Find the area of the region bounded by the following curves 5. \( y=\cos x \quad y=\sin x \) and the \( x \)-axis on \( [0, \pi] \)

To find the area of the region bounded by the curves \( y = \cos x \), \( y = \sin x \), and the \( x \)-axis on the interval \([0, \pi]\), we need to first identify the points of intersection of \( y = \cos x \) and \( y = \sin x \). They intersect when \( \cos x = \sin x \), which occurs at \( x = \frac{\pi}{4} \). Next, we determine which function lies above the other in the interval \([0, \frac{\pi}{4}]\) and \([\frac{\pi}{4}, \pi]\): - In \([0, \frac{\pi}{4}]\), \( \cos x \) is greater than \( \sin x \) (for instance, at \( x = 0, \cos(0) = 1\) and \( \sin(0) = 0\)). - In \([\frac{\pi}{4}, \pi]\), \( \sin x \) is greater than \( \cos x \) (at \( x = \frac{\pi}{2}, \sin(\frac{\pi}{2}) = 1\) and \( \cos(\frac{\pi}{2}) = 0\)). Now we compute the area: \[ \text{Area} = \int_0^{\frac{\pi}{4}} (\cos x - 0) \, dx + \int_{\frac{\pi}{4}}^{\pi} (\sin x - 0) \, dx \] Calculating the first integral: \[ \int_0^{\frac{\pi}{4}} \cos x \, dx = [\sin x]_0^{\frac{\pi}{4}} = \sin\left(\frac{\pi}{4}\right) - \sin(0) = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2} \] Calculating the second integral: \[ \int_{\frac{\pi}{4}}^{\pi} \sin x \, dx = [-\cos x]_{\frac{\pi}{4}}^{\pi} = (-\cos(\pi)) - (-\cos(\frac{\pi}{4})) = 1 - \left(-\frac{\sqrt{2}}{2}\right) = 1 + \frac{\sqrt{2}}{2} \] Adding both results gives us the total area: \[ \text{Total Area} = \frac{\sqrt{2}}{2} + \left(1 + \frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2} \] Thus, the area of the region bounded by the curves is: \[ \boxed{1 + \sqrt{2}} \]