Question:

The area enclosed by the curves $y=\sin x+\cos x$ and $y=|\cos x-\sin x|$ over the interval $\left[0,\dfrac{\pi}{2}\right]$ is

(A) $4(\sqrt{2}-1)$

(B) $2\sqrt{2}(\sqrt{2}-1)$

(D) $2(\sqrt{2}+1)$

Step-by-Step Solution

Key Concept: On $[0,\pi/4]$: $|\cos x-\sin x|=\cos x-\sin x$. On $[\pi/4,\pi/2]$: $=\sin x-\cos x$. Compute two integrals.

Area $=2\sqrt{2}(\sqrt{2}-1)$.

Correct Answer: (B) $2\sqrt{2}(\sqrt{2}-1)$