The area of the region bounded by the curves $y = |x-1|$ and $y = 3 - |x|$ is Options: (A) 6 sq. units (B) 2 sq. units (C) 3 sq. units (D) 4 sq. units

### Analyzing the Setup The problem asks for the area bounded by the curves $y = |x-1|$ and $y = 3 - |x|$. These functions represent two V-shaped graphs interacting on the Cartesian plane. The first curve, $y = |x-1|$, is an upright V-shape with its vertex at $(1, 0)$. The second curve, $y = 3 - |x|$, is an inverted V-shape with its vertex at $(0, 3)$. ### Finding the Intersection Points To determine the boundaries of the region, we set the functions equal to each other: $$|x-1| = 3 - |x|$$ We analyze this across different intervals: For $x < 0$: $$-(x-1) = 3 - (-x) \Rightarrow 1 - x = 3 + x \Rightarrow 2x = -2 \Rightarrow x = -1$$ At $x = -1$, the height is $y = 2$. For $x \ge 1$: $$(x-1) = 3 - x \Rightarrow 2x = 4 \Rightarrow x = 2$$ At $x = 2$, the height is $y = 1$. The region of interest is trapped between $x = -1$ and $x = 2$. ### The Calculus of Slicing The area is defined by the integral of the upper curve minus the lower curve: $$Area = \int_{-1}^{2} (y_{upper} - y_{lower}) dx$$ Because the modulus functions change behavior at $x = 0$ and $x = 1$, we split the integral into three segments: **Segment 1 ($x \in [-1, 0]$):** $$A_1 = \int_{-1}^{0} ((3+x) - (1-x)) dx = \int_{-1}^{0} (2+2x) dx = [2x + x^2]_{-1}^{0} = 0 - (-2 + 1) = 1$$ **Segment 2 ($x \in [0, 1]$):** $$A_2 = \int_{0}^{1} ((3-x) - (1-x)) dx = \int_{0}^{1} 2 dx = [2x]_{0}^{1} = 2$$ **Segment 3 ($x \in [1, 2]$):** $$A_3 = \int_{1}^{2} ((3-x) - (x-1)) dx = \int_{1}^{2} (4-2x) dx = [4x - x^2]_{1}^{2} = (8-4) - (4-1) = 1$$ ### Final Calculation Summing the areas of these three segments, we obtain the total area: $$Area = 1 + 2 + 1 = 4$$ The total area bounded by the two curves is **4 square units**.

JEE Main 2003

LEVELJEE Main

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

Q: Wait, why do we have to split the integral into three parts? Can't we just integrate from -1 to 2 in one go?

That is a fantastic question! The reason we cannot integrate in one go is that the algebraic definition of the functions changes at the critical points x = 0 and x = 1. The modulus function is not a single polynomial; it is a piecewise function. If you try to integrate without splitting, you are essentially ignoring the fact that the 'slope' of the function changes at these points. By splitting, we ensure we are using the correct linear expression for every segment of the curve.

Q: I am confused about the negative sign in 3 - |x|. Does it flip the graph?

Yes, absolutely! Think of the standard modulus function y = |x| as a V-shape opening upwards. When you multiply it by -1, it becomes y = -|x|, which is an inverted V-shape opening downwards. Adding 3 to it, y = 3 - |x|, simply shifts that inverted V-shape upwards by 3 units. It is a beautiful transformation of the parent function!

Q: Is there a faster way to solve this without integration? The exam is time-constrained!

You have a sharp eye! Yes, there is. If you plot the points carefully, you will realize that the bounded region is actually a rectangle (or a combination of geometric shapes). By calculating the distance between the intersection points and the vertices, you can find the side lengths and use the area formula for a rectangle. However, integration is the universal tool that works even when the shapes are not perfect polygons, so mastering the integral method is crucial for JEE Advanced.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the two curves. One is a standard V-shape with its vertex at (1, 0), and the other is an inverted V-shape with its vertex at (0, 3). The region bounded by them is the space trapped between these two 'V's. Can you identify the intersection points by solving the equation |x-1| = 3 - |x|?

Hint 2

Hint 2: The Arsenal. Remember that the modulus function |x| changes its definition at x = 0, and |x-1| changes at x = 1. To solve the integral, you must partition the x-axis into intervals based on these critical points: (-infinity, 0), [0, 1), and [1, infinity). This allows you to remove the modulus bars and work with simple linear expressions.

