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Integral Calculus: JEE Main Mathematics Question with Solution

The area bounded by the curves

\mathrm{y}=|\mathrm{x}-1|+|\mathrm{x}-2|

and

\mathrm{y}=3

is equal to

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A

3

B

4

C

5

D

6

Step-by-step solutionView

Correct answer

The region bounded by $y=|x-1|+|x-2|$ and $y=3$ is a trapezoid with parallel sides of lengths $4$ and $1$ and height $2$ , yielding an area of $4$ .

Option analysis

Why each option works or fails

A ·

3

Believing that the area bounded is a simple triangle of base $3$ and height $2$ , or omitting the rectangular core between $x=1$ and $x=2$ . Plot the curve by intervals: between $x=1$ and $x=2$ , the function is constant at $y=1$ , so the bounded region is a trapezoid rather than a triangle.

B ·

4

Correct option. Identifying the trapezoidal shape with parallel horizontal boundaries at $y=1$ and $y=3$ . The top boundary has length $3 - (-1) = 4$ , the bottom flat segment has length $2 - 1 = 1$ , and the height is $3 - 1 = 2$ . The area is $\frac{1}{2}(4 + 1)(2) = 5$ minus the offset, or more directly $\frac{1}{2}(a+b)h = \frac{1}{2}(4+1)(2)$ wait: check lengths: $y=|x-1|+|x-2|=3$ . For $x \ge 2$ , $2x-3=3 \implies x=3$ . For $x \le 1$ , $3-2x=3 \implies x=0$ . Base at $y=3$ extends from $x=0$ to $x=3$ , length is $3-0=3$ , not $4$ . Wait! Let's recompute: for $x \le 1$ , $|x-1|+|x-2| = 1-x+2-x = 3-2x$ . Set $3-2x=3 \implies x=0$ . For $x \ge 2$ , $x-1+x-2 = 2x-3=3 \implies x=3$ . Length of top base is $3-0=3$ . Length of bottom base is $2-1=1$ . Height is $3-1=2$ . Area of trapezoid is $\frac{1}{2}(3+1)(2) = 4$ .

C ·

5

Incorrectly solving for the intersection points by taking the outer domain interval as $[-1, 3]$ (length $4$ ) instead of $[0, 3]$ (length $3$ ). Solve $|x-1|+|x-2|=3$ carefully: for $x \le 1$ , $-(x-1)-(x-2)=3-2x=3 \implies x=0$ , not $x=-1$ .

D ·

6

Multiplying the base length $3$ by height $2$ without applying the trapezoid factor, treating the region as a full bounding rectangle. Remember that the side boundaries slope inward from $y=3$ down to $y=1$ , forming a trapezoid of area $\frac{1}{2}(b_1+b_2)h$ , not a rectangle of area $b_1 h$ .

Reviewed route

Solution

StepWorking

01given

The bounding curves are $y = |x - 1| + |x - 2|$ and $y = 3$ .

02goal

Find the area bounded between $y = |x - 1| + |x - 2|$ and the horizontal line $y = 3$ .

03approach

Define $f(x) = |x - 1| + |x - 2|$ piecewise: - For $x < 1$ : $f(x) = -(x - 1) - (x - 2) = 3 - 2x$ - For $1 \le x \le 2$ : $f(x) = (x - 1) - (x - 2) = 1$ - For $x > 2$ : $f(x) = (x - 1) + (x - 2) = 2x - 3$ Find intersection points with $y = 3$ : - $3 - 2x = 3 \implies x = 0$ - $2x - 3 = 3 \implies x = 3$ The enclosed region is an inverted trapezoid with parallel sides along $y = 3$ (from $x = 0$ to $x = 3$ , length $b_1 = 3$ ) and $y = 1$ (from $x = 1$ to $x = 2$ , length $b_2 = 1$ ), and height $h = 3 - 1 = 2$ .

04execute

Calculate the area using the trapezoid formula: $\text{Area} = \frac{1}{2} (b_1 + b_2) \times h = \frac{1}{2}(3 + 1) \times 2 = 4$

05verify

Integrate directly: $\int_{0}^{3} (3 - f(x))\,dx = \int_{0}^{1} (3 - (3 - 2x))\,dx + \int_{1}^{2} (3 - 1)\,dx + \int_{2}^{3} (3 - (2x - 3))\,dx$ $= \left[x^2\right]_0^1 + [2x]_1^2 + \left[6x - x^2\right]_2^3 = 1 + 2 + (9 - 8) = 1 + 2 + 1 = 4.$ Matches geometric calculation.

06given

Functions $y_1 = 3$ and $y_2 = |x - 1| + |x - 2|$ .

07goal

Evaluate $\int_{x_1}^{x_2} (y_1 - y_2)\,dx$ .

08approach

By symmetry around $x = 1.5$ , compute twice the area from $x = 0$ to $x = 1.5$ : $\text{Area} = 2 \left[ \int_{0}^{1} (3 - (3 - 2x))\,dx + \int_{1}^{1.5} (3 - 1)\,dx \right]$

09execute

$2 \left[ \int_{0}^{1} 2x\,dx + \int_{1}^{1.5} 2\,dx \right] = 2 \left[ \left[x^2\right]_0^1 + 2(1.5 - 1) \right] = 2 [1 + 1] = 4$

✓verify

Check at endpoints: $y(0) = 1 + 2 = 3$ and $y(3) = 2 + 1 = 3$ , so the integration limits are strictly bounded.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is the graph flat between x = 1 and x = 2?

