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

A Journey into Symmetry

Imagine you are standing before a vast, blank coordinate plane. Your task is to map the behavior of the function $y=||x-1|-2|$.

At first glance, the nested absolute value bars might look intimidating, like a fortress guarding a secret. But in mathematics, as in life, the most complex structures are often built from the simplest foundations. Let us dismantle this fortress, brick by brick.

The Inner Core

We begin with the innermost part: $f(x)=|x-1|$. This is the classic 'V' shape, a fundamental building block of coordinate geometry, shifted one unit to the right. It is perfectly symmetric about the vertical line $x=1$.

Now, we subtract 2 from this function, creating $g(x)=|x-1|-2$. This shifts our 'V' downward by two units. The vertex of our 'V', which was at $(1,0)$, now plunges to $(1,-2)$.

The Reflection

Now, we apply the outer modulus: $y=|g(x)|=||x-1|-2|$. The absolute value operator is a mirror. It takes everything below the x-axis and flips it upward into the positive realm.

The portion of our 'V' that dipped down to $-2$ is now reflected, creating a peak at $(1,2)$. The points where the graph previously crossed the x-axis—at $x=-1$ and $x=3$—remain fixed. We have successfully transformed a downward-pointing 'V' into an upward-pointing 'V' with a peak at $(1,2)$.

The Geometric Revelation

Look closely at the region bounded by this curve and the x-axis. It is not a chaotic mess; it is a perfect triangle with vertices at $P(-1,0)$, $Q(1,2)$, and $R(3,0)$.

The base of this triangle lies along the x-axis, stretching from $x=-1$ to $x=3$. The length of this base is $3-(-1)=4$. The height of the triangle is simply the y-coordinate of the peak, which is $2$.

Using the elementary formula for the area of a triangle, $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$, we calculate:

The Rigor of Integration

If you prefer the analytical path, we can verify this using calculus. We split the integral into two parts based on the symmetry of the function:

Evaluating these integrals, we find:

We have arrived at the value 4. However, you might notice that some official sources cite 8.

Why the discrepancy? It is a reminder that in competitive exams, the definition of the 'bounded region' is paramount. If the problem intended to bound the area between the curve and the line $y=2$, the geometry would change, and the area would indeed double to 8.

Never let a discrepancy discourage you. Instead, let it sharpen your intuition. You have mastered the absolute value, you have visualized the reflection, and you have bridged the gap between geometry and calculus. That, my friend, is the true victory.