Question

Find the area (in square units) of the region bounded by the lines \( x=0 \), \( x=\frac{\pi}{2} \), and the curves \( f(x) = \sin x \), \( g(x) = \cos x \):

• A. \( 2(\sqrt{2}-1) \)
• B. \( 2(\sqrt{3}-1) \)
• C. \( 2(\sqrt{2}+1) \)
• D. \( 3\sqrt{2}+1 \)

Hint:

For areas between two curves, always integrate the difference: \[ \int_{a}^{b} [f(x) - g(x)] dx \] and split into regions where their ordering changes.

Answer 1

The Correct Option is A

The area enclosed between the curves \( \sin x \) and \( \cos x \) is determined by: \[ A = \int_0^{\pi/4} (\cos x - \sin x) dx + \int_{\pi/4}^{\pi/2} (\sin x - \cos x) dx \] Computing both integrals and simplifying: \[ A = 2(\sqrt{2} - 1) \] Thus, the correct answer is: \[ 2(\sqrt{2} - 1) \]