The area, enclosed by the curves y=sin x+cos x and y=|cos x-sin x| and the lines x=0,x=π/2, is :

Solution

Maths sir · 4 steps

Split at x=π/4 where the modulus switches sign; the two halves turn out equal.

1. Find the top curve. On [0,π/2] both sin x and cos x are non-negative, so sin x+cos x ≥ |cos x-sin x|. The top curve is always y=sin x+cos x.
2. Open the modulus. For 0≤ x≤π/4, cos x≥sin x, so the gap is (sin x+cos x)-(cos x-sin x)=2 sin x. For π/4≤ x≤π/2, sin x≥cos x, so the gap is (sin x+cos x)-(sin x-cos x)=2 cos x.

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

3. Evaluate. The first integral is 2(1-1/√2)=2-√2. The second is 2(1-1/√2)=2-√2 as well, which you expect, because replacing x by π/2-x swaps sin x and cos x.
4. So. A=2(2-√2)=4-2√2=2√2(√2-1), which is option (a).

The idea to keep. With a modulus inside the area, first find where the expression inside changes sign and split the integral there. Then check which curve is on top in each piece before subtracting.

Answer from the NTA final answer key. Solution by PrepWiser.