x^2=(1-x{)}^2$, we find the intersection of the first two at $x=\frac{1}{2}$. Equating the second and third, $(1-x{)}^2=2x(1-x)$, gives us $x=\frac{1}{3}$.

Finally, the first and third meet at $x^2=2x(1-x)$, which yields $x=\frac{2}{3}$. These three points—$x=\frac{1}{3}$, $x=\frac{1}{2}$, and $x=\frac{2}{3}$—are the vertices of our region and define the boundaries of our integration.

Symmetry

Now, we utilize symmetry to save precious time. The region is perfectly symmetric about the vertical line $x=\frac{1}{2}$.

Instead of integrating across the entire range from $\frac{1}{3}$ to $\frac{2}{3}$, we can calculate the area of the left half—from $x=\frac{1}{3}$ to $x=\frac{1}{2}$—and simply multiply by two. This is the kind of insight that separates the top rankers from the rest.

The area $A$ is given by the integral:

In our interval of $[\frac{1}{3},\frac{1}{2}]$, the upper curve is the downward parabola $y=2x(1-x)$, and the lower curve is $y=(1-x{)}^2$. Our integrand becomes:

Integrating this expression term by term, we obtain:

We evaluate the integral from $\frac{1}{3}$ to $\frac{1}{2}$. At the upper limit $x=\frac{1}{2}$:

At the lower limit $x=\frac{1}{3}$:

The difference is:

Multiplying by our symmetry factor of $2$, we get $A=\frac{5}{108}$. Finally, calculating $540A$:

The final result is 25. You have navigated the curves, utilized symmetry, and conquered the calculus.