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Question:

The area enclosed by the curves y = sinx + cosx and y = |cosx - sinx| over the interval [0, π/2] is

2√2(√2 - 1)

2(√2 + 1)

4(√2 - 1)

2√2(√2 + 1)

Solution:

cosx > sinx, ∀x ∈ (0, π/4), and cosx < sinx, ∀x ∈ (π/4, π/2)
y1 = sinx + cosx
y2 = |cosx - sinx|
Area = ∫₀^(π/4) (sinx + cosx - (cosx - sinx))dx + ∫(π/4)^(π/2) (sinx + cosx - (sinx - cosx))dx
= ∫₀^(π/4) 2sinx dx + ∫
(π/4)^(π/2) 2cosx dx
= [-2cosx]₀^(π/4) + [2sinx]_(π/4)^(π/2)
= -2(cos(π/4) - cos0) + 2(sin(π/2) - sin(π/4))
= -2(√2/2 - 1) + 2(1 - √2/2)
= -√2 + 2 + 2 - √2
= 4 - 2√2
= 2(2 - √2)
= 2√2(√2 - 1)