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NVAMediumJEE 2023Applications of Integrals (Area)

JEE Mathematics 2023 Question with Solution

Let AA be the area of the region {(x,y):y≥x2,y≥(1−x)2,y≤2x(1−x)}\{ (x, y) : y \geq x^2, y \geq (1 - x)^2, y \leq 2x(1 - x) \} .Then 540A540A is equal to .....

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: The region is defined by

{(x,y):y≥x2,  y≥(1−x)2,  y≤2x(1−x)}\{(x,y): y \ge x^2,\; y \ge (1-x)^2,\; y \le 2x(1-x)\}

Find: The value of 540A540A, where AA is the enclosed area.

Graph of the parabolas y equals x squared, y equals one minus x whole squared, and y equals 2x(1 minus x), with the shaded common region between x equals one third and x equals two thirds.

The lower boundary is the upper envelope of x2x^2 and (1−x)2(1-x)^2. By symmetry, the area can be written as

A=2∫1/22/3(2x(1−x)−x2) dxA = 2\int_{1/2}^{2/3} \left(2x(1-x) - x^2\right) \, dx

Now simplify the integrand:

2x(1−x)−x2=2x−2x2−x2=2x−3x22x(1-x) - x^2 = 2x - 2x^2 - x^2 = 2x - 3x^2

So,

A=2∫1/22/3(2x−3x2) dxA = 2\int_{1/2}^{2/3} (2x - 3x^2) \, dx

Integrating,

A=2[x2−x3]1/22/3A = 2\left[ x^2 - x^3 \right]_{1/2}^{2/3}

Evaluate the bracket:

[x2−x3]1/22/3=(49−827)−(14−18)\left[ x^2 - x^3 \right]_{1/2}^{2/3} = \left(\frac{4}{9} - \frac{8}{27}\right) - \left(\frac{1}{4} - \frac{1}{8}\right) =427−18=32−27216=5216= \frac{4}{27} - \frac{1}{8} = \frac{32 - 27}{216} = \frac{5}{216}

Therefore,

A=2×5216=5108A = 2 \times \frac{5}{216} = \frac{5}{108}

Hence,

540A=540×5108=25540A = 540 \times \frac{5}{108} = 25

Therefore, the required value is 2525.

Piecewise Boundary Method

Given: The same bounded region with upper curve y=2x(1−x)y = 2x(1-x) and lower boundary equal to max⁡{x2,(1−x)2}\max\{x^2,(1-x)^2\}.

Find: 540A540A.

First find where the top curve meets the lower curves. For y=x2y = x^2 and y=2x(1−x)y = 2x(1-x),

x2=2x(1−x)x^2 = 2x(1-x) 3x2−2x=03x^2 - 2x = 0 x(3x−2)=0x(3x-2)=0

So x=0x=0 or x=23x=\frac{2}{3}.

For y=(1−x)2y = (1-x)^2 and y=2x(1−x)y = 2x(1-x),

(1−x)2=2x(1−x)(1-x)^2 = 2x(1-x)

This gives intersections at x=13x=\frac{1}{3} and x=1x=1. The bounded common region exists for

13≤x≤23\frac{1}{3} \le x \le \frac{2}{3}

Now the lower boundary changes at x=12x=\frac{1}{2}:

  • for x∈[13,12]x \in \left[\frac{1}{3},\frac{1}{2}\right], lower curve is (1−x)2(1-x)^2
  • for x∈[12,23]x \in \left[\frac{1}{2},\frac{2}{3}\right], lower curve is x2x^2

So,

A=∫1/31/2(2x(1−x)−(1−x)2) dx+∫1/22/3(2x(1−x)−x2) dxA = \int_{1/3}^{1/2} \left(2x(1-x) - (1-x)^2\right) \, dx + \int_{1/2}^{2/3} \left(2x(1-x) - x^2\right) \, dx

For the first integral,

2x(1−x)−(1−x)2=(2x−2x2)−(1−2x+x2)=4x−3x2−12x(1-x) - (1-x)^2 = (2x - 2x^2) - (1 - 2x + x^2) = 4x - 3x^2 - 1

Hence,

A1=∫1/31/2(4x−3x2−1) dxA_1 = \int_{1/3}^{1/2} (4x - 3x^2 - 1) \, dx A1=[2x2−x3−x]1/31/2=5216A_1 = \left[2x^2 - x^3 - x\right]_{1/3}^{1/2} = \frac{5}{216}

For the second integral,

2x(1−x)−x2=2x−3x22x(1-x) - x^2 = 2x - 3x^2

Hence,

A2=∫1/22/3(2x−3x2) dxA_2 = \int_{1/2}^{2/3} (2x - 3x^2) \, dx A2=[x2−x3]1/22/3=5216A_2 = \left[x^2 - x^3\right]_{1/2}^{2/3} = \frac{5}{216}

Therefore,

A=A1+A2=5216+5216=5108A = A_1 + A_2 = \frac{5}{216} + \frac{5}{216} = \frac{5}{108}

Thus,

540A=540⋅5108=25540A = 540 \cdot \frac{5}{108} = 25

Therefore, the required value is 2525.

Common mistakes

  • Taking the lower boundary as x2x^2 or (1−x)2(1-x)^2 throughout the interval is incorrect. The condition y≥x2y \ge x^2 and y≥(1−x)2y \ge (1-x)^2 means the lower boundary is max⁡{x2,(1−x)2}\max\{x^2,(1-x)^2\}, which changes at x=12x=\frac{1}{2}.

  • Using the wrong interval of integration is a common error. The common region exists only where 2x(1−x)≥max⁡{x2,(1−x)2}2x(1-x) \ge \max\{x^2,(1-x)^2\}, which leads to x∈[13,23]x \in \left[\frac{1}{3},\frac{2}{3}\right].

  • While using symmetry, students sometimes double the wrong half-interval. Symmetry is valid here, but it must be applied to an interval such as [12,23]\left[\frac{1}{2},\frac{2}{3}\right] where the lower curve is clearly x2x^2. Otherwise the integrand becomes incorrect.

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