SOLUTION 27: Compute the area of the region enclosed by the graphs of the equations $ y= \ln x \ \ (or \ \ x=e^y)$, $ y= 1-x \ \ (or \ \ x= 1-y) $, and $y=2$ . Begin by finding the points of intersection of the the three graphs. From $ y= \ln x $ and $ y= 1-x $ we get that $$ x=1 $$ From $ y=1-x $ and $ y=2 $ we get that $$ 1-x = 2 \ \ \longrightarrow \ \ x=-1 $$ From $ y= \ln x $ and $ y=2 $ we get that $$ \ln x = 2 \ \ \longrightarrow \ \ x=e^2 $$ Now see the given graph of the enclosed region.

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a.) Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that $$ -1 \le x \le 1 \ \ and \ \ e^{-x} \le y \le e^x \ $$ in addition to $$ -1 \le x \le 1 \ \ and \ \ e^{-x} \le y \le e^x \ , $$ so that the area of this region is $$ AREA = \displaystyle{ \int_{-1}^{1} ( Top \ - \ Bottom ) \ dx + \int_{1}^{e^2} ( Top \ - \ Bottom ) \ dx } $$ $$ = \displaystyle{ \int_{-1}^{1} ( e^x - e^{-x} ) \ dx + \int_{-1}^{1} ( e^x - e^{-x} ) \ dx } $$ $ \Big( $ Recall that $ \displaystyle{ \int e^{kx} \ dx } = \displaystyle{ \frac{1}{k} e^{kx} + C } $. $ \Big) $ $$ \displaystyle { = ( e^x - ( - e^{-x} )) \Big\vert_{0}^{2} } $$ $$ \displaystyle { = ( e^x + e^{-x} ) \Big\vert_{0}^{2} } $$ $$ \displaystyle { = ( e^{(2)} + e^{-(2)} ) - ( e^{(0)} + e^{-(0)} ) } $$ $$ \displaystyle { = e^{2} + e^{-2} - ( 1 + 1 ) } $$ $$ \displaystyle { = e^{2} + e^{-2} - 2 } $$ b.) Using horizontal cross-sections to describe this region, which is made up of two smaller regions, we get that $$ e^{-2} \le y \le 1 \ \ and \ \ - \ln y \le x \le 2 \ , $$ in addition to $$ 1 \le x \le e^2 \ \ and \ \ \ln y \le y \le 2 \ , $$ so that the area of this region is $$ AREA = \displaystyle{ \int_{e^{-2}}^{1} ( Right \ - \ Left ) \ dy + \int_{1}^{e^2} ( Right \ - \ Left ) \ dy } $$ $$ = \displaystyle{ \int_{e^{-2}}^{1} ( 2 - (- \ln y) ) \ dy + \int_{1}^{e^2} ( 2 - \ln y ) \ dy } $$ $$ = \displaystyle{ \int_{e^{-2}}^{1} ( 2 + \ln y ) \ dy + \int_{1}^{e^2} ( 2 - \ln y ) \ dy } $$ $ \Big( $ Use Integration by Parts to integrate $ \ \displaystyle{ \int \ln y \ dy } $. Recall that the Integration by Parts formula is $ \displaystyle{ \ \int u \ dv } = uv - \displaystyle{ \int v \ du } $. Let $ \ u = \ln y \ $ and $ \ dv = dy \ $, so that $ \displaystyle{ \ du = \frac{1}{y} dy \ } $ and $ \ v=y \ $. Then $$ \displaystyle{ \int \ln y \ dy = y \ln y - \int y \cdot \frac{1}{y} \ dy } $$ $$ \displaystyle{ = y \ln y - \int 1 \ dy } $$ $$ \displaystyle{ = y \ln y - y + C } \ \Big) $$ Continuing with the definite integral, we get that $$ = \displaystyle{ \int_{e^{-2}}^{1} ( 2 + \ln y ) \ dy + \int_{1}^{e^2} ( 2 - \ln y ) \ dy } $$ $$ \displaystyle{ = ( 2y + (y \ln y - y) ) \Big\vert_{e^{-2}}^{1} + ( 2y - (y \ln y - y) ) \Big\vert_{1}^{e^{2}} } $$ $$ \displaystyle{ = ( y + y \ln y ) \Big\vert_{e^{-2}}^{1} + ( 3y - y \ln y ) \Big\vert_{1}^{e^{2}} } $$ $$ \displaystyle{ = \Big( \Big( (1) + (1) \ln (1) \Big) - \Big( (e^{-2}) + (e^{-2}) \ln (e^{-2}) \Big) + \Big( \Big( 3(e^2) - (e^2) \ln (e^2) \Big) - \Big( 3(1) - (1) \ln (1) \Big) \Big) } $$ $$ = (1+0) - (e^{-2} + e^{-2}(-2)) + (3e^2 - e^2(2) ) - (3-0) $$ $$ = e^{2} + e^{-2} - 2 $$

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