SOLUTION 10: Compute the area of the
enclosed region bounded by the graphs of the equations $ y =
\tan^2 x $, $ y = 0 $, and $ x =1 $ .
Now see the given graph of the enclosed region.
Using vertical cross-sections to describe this region, we get that
$$ 0 \le x \le 1 \ \ and \ \ 0 \le y \le \tan^2 x \ , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{0}^{1} (Top \ - \ Bottom) \ dx } $$
$$ = \displaystyle { \int_{0}^{1} (\tan^{2} x - 0) \ dx } $$
$$ = \displaystyle { \int_{0}^{1} \tan^{2} x \ dx } $$
(Recall that $ 1 + \tan^2 x = \sec^2 x $, so that $ \tan^2 x = \sec^2 x - 1 $.)
$$ = \displaystyle { \int_{0}^{1} ( \sec^{2} x - 1 ) \ dx } $$
$$ = \displaystyle { ( \tan x - x ) \Big\vert_{0}^{1} } $$
$$ = \displaystyle { ( \tan 1 - 1) - ( \tan 0 - 0 ) } $$
$$ = \displaystyle { ( \tan 1 - 1) - ( 0 - 0 ) } $$
$$ = \displaystyle { \tan 1 - 1 } $$
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