Area Solution 10

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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.

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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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