Animated Solution for Mathematics - Definite Integration: The area bounded by the curve $y=|x^2-9|$ and the line $y=3$ is :

Welcome, future engineer. Today, we are going to tackle a problem that might look like a standard calculus exercise, but it is actually a beautiful lesson in geometric intuition and symmetry. We are tasked with finding the area bounded by the curve $y=|x^2-9|$ and the line $y=3$.

Before we touch a single integral sign, let us paint a picture in our minds.

Imagine the standard parabola $y=x^2-9$. It opens upwards, crossing the x-axis at $x=3$ and $x=-3$, with its vertex at $(0,-9)$.

Now, apply the modulus operator. The modulus function is a 'reflector' that takes everything below the x-axis and flips it upwards. That deep valley between $x=-3$ and $x=3$ is reflected, creating a sharp peak at $(0,9)$.

The graph now looks like a 'W'. This is the 'split personality' of the modulus function: For $|x|\ge 3$, the curve is $y=x^2-9$. For $|x|<3$, the curve is $y=9-x^2$.

We introduce the line $y=3$. This is a horizontal line cutting across our 'W' shape. We need to find the area trapped between this line and the curve by identifying their intersection points.

We set $|x^2-9|=3$. This splits into two cases: 1. $x^2-9=3\Rightarrow x^2=12\Rightarrow x=\pm 2\sqrt{3}$ 2. $x^2-9=-3\Rightarrow x^2=6\Rightarrow x=\pm \sqrt{6}$

These four points, $x=\pm 2\sqrt{3}$ and $x=\pm \sqrt{6}$, define the boundaries of our 'pockets' of area.

Notice that our function $f(x)=|x^2-9|$ is an even function, meaning it is perfectly symmetric about the y-axis. The area on the right side of the y-axis is a perfect mirror image of the area on the left.

Instead of calculating the entire area, let us focus solely on the right side ($x>0$) and simply multiply our final result by $2$. We are now focusing on the interval from $x=\sqrt{6}$ to $x=2\sqrt{3}$.

We cannot simply integrate from $x=\sqrt{6}$ to $x=2\sqrt{3}$ in one go because the curve changes its definition at $x=3$. We must split our integral into two distinct parts: 1. From $x=\sqrt{6}$ to $x=3$, the curve is $y=9-x^2$. 2. From $x=3$ to $x=2\sqrt{3}$, the curve is $y=x^2-9$.

Our total area is given by:

Let us simplify the integrands. The first part becomes ${\int }_{\sqrt{6}}^3(x^2-6)dx$, and the second part becomes ${\int }_3^{2\sqrt{3}}(12-x^2)dx$.

For the first integral, the antiderivative is $\frac{x^3}{3}-6x$. Evaluating from $\sqrt{6}$ to $3$:

For the second integral, the antiderivative is $12x-\frac{x^3}{3}$. Evaluating from $3$ to $2\sqrt{3}$:

Now, we combine these results. The total area is $2\times (I_1+I_2)$:

Factoring out a $4$, we arrive at the final result:

We have navigated the modulus, respected the symmetry, and executed the integration with precision. This is the essence of JEE Advanced mathematics—not just grinding through numbers, but understanding the geometry of the problem.