Question:

<div class='solution'> <p>The enclosed region between the two curves and the x-axis:</p> <p>On \([0,\pi/4]\): sin x ≤ cos x, both above x-axis. On \([\pi/4,\pi/2]\): cos x ≤ sin x.</p> <p>Area = \(\int_0^{\pi/4}\cos x\,dx + \int_{\pi/4}^{\pi/2}\sin x\,dx - \int_0^{\pi/4}\sin x\,dx - \int_{\pi/4}^{\pi/2}\cos x\,dx\)</p> <p>Wait — this double-counts. The area between the two curves and x-axis is the area under the "outer envelope":</p> <p>\(A=\int_0^{\pi/4}\cos x\,dx+\int_{\pi/4}^{\pi/2}\sin x\,dx=[\sin x]_0^{\pi/4}+[-\cos x]_{\pi/4}^{\pi/2}\)</p> <p>\(=\frac{1}{\sqrt2}+(0+\frac{1}{\sqrt2})=\frac{2}{\sqrt2}=\sqrt{2}\).</p> <p>But that's the area under each branch — need to subtract the triangle between them. Area between the two curves (above x-axis):</p> <p>\(\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx+\text{area of lower curve}\). Final answer \(=2-\sqrt{2}\).</p> </div>