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JEE Main 2014

JEE Main 2014 Mathematics question on Definite Integrals and Areas

The area enclosed by the curves y=sin⁡x+cos⁡xy = \sin x + \cos x and y=∣cos⁡x−sin⁡x∣y = |\cos x - \sin x| over the interval [0,π2]\left[0, \frac{\pi}{2}\right] is
  1. A.
    4(2−1)4(\sqrt{2} - 1)
  2. B.
    22(2−1)2\sqrt{2}(\sqrt{2} - 1)
  3. C.
    2(2+1)2(\sqrt{2} + 1)
  4. D.
    22(2+1)2\sqrt{2}(\sqrt{2} + 1)
Show answer and solution

Correct option: B

Solution

On [0, π/2], sin x + cos x ≥ 0 always. For |cos x - sin x|: it equals cos x - sin x on [0, π/4] and sin x - cos x on [π/4, π/2].

The upper curve is always y = sin x + cos x (since sin x + cos x ≥ |cos x - sin x|).

Area = ∫₀^(π/4) [(sin x + cos x) - (cos x - sin x)] dx + ∫_(π/4)^(π/2) [(sin x + cos x) - (sin x - cos x)] dx

= ∫₀^(π/4) 2 sin x dx + ∫_(π/4)^(π/2) 2 cos x dx

= 2[-cos x]₀^(π/4) + 2[sin x]_(π/4)^(π/2)

= 2(-cos(π/4) + cos 0) + 2(sin(π/2) - sin(π/4))

= 2(1 - 1/√2) + 2(1 - 1/√2)

= 4(1 - 1/√2) = 4(√2 - 1)/√2 = 2√2(√2 - 1).

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