Find the area of the region bounded by the curves f(x) = sin(x), g(x) = cos(x), x = 0, and x = π/2.

Steps to Solve

1. Identify the curves and boundaries

The curves are ( f(x) = \sin x ) and ( g(x) = \cos x ). The vertical lines are at ( x = 0 ) and ( x = \frac{\pi}{2} ).

2. Determine the area to be calculated

The area between the curves from ( x = 0 ) to ( x = \frac{\pi}{2} ) is given by the integral: $$ A = \int_{0}^{\frac{\pi}{2}} (f(x) - g(x)) , dx $$

3. Find the points of intersection

Set ( \sin x = \cos x ) to find the intersection points within the interval. This occurs at: $$ x = \frac{\pi}{4} $$

4. Set up the integral

Since ( \sin x ) is above ( \cos x ) in the interval ( \left[0, \frac{\pi}{4}\right] ) and below in ( \left[\frac{\pi}{4}, \frac{\pi}{2}\right] ), we calculate the area as: $$ A = \int_{0}^{\frac{\pi}{4}} (\sin x - \cos x) , dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\cos x - \sin x) , dx $$

5. Calculate the first integral

Calculate the integral from ( 0 ) to ( \frac{\pi}{4} ): $$ \int (\sin x - \cos x) , dx = -\cos x - \sin x $$

Evaluate it from ( 0 ) to ( \frac{\pi}{4} ): $$ \left[-\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right] - \left[-\cos(0) - \sin(0)\right] $$

This simplifies to: $$ \left[-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right] - [-1 - 0] = -\sqrt{2} + 1 $$

6. Calculate the second integral

Calculate the integral from ( \frac{\pi}{4} ) to ( \frac{\pi}{2} ): $$ \int (\cos x - \sin x) , dx = \sin x + \cos x $$

Evaluate it from ( \frac{\pi}{4} ) to ( \frac{\pi}{2} ): $$ \left[\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right] - \left[\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right] $$

This simplifies to: $$ [1 + 0] - \left[\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right] = 1 - \sqrt{2} $$

7. Combine the results

Add both areas: $$ A = (-\sqrt{2} + 1) + (1 - \sqrt{2}) = 2 - 2\sqrt{2} $$