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 $\tan^2 x = \sec^2 x - 1$.) $$= \displaystyle { \int_{0}^{1} ( \sec^{2} x - 1 ) \ dx }$$ $$= \displaystyle { ( \tan x - x ) \Big\vert_{0}^{1} }$$ $$= \displaystyle { ( \tan 1 - 1) - ( \tan 0 - 0 ) }$$ $$= \displaystyle { ( \tan 1 - 1) - ( 0 - 0 ) }$$ $$= \displaystyle { \tan 1 - 1 }$$