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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Area Under Curves

The area, enclosed by the curves

y = sin x + cos x

and

y = ∣ cos x - sin x ∣

and the lines

x = 0, x = \frac{π}{2}

, is:

(A) $22 (2 + 1)$
(B) $22 (2 - 1)$ ✓ Correct
(C) $2 (2 + 1)$
(D) $4 (2 - 1)$

Answer: (B)

Solution

In

[0, π /4]

,

cos x > sin x

, so

∣ cos x - sin x ∣ = cos x - sin x

.

In

[π /4, π /2]

,

sin x > cos x

, so

∣ cos x - sin x ∣ = sin x - cos x

.

The area is:

A = \int_{0}^{π /4} [(sin x + cos x) - (cos x - sin x)] d x + \int_{π /4}^{π /2} [(sin x + cos x) - (sin x - cos x)] d x

= \int_{0}^{π /4} 2 sin x d x + \int_{π /4}^{π /2} 2 cos x d x

= 2 [- cos x]_{0}^{π /4} + 2 [sin x]_{π /4}^{π /2}

= 2 (- \frac{1}{2} + 1) + 2 (1 - \frac{1}{2})

= 4 (1 - \frac{1}{2}) = 4 \cdot \frac{2 - 1}{2} = 22 (2 - 1)

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option

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