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
$$ AREA = \displaystyle{ \int_{0}^{\pi / 4} (Top \ - \ Bottom) \ dx
} $$
$$ = \displaystyle { \int_{0}^{\pi / 4} ( \cos x - \sin x ) \ dx } $$
$$ = \displaystyle { \Big( \sin x - (-\cos x) \Big)
\Big\vert_{0}^{\pi / 4} } $$
$$ = \displaystyle { \Big( \sin x + \cos x \Big) \Big\vert_{0}^{\pi
/ 4} } $$
$$ = \displaystyle { \Big( \sin \frac{\pi}{4} + \cos \frac{\pi}{4}
\Big) - \Big( \sin 0 + \cos 0 \Big) } $$
$$ = \displaystyle { \Big( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}
\Big) - \Big( 0 + 1 \Big) } $$
$$ = \displaystyle { \sqrt{2} - 1 } $$
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