SOLUTION 6: Compute the area of the
region enclosed by the graphs of the equations $ y=\ln x $, $ y=1 $
and $ x=e^{2} $ . Begin by finding the points of intersection of
the two graphs. From $ y=\ln x $ and $ y=1 $ we get that
$$ \ln x = 1 \ \ \longrightarrow
\ \ \ x=e $$
Now see the given graph of the enclosed region.
Using vertical cross-sections to describe this region, we get that
$$ e \le x \le e^2 \ \ and \ \ 1 \le y \le \ln x , $$
so that the area of this region is
$$ AREA = \displaystyle{ \int_{e}^{e^{2}} (Top \ - \ Bottom) \ dx
} $$
$$ = \displaystyle { \int_{e}^{e^{2}} (\ln x - 1) \ dx } $$
$$ = \displaystyle { \int_{e}^{e^{2}} \ln x \ dx - \int_{e}^{e^{2}}
1 \ dx } $$
$\Big($For $ \displaystyle { \int \ln x \ dx } \ $ use Integration by Parts. Recall that the Integration by Parts Formula is
$ \ \displaystyle { \int u \ dv = uv - \int v \ du } $ . Let $ \ u = \ln x \ $ and $ \ dv =
dx \ $, so that $ \ du = \displaystyle{ 1 \over x } \ dx \ $ and $ \ v = x \ $. Then
$ \displaystyle{ \int \ln x \ dx } = \displaystyle { x \ln x - \int \Big(\frac{1}{x}\Big)(x) \ dx }
= \displaystyle { x \ln x - \int 1 \ dx }
= \displaystyle { x \ln x - x + C } . \Big) $
$$ = \displaystyle { \Big( (x \ln x - x) - x \Big)
\Big\vert_{e}^{e^{2}}
} $$
$$ = \displaystyle { \Big( x \ln x - 2x \Big)
\Big\vert_{e}^{e^{2}} } $$
$$ = \displaystyle { \Big( e^{2} \ln e^{2} - 2e^{2} \Big) - \Big( e
\ln e - 2e \Big) } $$
$$ = \displaystyle { \Big( e^{2}(2) - 2e^{2} \Big) - \Big( e(1) - 2e
\Big) } $$
$$ = \displaystyle { \Big( 0 \Big) - \Big( -e \Big) } $$
$$ = \displaystyle { e } $$
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