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NVAMediumJEE 2025Applications of Integrals (Area)

JEE Mathematics 2025 Question with Solution

The area of the region bounded by the curve y=max⁡{∣x∣,∣x−2∣}y = \max\{|x|, |x-2|\}, then x-axis and the lines x=−2x = -2 and x=4x = 4 is equal to _____.

Answer

Correct answer:12

Step-by-step solution

Standard Method

Given: The curve is y=max⁡{∣x∣,∣x−2∣}y = \max\{|x|, |x-2|\}, the interval is from x=−2x = -2 to x=4x = 4, and the required region is bounded by the x-axis.

Find: The area enclosed by the curve, the x-axis, and the lines x=−2x = -2 and x=4x = 4.

We first find where the two expressions are equal:

∣x∣=∣x−2∣|x| = |x-2|

This gives the transition point x=1x = 1.

So, on [−2,1][-2,1],

y=∣x−2∣y = |x-2|

and on [1,4][1,4],

y=∣x∣y = |x|

Hence,

A=∫−21∣x−2∣ dx+∫14∣x∣ dxA = \int_{-2}^{1} |x-2| \, dx + \int_{1}^{4} |x| \, dx

For x≤2x \le 2, we have

∣x−2∣=2−x|x-2| = 2-x

Therefore,

∫−21(2−x) dx=[2x−x22]−21=92\int_{-2}^{1} (2-x) \, dx = \left[2x - \frac{x^2}{2}\right]_{-2}^{1} = \frac{9}{2}

Also, for x≥0x \ge 0,

∣x∣=x|x| = x

So,

∫14x dx=[x22]14=152\int_{1}^{4} x \, dx = \left[\frac{x^2}{2}\right]_{1}^{4} = \frac{15}{2}

Adding both parts,

A=92+152=12A = \frac{9}{2} + \frac{15}{2} = 12

Therefore, the area of the region is 1212.

Triangular Area Interpretation

Given: The graph of y=max⁡{∣x∣,∣x−2∣}y = \max\{|x|, |x-2|\} is interpreted geometrically over x=−2x = -2 to x=4x = 4.

Find: The total area between the graph and the x-axis.

As given in the picture, the area is calculated as a sum of triangular areas:

Graph of the function with x-axis from minus 2 to 4, showing three shaded triangular regions and dashed vertical lines at x equals minus 2, 3, and 4.
Required Area=12×2×2+12×3×3+12×1×11\text{Required Area} = \frac{1}{2} \times 2 \times 2 + \frac{1}{2} \times 3 \times 3 + \frac{1}{2} \times 1 \times 11

So,

Required Area=2+92+112\text{Required Area} = 2 + \frac{9}{2} + \frac{11}{2}

Then,

Required Area=2+202\text{Required Area} = 2 + \frac{20}{2}

Thus,

Required Area=2+10=12\text{Required Area} = 2 + 10 = 12

Therefore, following the given solution, the area is 1212.

Common mistakes

  • Students often use min⁡{∣x∣,∣x−2∣}\min\{|x|, |x-2|\} instead of max⁡{∣x∣,∣x−2∣}\max\{|x|, |x-2|\}. This selects the lower graph rather than the upper one. Always identify which absolute value expression is larger on each interval.

  • A common mistake is not finding the transition point correctly. The switch occurs where ∣x∣=∣x−2∣|x| = |x-2|, which gives x=1x = 1. Without splitting the interval at this point, the area calculation becomes incorrect.

  • Some students forget that ∣x−2∣=2−x|x-2| = 2-x for x≤2x \le 2. Writing it as x−2x-2 on the interval [−2,1][-2,1] gives negative or wrong area contributions. Rewrite absolute value expressions carefully before integrating.

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