Question

Find area bounded by the curves \(y\) = \(max\{sin x, cos x\}\) and \(x-axis\) between \(x = -  \pi  \)and \(x =   \pi\)

• A. \(2 + \sqrt{2}\)
• B. \(\sqrt{2}\)
• C. \(1 + \sqrt{2}\)
• D. \(2\sqrt{2}\)

Answer 1

The Correct Option is D

To find the area bounded by the curves \(y = \max\{\sin x, \cos x\}\) and the x-axis between \(x = -\pi\) and \(x = \pi\), we need to analyze where each function \(\sin x\) and \(\cos x\) dominates.

Step 1: Analyze the functions \(\sin x\) and \(\cos x\)

The functions \(\sin x\) and \(\cos x\) have the following important points in one period from \(-\pi\) to \(\pi\):

• \(x = -\pi\): \(\sin(-\pi) = 0\) and \(\cos(-\pi) = -1\)
• \(x = -\frac{\pi}{4}\): \(\sin(-\frac{\pi}{4}) = -\frac{\sqrt{2}}{2}\) and \(\cos(-\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)
• \(x = 0\): \(\sin(0) = 0\) and \(\cos(0) = 1\)
• \(x = \frac{\pi}{4}\): \(\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\) and \(\cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}\)
• \(x = \frac{\pi}{2}\): \(\sin(\frac{\pi}{2}) = 1\) and \(\cos(\frac{\pi}{2}) = 0\)
• \(x = \pi\): \(\sin(\pi) = 0\) and \(\cos(\pi) = -1\)

From these values, you can see:

• From \(x = -\pi\) to \(x = -\frac{\pi}{4}\), \(\cos x\) dominates.
• From \(x = -\frac{\pi}{4}\) to \(x = \frac{\pi}{4}\), both functions are equal; thus either can be used.
• From \(x = \frac{\pi}{4}\) to \(x = \frac{3\pi}{4}\), \(\sin x\) dominates.
• From \(x = \frac{3\pi}{4}\) to \(x = \pi\), \(\cos x\) dominates.

Step 2: Calculate the area using integration

• Area under \(\cos x\) from \(x = -\pi\) to \(x = -\frac{\pi}{4}\): \(\int_{-\pi}^{-\frac{\pi}{4}} \cos x \, dx = \left[\sin x\right]_{-\pi}^{-\frac{\pi}{4}} = \sin\left(-\frac{\pi}{4}\right) - \sin(-\pi) = -\frac{\sqrt{2}}{2} - 0 = -\frac{\sqrt{2}}{2}\)
• Area shared \((x = -\frac{\pi}{4} \text{ to } x = \frac{\pi}{4})\): \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos x \, dx = [\sin x]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \sin\left(\frac{\pi}{4}\right) - \sin\left(-\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right)= \sqrt{2}\)
• Area under \(\sin x\) from \(x = \frac{\pi}{4}\) to \(x = \frac{3\pi}{4}\): \(\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \sin x \, dx = [-\cos x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}} = -\cos\left(\frac{3\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}\)
• Area under \(\cos x\) from \(x = \frac{3\pi}{4}\) to \(x = \pi\): \(\int_{\frac{3\pi}{4}}^{\pi} \cos x \, dx = [\sin x]_{\frac{3\pi}{4}}^{\pi} = \sin(\pi) - \sin\left(\frac{3\pi}{4}\right) = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}\)

Step 3: Add the areas to find the total area

The total area bounded by the curves and the x-axis is calculated by summing these areas:

Total Area = \(-\frac{\sqrt{2}}{2} + \sqrt{2} + \sqrt{2} - \frac{\sqrt{2}}{2} = 2\sqrt{2}\)

Hence, the correct answer is \(\boxed{2\sqrt{2}}\).