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

Welcome, fellow traveler on the path of JEE mastery! Today, we are going to unravel a problem that might look like a standard 'area under a curve' question, but it hides a beautiful, piecewise secret.

When you first look at the region $A=\{(x,y):0\le y\le 1+x|x|\text{ and }-1\le x\le 1\}$, you might be tempted to jump straight into a single integral. But pause—take a breath. The true beauty of calculus lies in recognizing when a function is not what it seems.

The Geometry of the Absolute Value

Imagine you are walking along the $x$-axis, starting from $x=-1$ and heading toward $x=1$. As you pass the origin ($x=0$), something changes. The function governing the height of our region, $y=1+x|x|$, behaves differently because of that absolute value sign.

When $x$ is negative, $|x|$ becomes $-x$. So, our function transforms into $y=1+x(-x)=1-x^2$. This is a downward-opening parabola.

But the moment you cross into positive territory ($x\ge 0$), $|x|$ becomes $x$, and our function transforms into $y=1+x(x)=1+x^2$. This is an upward-opening parabola. We are essentially stitching two different curves together at the $y$-axis.

The Art of Splitting the Integral

Because the 'roof' of our region changes its shape at $x=0$, we cannot use a single, monolithic integral. We must respect the boundary. We split our total area $A$ into two distinct parts: $A_1$ for the left side and $A_2$ for the right side.

This is the fundamental strategy in calculus: when the rule changes, the integral must change too. It is like navigating a river that changes its current halfway through—you must adjust your strategy accordingly.

The Calculation

Let's tackle $A_1$ first. We integrate $1-x^2$ from $-1$ to $0$. Using the power rule, the antiderivative is $x-\frac{x^3}{3}$.

Evaluating this from $-1$ to $0$ gives us:

Now, for $A_2$, we integrate $1+x^2$ from $0$ to $1$. The antiderivative is $x+\frac{x^3}{3}$.

Evaluating this from $0$ to $1$ gives us:

The Elegance of the Result

Finally, we bring them together. The total area is simply the sum of these two parts:

Two square units. A clean, integer result that emerges from the complexity of piecewise functions. This is the magic of mathematics—no matter how jagged or complex the path, the fundamental laws of calculus guide us to a precise, elegant solution.