Question

The area enclosed between the curve \( y = \sin x, y = \cos x \), \(\text{ for }\) \( 0 \leq x \leq \frac{\pi}{2} \) \(\text{ is:}\)

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

Hint:

When calculating the area between curves, split the integral where the curves intersect or change dominance.

Answer 1

The Correct Option is D

We are given the curves \( y = \sin x \) and \( y = \cos x \) for the interval \( 0 \leq x \leq \frac{\pi}{2} \). We need to find the area enclosed between these two curves.

Step 1: Set up the integral for the area.
The area between two curves is given by: \[ \text{Area} = \int_{0}^{\frac{\pi}{2}} \left| \sin x - \cos x \right| dx \] Since \( \sin x \) is greater than \( \cos x \) for \( 0 \leq x \leq \frac{\pi}{4} \) and \( \cos x \) is greater for \( \frac{\pi}{4} \leq x \leq \frac{\pi}{2} \), we split the integral as follows: \[ \text{Area} = \int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) dx \]

Step 2: Evaluate the integrals.
The first integral is: \[ \int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) dx = \left[ -\cos x - \sin x \right]_{0}^{\frac{\pi}{4}} = (-\cos \frac{\pi}{4} - \sin \frac{\pi}{4}) - (-\cos 0 - \sin 0) = -\sqrt{2}/2 - \sqrt{2}/2 + 1 = 1 - \sqrt{2} \] The second integral is: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) dx = \left[ \sin x + \cos x \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = (\sin \frac{\pi}{2} + \cos \frac{\pi}{2}) - (\sin \frac{\pi}{4} + \cos \frac{\pi}{4}) = 1 + 0 - \sqrt{2}/2 - \sqrt{2}/2 = 1 - \sqrt{2} \]

Step 3: Add the results.
Adding both parts of the integral gives the total area: \[ \text{Area} = (1 - \sqrt{2}) + (1 - \sqrt{2}) = 2(\sqrt{2} - 1) \] Thus, the correct answer is \( 2(\sqrt{2} - 1) \), corresponding to option (d).