The area of the region bounded by the curves y = |x-2|, x=1, x=3 and the x-axis is
(a) 4
(b) 2
(c) 3

Q: Wait, why do we have to split the integral at $x=2$? Can't we just integrate $|x-2|$ directly?

The modulus function $|x-2|$ is defined piecewise. Because the derivative is discontinuous at $x=2$, the fundamental theorem of calculus requires us to handle the two linear segments separately. Think of it as two different lines meeting at a point; you wouldn't try to describe both with one equation, so don't try to integrate them as one!

Q: Is it always better to use geometry instead of integration for these problems?

Whenever the function represents a simple geometric shape like a triangle, rectangle, or circle, geometry is almost always faster and less prone to calculation errors. Integration is your heavy artillery—use it when the shape is irregular, but always keep an eye out for the elegant geometric shortcut.

Q: What if I forgot the definition of the modulus function? How do I handle the negative sign?

The modulus $|f(x)|$ is simply a 'distance' operator. It forces the output to be positive. If $x-2$ is negative (which happens when $x < 2$), we multiply it by $-1$ to make it positive. That is why $|x-2|$ becomes $-(x-2)$ for $x < 2$.

Q: Why is the height of the triangle 1? I'm confused about the coordinates.

At the boundaries $x=1$ and $x=3$, plug the values into $y = |x-2|$. For $x=1$, $y = |1-2| = |-1| = 1$. For $x=3$, $y = |3-2| = 1$. Since the base is the distance from the vertex ($x=2$) to the boundary ($x=1$ or $x=3$), the base is $1$ and the height is $1$.

Q: Does the x-axis boundary change anything?

The x-axis is simply the line $y=0$. Since our function $y = |x-2|$ is always greater than or equal to zero, the area is naturally bounded by the x-axis. If the function had dipped below the x-axis, we would have had to take the absolute value of the integral, but here, the geometry is straightforward.

Question

The area of the region bounded by the curves $y = |x-2|, x=1, x=3$ and the x-axis is

Options:

(A) 4

(B) 2

(C) 3

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the V-Shape

Welcome, future engineer! Today, we are going to peel back the layers of a classic JEE problem. Often, when students see the modulus symbol $|x-2|$, they panic, thinking they need to perform complex calculus.

But I want you to pause. Take a deep breath. Mathematics is not about memorizing formulas; it is about seeing the hidden geometry in the numbers.

### Visualizing the V-Shape

Imagine you are standing on a coordinate plane. You are looking at the function $y = |x-2|$.

The modulus function is a master of symmetry. It takes any negative input and flips it into the positive realm. The 'vertex' or the tip of this V-shape occurs exactly where the expression inside the modulus vanishes.

Setting $x - 2 = 0$, we find our anchor point at $(2, 0)$. This is the heart of our graph. To the left of $x=2$, the line descends; to the right, it ascends. It is a perfect, sharp reflection.

### Defining the Boundaries

We are tasked with finding the area bounded by this V-shape, the vertical lines $x=1$ and $x=3$, and the x-axis ($y=0$).

If you sketch this, you will see two beautiful, identical right-angled triangles sitting side-by-side. One triangle is formed between $x=1$ and $x=2$, and the other between $x=2$ and $x=3$.

This is the 'Aha!' moment. We do not need to grind through heavy integration if we can see the geometry.

### The Calculus Approach

However, to be rigorous, let us look at the integral. The definition of the modulus function tells us that:

$$|x-2| = \begin{cases} -(x-2) & 	ext{if } x < 2 \ x-2 & 	ext{if } x \ge 2 \end{cases}$$

To find the total area $A$, we must split the integral at the vertex $x=2$:

$$A = \int_{1}^{3} |x-2| dx = \int_{1}^{2} (2-x) dx + \int_{2}^{3} (x-2) dx$$

Look at the first integral, $A_1 = \int_{1}^{2} (2-x) dx$. When you evaluate this, you are calculating the area of a triangle with base $1$ (the distance from $1$ to $2$) and height $1$ (the value of $y$ at $x=1$).

Using the formula for the area of a triangle:

$$A_1 = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 1 	imes 1 = 0.5$$

### The Final Synthesis

Now, look at the second integral, $A_2 = \int_{2}^{3} (x-2) dx$. By symmetry, this is identical to the first.

The base is the distance from $2$ to $3$, which is $1$, and the height at $x=3$ is $|3-2| = 1$. Thus, $A_2 = 0.5$.

Adding these together, we get the total area:

$$A = A_1 + A_2 = 0.5 + 0.5 = 1$$

Isn't that elegant? We started with a function that looked intimidating, split it into its logical components, and found that the entire region is just two simple triangles.

You have successfully navigated the modulus trap! Remember, in JEE Advanced, the most complex-looking problems often have the most beautiful, simple geometric souls. Keep looking for that simplicity, and you will conquer any problem they throw at you. The final answer is **1**.

Animated Solution for Mathematics - Definite Integration: The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

Select Answer:

Visualized Solution

Visualizing the Curve $y = ∣ x - 2∣$

Locating the Vertex

Identifying the Boundaries

Splitting the Modulus Function

Setting up the Integral

Evaluating the First Area

Evaluating the Second Area

Summing the Areas

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry of the V-Shape

Visualizing the V-Shape

Defining the Boundaries

The Calculus Approach

The Final Synthesis

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region bounded by the curves y=∣x−2∣,x=1,x=3 and the x-axis is