Hint 3

Hint 3: The Execution. The area is the integral of (Upper Curve - Lower Curve) dx. Set up your integral as the sum of three parts: the integral from -1 to 0, 0 to 1, and 1 to 2. For the first part, the upper curve is 3 - (-x) and the lower is -(x-1). Proceed with the integration carefully.

Visualized Solution

Visualizing the Curves

Given curves: $y = ∣ x - 1∣$ and $y = 3 - ∣ x ∣$
We need to find the area bounded by these two curves.

Graphing $y = ∣ x - 1∣$

$y = ∣ x - 1∣$ is a standard V-shaped curve.
The vertex is at $x - 1 = 0 ⟹ x = 1$ .
Vertex coordinates: $(1, 0)$ .

Graphing $y = 3 - ∣ x ∣$

$y = 3 - ∣ x ∣$ involves a negative modulus, so it's an inverted V-shape.
The vertex is at $x = 0$ .
Maximum value is $y = 3$ , so vertex is at $(0, 3)$ .

Finding Intersection Points

To find where the curves meet, equate them:
$∣ x - 1∣ = 3 - ∣ x ∣$
We need to solve this equation for different intervals of $x$ .

Solving for $x < 0$

For $x < 0$ , both $∣ x - 1∣$ and $∣ x ∣$ open with a negative sign.
$- (x - 1) = 3 - (- x)$
$1 - x = 3 + x ⟹ 2 x = - 2 ⟹ x = - 1$
At $x = - 1$ , $y = 3 - ∣ - 1∣ = 2$ . Point: $(- 1, 2)$ .

Solving for $x \geq 1$

For $x \geq 1$ , both $∣ x - 1∣$ and $∣ x ∣$ open with a positive sign.
$(x - 1) = 3 - (x)$
$x - 1 = 3 - x ⟹ 2 x = 4 ⟹ x = 2$
At $x = 2$ , $y = 3 - ∣2∣ = 1$ . Point: $(2, 1)$ .

The Bounded Region

The required area is enclosed between $x = - 1$ and $x = 2$ .
Area $= \int_{- 1}^{2} (y_{u pp er} - y_{l o w er}) d x$
Upper curve: $y = 3 - ∣ x ∣$
Lower curve: $y = ∣ x - 1∣$

Splitting the Integral

Modulus functions change behavior at their critical points.
Critical points in $[- 1, 2]$ are $x = 0$ and $x = 1$ .
We must split the integral at these points:
$\int_{- 1}^{0} \dots d x + \int_{0}^{1} \dots d x + \int_{1}^{2} \dots d x$

Formulating the Integrals

$A_{1} = \int_{- 1}^{0} ((3 + x) - (1 - x)) d x$
$A_{2} = \int_{0}^{1} ((3 - x) - (1 - x)) d x$
$A_{3} = \int_{1}^{2} ((3 - x) - (x - 1)) d x$

Evaluating First Segment

$A_{1} = \int_{- 1}^{0} (2 + 2 x) d x$
$A_{1} = [2 x + x^{2}]_{- 1}^{0}$
$A_{1} = 0 - (- 2 + 1) = 1$ sq. unit

Evaluating Second Segment

$A_{2} = \int_{0}^{1} ((3 - x) - (1 - x)) d x$
$A_{2} = \int_{0}^{1} 2 d x$
$A_{2} = [2 x]_{0}^{1} = 2 (1) - 2 (0) = 2$ sq. units

Evaluating Third Segment

$A_{3} = \int_{1}^{2} ((3 - x) - (x - 1)) d x$
$A_{3} = \int_{1}^{2} (4 - 2 x) d x$
$A_{3} = [4 x - x^{2}]_{1}^{2} = (8 - 4) - (4 - 1) = 1$ sq. unit

Total Area

Total Area $= A_{1} + A_{2} + A_{3}$
Total Area $= 1 + 2 + 1 = 4$ sq. units
Pro Tip: The bounded region is actually a rectangle with side lengths $2$ and $22$ . Area $= 2 \times 22 = 4$ .

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Setup

The problem asks for the area bounded by the curves

y = ∣ x - 1∣

and

y = 3 - ∣ x ∣

. These functions represent two V-shaped graphs interacting on the Cartesian plane.

