Animated Solution for Mathematics - Definite Integration: The area of the region $\{(x,y):|x-y|\le y\le 4\sqrt{x}\}$ is

Welcome, future engineers. Today, we are going to dissect a problem that often intimidates students at first glance: the area of a region defined by a compound inequality involving a modulus function.

When you see $\{(x,y):|x-y|\le y\le 4\sqrt{x}\}$, your instinct might be to panic. But I want you to take a deep breath. In JEE Advanced, the complexity is often just a mask for a very elegant, simple geometric truth. Let us peel back that mask together.

The first hurdle is the expression $|x-y|\le y$. Many students try to square this immediately, but that is a trap. Instead, let us rely on the fundamental definition of the modulus: $|a|\le b$ is equivalent to $-b\le a\le b$.

Applying this to our problem, we get:

Now, let us split this into two manageable linear inequalities. First, take the left side: $-y\le x-y$. If we add $y$ to both sides, we get $0\le x$, or simply $x\ge 0$. This is a massive revelation! It tells us our entire region is confined to the right half of the Cartesian plane.

Next, take the right side: $x-y\le y$. Adding $y$ to both sides gives $x\le 2y$, which rearranges to $y\ge \frac{x}{2}$. Suddenly, the intimidating modulus has vanished, leaving us with a clear geometric instruction: our region lies above the line $y=\frac{x}{2}$.

Now we look at the second part of the condition: $y\le 4\sqrt{x}$. If we square both sides, we get $y^2\le 16x$. This is the classic equation of a right-opening parabola.

Because $y$ is defined as a positive square root, we are only dealing with the upper branch. So, what do we have? We have a region bounded below by the line $y=\frac{x}{2}$ and bounded above by the parabola $y=4\sqrt{x}$.

Imagine the graph: the parabola starts at the origin and curves upwards, while the line starts at the origin and cuts through the first quadrant. The region trapped between them is the area we need to calculate.

Before we can integrate, we need to know where this region begins and ends. We know it starts at the origin $(0,0)$. Where does it close? We find this by setting the two curves equal to each other:

Squaring both sides gives us $16x=\frac{x^2}{4}$, which simplifies to $x^2=64x$. Solving $x^2-64x=0$ gives us $x=0$ and $x=64$. Our limits of integration are locked in: from $0$ to $64$.

Now, we set up our definite integral. The area $A$ is the integral of the upper curve minus the lower curve:

Let us break this into two parts. First, the integral of $4\sqrt{x}$, which is $4x^{1/2}$. Using the power rule, we get:

Second, the integral of $\frac{x}{2}$ is straightforward:

Now, we evaluate from $0$ to $64$. Plugging in $64$:

Since $64^{3/2}=(8^2{)}^{3/2}=8^3=512$, the first term becomes:

And the second term is:

Finally, we subtract:

There it is. The area is $\frac{1024}{3}$. You see? By breaking the problem down, respecting the definitions, and trusting your calculus, you turned a terrifying inequality into a beautiful, solvable area.