SOLUTION 7: Compute the area of the region enclosed by the graphs of the equations $y = \cos x$, $y = \sin x$ and $x = 0$ . Begin by finding the points of intersection of the two graphs. From $y= \cos x$ and $y= \sin x$ we get that $$\cos x = \sin x \ \ \longrightarrow$$ $$\displaystyle \frac{\sin x}{\cos x} = 1 \ \ \ \ \longrightarrow$$ $$\tan x = 1 \ \ \longrightarrow \ \ x = \displaystyle \frac{\pi}{4}$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, we get that $$0 \le x \le \displaystyle \frac{\pi}{4} \ \ and \ \ \sin x \le y \le \cos x ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{\pi / 4} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{0}^{\pi / 4} ( \cos x - \sin x ) \ dx }$$ $$= \displaystyle { \Big( \sin x - (-\cos x) \Big) \Big\vert_{0}^{\pi / 4} }$$ $$= \displaystyle { \Big( \sin x + \cos x \Big) \Big\vert_{0}^{\pi / 4} }$$ $$= \displaystyle { \Big( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \Big) - \Big( \sin 0 + \cos 0 \Big) }$$ $$= \displaystyle { \Big( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \Big) - \Big( 0 + 1 \Big) }$$ $$= \displaystyle { \sqrt{2} - 1 }$$