Animated Solution for Mathematics - Definite Integration: The area of the region $x,y:x^2\le y\le |x^2-4|,y\ge 1$ is

A Journey Through Area

Welcome, fellow explorer of the mathematical landscape. Today, we are not just solving an integral; we are mapping a territory.

When you look at a problem involving inequalities like $x^2\le y\le |x^2-4|$ and $y\ge 1$, do not see them as abstract symbols. See them as boundaries on a map.

You are standing in a region defined by the interplay of three distinct curves: the standard upward-opening parabola $y=x^2$, the horizontal floor $y=1$, and the reflected 'mountain' of the modulus function $y=|x^2-4|$.

Visualizing the Battlefield

First, let us sketch our terrain. The parabola $y=x^2$ is our familiar friend, starting at the origin and opening upwards.

The function $y=|x^2-4|$ is more interesting. Between $x=-2$ and $x=2$, the expression $x^2-4$ is negative, so the modulus reflects it upwards, creating a peak at $(0,4)$.

Outside this interval, it behaves like the standard parabola $y=x^2-4$. Our region is the space trapped between the upward parabola $y=x^2$ and this reflected mountain, but with a strict condition: we must stay above the line $y=1$. This line acts as a floor, cutting off the bottom of our shape.

The Power of Symmetry

In JEE Advanced, time is your most precious resource. Look at the region; it is perfectly symmetric about the $y$-axis.

This is a gift! We do not need to integrate from $-\sqrt{2}$ to $\sqrt{2}$.

We can simply calculate the area for the right half, where $x\ge 0$, and multiply the result by $2$. This simple realization cuts our algebraic labor in half and significantly reduces the risk of a sign error.

The Critical Junctions

Before we integrate, we must know where our boundaries meet.

The intersection of the upper boundary $y=4-x^2$ and the lower boundary $y=x^2$ occurs when $4-x^2=x^2$, which simplifies to $2x^2=4$, or $x=\pm \sqrt{2}$.

The intersection of the floor $y=1$ and the parabola $y=x^2$ occurs at $x^2=1$, or $x=\pm 1$. These points, $x=1$ and $x=\sqrt{2}$, are the critical junctions where the nature of our region changes.

The Integral Setup

Now, let us construct our integral. For the right half ($x>0$), the region starts at $x=0$.

From $x=0$ to $x=1$, the upper boundary is $y=4-x^2$ and the lower boundary is the line $y=1$. The height of our vertical strip is $(4-x^2)-1=3-x^2$.

From $x=1$ to $x=\sqrt{2}$, the upper boundary remains $y=4-x^2$, but the lower boundary is now the parabola $y=x^2$. The height of our strip is $(4-x^2)-x^2=4-2x^2$.

Thus, the total area $A$ is given by:

The Final Triumph

Let us execute the integration with precision. For the first part, the integral of $3-x^2$ is $3x-\frac{x^3}{3}$.

Evaluating from $0$ to $1$, we get $(3-\frac{1}{3})-0=\frac{8}{3}$.

For the second part, the integral of $4-2x^2$ is $4x-\frac{2x^3}{3}$. Evaluating from $1$ to $\sqrt{2}$, we get:

Since $(\sqrt{2}{)}^3=2\sqrt{2}$, this becomes:

Adding these together, we have:

Finally, multiplying by our symmetry factor of $2$, we arrive at the elegant result:

You have successfully navigated the curves, respected the constraints, and arrived at the truth. This is the essence of JEE Advanced mathematics—not just calculation, but the art of structured thinking.