Animated Solution for Mathematics - Definite Integration: The area bounded by the curves $y=|x^2-1|$ and $y=1$ is

Welcome, future engineers. Today, we are going to dissect a problem that is a classic in the JEE Advanced repertoire. It is not just about finding an area; it is about understanding the soul of a function.

We are looking at the area bounded by $y=|x^2-1|$ and $y=1$. Imagine you are standing on a coordinate plane. You have a standard parabola, $y=x^2-1$, which dips below the x-axis.

But then, the modulus operator arrives like a mirror, reflecting that negative dip back up into the positive realm. The result is a beautiful, W-shaped curve. Our mission is to calculate the area trapped between this W-curve and the horizontal line $y=1$.

Before we can calculate any area, we must know where these two curves intersect. We set $|x^2-1|=1$.

This equation splits into two scenarios. First, $x^2-1=1$, which leads to $x^2=2$, giving us $x=\pm \sqrt{2}$.

Second, $x^2-1=-1$, which simplifies to $x^2=0$, giving us $x=0$. These points—$0$, $\sqrt{2}$, and $-\sqrt{2}$—are the boundaries of our integration. They define the limits of our battlefield.

Here is where the elite student separates themselves from the crowd. Look at the graph. Do you see the symmetry?

Both $y=|x^2-1|$ and $y=1$ are symmetric about the y-axis. This means the area on the left is identical to the area on the right.

Instead of calculating the integral from $-\sqrt{2}$ to $\sqrt{2}$, we can simply calculate the area in the first quadrant (where $x\ge 0$) and multiply it by two. This is not just a trick; it is a mindset. Always look for symmetry to simplify your life.

Now, we must handle the modulus. The expression $x^2-1$ changes sign at $x=1$. This is our critical point.

For $0\le x<1$, $x^2$ is less than $1$, so $x^2-1$ is negative. The modulus flips it, making the curve $y=1-x^2$.

For $1\le x\le \sqrt{2}$, $x^2$ is greater than $1$, so the modulus drops, and the curve is $y=x^2-1$. We have two distinct regions to integrate.

Let us set up our integrals. The area is the integral of (Upper Curve - Lower Curve).

In the first region, from $0$ to $1$, the upper curve is $y=1$ and the lower curve is $y=1-x^2$. The integrand is $1-(1-x^2)=x^2$.

In the second region, from $1$ to $\sqrt{2}$, the upper curve is still $y=1$, but the lower curve is now $y=x^2-1$. The integrand is $1-(x^2-1)=2-x^2$.

Our total area is given by:

Now, we integrate. The integral of $x^2$ is $\frac{x^3}{3}$. The integral of $2-x^2$ is $2x-\frac{x^3}{3}$.

Evaluating the first part from $0$ to $1$ gives us $\frac{1}{3}$. Evaluating the second part from $1$ to $\sqrt{2}$ requires careful arithmetic:

Combining these, we get:

Factoring out the $4$, we arrive at our final answer:

You have successfully navigated the modulus, the symmetry, and the integration. This is the essence of JEE Advanced mathematics—systematic, logical, and deeply satisfying.