Animated Solution for Mathematics - Definite Integration: Let the area of the region $\{(x,y):2y\le x^2+3,y+|x|\le 3,y\ge |x-1|\}$ be A. Then 6A is equal to :

We are tasked with finding the area of a region defined by three distinct inequalities: $2y\le x^2+3$, $y+|x|\le 3$, and $y\ge |x-1|$.

This is a classic JEE Advanced problem that tests your ability to translate algebraic constraints into a physical, bounded shape.

Imagine standing on the Cartesian plane. We have three 'fences' enclosing our territory:

1. The first, $y\le \frac{x^2+3}{2}$, is an upward-opening parabola. 2. The second, $y\le 3-|x|$, is a downward-pointing V-shape with a peak at $(0,3)$. 3. The third, $y\ge |x-1|$, is an upward-pointing V-shape with its vertex at $(1,0)$.

The first step is to find the 'vertices' of our region—the points where these curves intersect. By solving the equations $y=3-|x|$ and $y=|x-1|$, we find the intersection points at $(2,1)$ and $(-1,2)$. These are the anchors of our region.

Is the upper boundary the same throughout the entire region? Let's check the intersection of the parabola $y=\frac{x^2+3}{2}$ and the line $y=3-|x|$.

For $x\ge 0$, setting $\frac{x^2+3}{2}=3-x$ leads to $x^2+2x-3=0$, which gives us $x=1$.

This is a critical realization! At $x=1$, the parabola and the line meet. Before $x=1$, the parabola is the 'ceiling' of our region. After $x=1$, the line $y=3-x$ becomes the 'ceiling.'

Because the ceiling changes, we cannot use a single integral. We must split our journey into two parts.

We define our total area $A$ as the sum of two regions: $A_1$ (from $x=-1$ to $x=1$) and $A_2$ (from $x=1$ to $x=2$).

For $A_1$, the upper bound is the parabola $\frac{x^2+3}{2}$ and the lower bound is the line $y=1-x$. The integral becomes:

Simplifying this, we get:

Now for $A_2$, the region from $x=1$ to $x=2$. Here, the upper bound is $y=3-x$ and the lower bound is $y=x-1$. The integral is:

Calculating this, we find:

We have our two pieces: $A_1=\frac{4}{3}$ and $A_2=1$. Adding them together, the total area is:

The question asks for $6A$. Thus:

We have navigated the curves, respected the boundaries, and arrived at the solution. Remember, in JEE Advanced, the math is just the language; the true skill is in the visualization.