Animated Solution for Mathematics - Definite Integration: The area (in sq. units) of the region $\{(x,y):0\le y\le 2|x|+1,0\le y\le x^2+1,|x|\le 3\}$ is

A Journey Through Symmetry

Welcome, fellow traveler of the JEE path. Today, we aren't just solving an area problem; we are learning to see the hidden architecture of functions.

When you look at the region defined by $y\le 2|x|+1$ and $y\le x^2+1$, it is easy to feel overwhelmed by the absolute value or the overlapping boundaries. But take a deep breath. Let us peel back the layers of this problem together.

The Geometry of Symmetry

First, let us look at our players. We have $y=x^2+1$, a classic parabola shifted one unit upward, and $y=2|x|+1$, a sharp V-shape also starting at $(0,1)$.

Notice something beautiful? Both functions are even. If you replace $x$ with $-x$, the equations remain unchanged.

This tells us that the graph is perfectly symmetric about the $y$-axis. In the world of JEE, symmetry is your best friend. It means we only need to solve for the right side ($x\ge 0$) and then simply double our result. We have effectively halved our workload before even picking up the pen.

The Intersection of Paths

Now, we face the constraint: $y\le \min (2|x|+1,x^2+1)$. This 'min' condition is the heart of the problem. It tells us that the upper boundary of our region is not just one curve, but a composite of two.

To find where the boundary switches from the parabola to the line, we set them equal to each other for $x\ge 0$:

Subtracting $1$ from both sides gives us $x^2=2x$, which simplifies to $x(x-2)=0$. We find intersections at $x=0$ and $x=2$.

This is our 'switch' point. From $x=0$ to $x=2$, the parabola $y=x^2+1$ is the lower curve. Beyond $x=2$, the linear function $y=2x+1$ takes over as the lower boundary.

The Calculus of Areas

With our boundaries defined, we move to the integration. We split the area of the right half, let's call it $A_{right}$, into two distinct integrals.

First, the parabolic region:

Next, we handle the linear region from $x=2$ to $x=3$:

The Final Synthesis

We are almost there. The area of the right half is the sum of these two parts:

Finally, we invoke the symmetry we identified at the very beginning. The total area of the region is twice the area of the right half:

Look at that result. It is elegant, precise, and derived from a clear understanding of how these functions behave. You didn't just calculate an area; you mapped the territory.

Keep this mindset—always visualize, always check for symmetry, and never fear the intersection of curves. You are doing great. The final answer is $\frac{64}{3}$.