The first curve,

y = ∣ x - 1∣

, is an upright V-shape with its vertex at

(1, 0)

. The second curve,

y = 3 - ∣ x ∣

, is an inverted V-shape with its vertex at

(0, 3)

Finding the Intersection Points

To determine the boundaries of the region, we set the functions equal to each other:

∣ x - 1∣ = 3 - ∣ x ∣

We analyze this across different intervals:

For

x < 0

- (x - 1) = 3 - (- x) \Rightarrow 1 - x = 3 + x \Rightarrow 2 x = - 2 \Rightarrow x = - 1

x = - 1

, the height is

y = 2

For

x \geq 1

(x - 1) = 3 - x \Rightarrow 2 x = 4 \Rightarrow x = 2

x = 2

, the height is

y = 1

The region of interest is trapped between

x = - 1

and

x = 2

The Calculus of Slicing

The area is defined by the integral of the upper curve minus the lower curve:

A r e a = \int_{- 1}^{2} (y_{u pp er} - y_{l o w er}) d x

Because the modulus functions change behavior at

x = 0

and

x = 1

, we split the integral into three segments:

Segment 1 ( $x \in [- 1, 0]$ ):

A_{1} = \int_{- 1}^{0} ((3 + x) - (1 - x)) d x = \int_{- 1}^{0} (2 + 2 x) d x = [2 x + x^{2}]_{- 1}^{0} = 0 - (- 2 + 1) = 1

Segment 2 ( $x \in [0, 1]$ ):

A_{2} = \int_{0}^{1} ((3 - x) - (1 - x)) d x = \int_{0}^{1} 2 d x = [2 x]_{0}^{1} = 2

Segment 3 ( $x \in [1, 2]$ ):

A_{3} = \int_{1}^{2} ((3 - x) - (x - 1)) d x = \int_{1}^{2} (4 - 2 x) d x = [4 x - x^{2}]_{1}^{2} = (8 - 4) - (4 - 1) = 1

Final Calculation

Summing the areas of these three segments, we obtain the total area:

A r e a = 1 + 2 + 1 = 4

The total area bounded by the two curves is 4 square units.

Common Traps & FAQs

Wait, why do we have to split the integral into three parts? Can't we just integrate from -1 to 2 in one go?

That is a fantastic question! The reason we cannot integrate in one go is that the algebraic definition of the functions changes at the critical points x = 0 and x = 1. The modulus function is not a single polynomial; it is a piecewise function. If you try to integrate without splitting, you are essentially ignoring the fact that the 'slope' of the function changes at these points. By splitting, we ensure we are using the correct linear expression for every segment of the curve.

I am confused about the negative sign in 3 - |x|. Does it flip the graph?

Yes, absolutely! Think of the standard modulus function y = |x| as a V-shape opening upwards. When you multiply it by -1, it becomes y = -|x|, which is an inverted V-shape opening downwards. Adding 3 to it, y = 3 - |x|, simply shifts that inverted V-shape upwards by 3 units. It is a beautiful transformation of the parent function!

Is there a faster way to solve this without integration? The exam is time-constrained!

You have a sharp eye! Yes, there is. If you plot the points carefully, you will realize that the bounded region is actually a rectangle (or a combination of geometric shapes). By calculating the distance between the intersection points and the vertices, you can find the side lengths and use the area formula for a rectangle. However, integration is the universal tool that works even when the shapes are not perfect polygons, so mastering the integral method is crucial for JEE Advanced.

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

Select Answer:

Visualized Solution

Visualizing the Curves

Graphing $y = ∣ x - 1∣$

Graphing $y = 3 - ∣ x ∣$

Finding Intersection Points

Solving for $x < 0$

Solving for $x \geq 1$

The Bounded Region

Splitting the Integral

Formulating the Integrals

Evaluating First Segment

Evaluating Second Segment

Evaluating Third Segment

Total Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Intersection Points

The Calculus of Slicing

Final Calculation

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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−1∣ and y=3−∣x∣ is

Select Answer:

Visualized Solution

Visualizing the Curves

Graphing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=∣x−1∣

Graphing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=3−∣x∣

Finding Intersection Points

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x<0

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Splitting the Integral

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Evaluating First Segment

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Evaluating Third Segment

Total Area

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Finding the Intersection Points

The Calculus of Slicing

Final Calculation

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