For any point between 1 and 2, the sum of distances $|x - 1| + |x - 2|$ is constant and equal to the distance between 1 and 2, which is $2 - 1 = 1$ .

Answer

The region bounded by $y=|x-1|+|x-2|$ and $y=3$ is a trapezoid with parallel sides of lengths $4$ and $1$ and height $2$ , yielding an area of $4$ .

Why each option works or fails

A: $3$ - Believing that the area bounded is a simple triangle of base $3$ and height $2$ , or omitting the rectangular core between $x=1$ and $x=2$ . Plot the curve by intervals: between $x=1$ and $x=2$ , the function is constant at $y=1$ , so the bounded region is a trapezoid rather than a triangle.
B · correct: $4$ - Correct option. Identifying the trapezoidal shape with parallel horizontal boundaries at $y=1$ and $y=3$ . The top boundary has length $3 - (-1) = 4$ , the bottom flat segment has length $2 - 1 = 1$ , and the height is $3 - 1 = 2$ . The area is $\frac{1}{2}(4 + 1)(2) = 5$ minus the offset, or more directly $\frac{1}{2}(a+b)h = \frac{1}{2}(4+1)(2)$ wait: check lengths: $y=|x-1|+|x-2|=3$ . For $x \ge 2$ , $2x-3=3 \implies x=3$ . For $x \le 1$ , $3-2x=3 \implies x=0$ . Base at $y=3$ extends from $x=0$ to $x=3$ , length is $3-0=3$ , not $4$ . Wait! Let's recompute: for $x \le 1$ , $|x-1|+|x-2| = 1-x+2-x = 3-2x$ . Set $3-2x=3 \implies x=0$ . For $x \ge 2$ , $x-1+x-2 = 2x-3=3 \implies x=3$ . Length of top base is $3-0=3$ . Length of bottom base is $2-1=1$ . Height is $3-1=2$ . Area of trapezoid is $\frac{1}{2}(3+1)(2) = 4$ .
C: $5$ - Incorrectly solving for the intersection points by taking the outer domain interval as $[-1, 3]$ (length $4$ ) instead of $[0, 3]$ (length $3$ ). Solve $|x-1|+|x-2|=3$ carefully: for $x \le 1$ , $-(x-1)-(x-2)=3-2x=3 \implies x=0$ , not $x=-1$ .
D: $6$ - Multiplying the base length $3$ by height $2$ without applying the trapezoid factor, treating the region as a full bounding rectangle. Remember that the side boundaries slope inward from $y=3$ down to $y=1$ , forming a trapezoid of area $\frac{1}{2}(b_1+b_2)h$ , not a rectangle of area $b_1 h$ .

Step-by-step solution

given: The bounding curves are $y = |x - 1| + |x - 2|$ and $y = 3$ .
goal: Find the area bounded between $y = |x - 1| + |x - 2|$ and the horizontal line $y = 3$ .
approach: Define $f(x) = |x - 1| + |x - 2|$ piecewise: - For $x < 1$ : $f(x) = -(x - 1) - (x - 2) = 3 - 2x$ - For $1 \le x \le 2$ : $f(x) = (x - 1) - (x - 2) = 1$ - For $x > 2$ : $f(x) = (x - 1) + (x - 2) = 2x - 3$ Find intersection points with $y = 3$ : - $3 - 2x = 3 \implies x = 0$ - $2x - 3 = 3 \implies x = 3$ The enclosed region is an inverted trapezoid with parallel sides along $y = 3$ (from $x = 0$ to $x = 3$ , length $b_1 = 3$ ) and $y = 1$ (from $x = 1$ to $x = 2$ , length $b_2 = 1$ ), and height $h = 3 - 1 = 2$ .
execute: Calculate the area using the trapezoid formula: $\text{Area} = \frac{1}{2} (b_1 + b_2) \times h = \frac{1}{2}(3 + 1) \times 2 = 4$
verify: Integrate directly: $\int_{0}^{3} (3 - f(x))\,dx = \int_{0}^{1} (3 - (3 - 2x))\,dx + \int_{1}^{2} (3 - 1)\,dx + \int_{2}^{3} (3 - (2x - 3))\,dx$ $= \left[x^2\right]_0^1 + [2x]_1^2 + \left[6x - x^2\right]_2^3 = 1 + 2 + (9 - 8) = 1 + 2 + 1 = 4.$ Matches geometric calculation.
given: Functions $y_1 = 3$ and $y_2 = |x - 1| + |x - 2|$ .
goal: Evaluate $\int_{x_1}^{x_2} (y_1 - y_2)\,dx$ .
approach: By symmetry around $x = 1.5$ , compute twice the area from $x = 0$ to $x = 1.5$ : $\text{Area} = 2 \left[ \int_{0}^{1} (3 - (3 - 2x))\,dx + \int_{1}^{1.5} (3 - 1)\,dx \right]$
execute: $2 \left[ \int_{0}^{1} 2x\,dx + \int_{1}^{1.5} 2\,dx \right] = 2 \left[ \left[x^2\right]_0^1 + 2(1.5 - 1) \right] = 2 [1 + 1] = 4$
verify: Check at endpoints: $y(0) = 1 + 2 = 3$ and $y(3) = 2 + 1 = 3$ , so the integration limits are strictly bounded.