Area Solution 9

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SOLUTION 9: Compute the area of the region enclosed by the graphs of the equations

$y=\ln x$ and

$y= (\ln x)^2$ . Begin by finding the points of intersection of the two graphs. From

$y=\ln x$ and

$y= (\ln x)^2$ we get that

$\ln x = (\ln x)^{2} \ \ \longrightarrow$

$\ln x - (\ln x)^{2} = 0 \ \ \longrightarrow$

$\ln x (1 - \ln x) = 0 \ \ \longrightarrow$

$\ln x = 0 \ \ or \ \ \ln x = 1 $ $ \ \ \longrightarrow \ \ x = 1 \ \ or \ \ x = e$

$\displaystyle 2x = 2 \ \ \longrightarrow \ \ x = 1$ Now see the given graph of the enclosed region.

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Using vertical cross-sections to describe this region, we get that

$1 \le x \le e \ \ and \ \ (\ln x)^{2} \le y \le \ln x ,$ so that the area of this region is

$AREA = \displaystyle{ \int_{1}^{e} (Top \ - \ Bottom) \ dx }$

$= \displaystyle { \int_{1}^{e} (\ln x - (\ln x)^{2}) \ dx }$

$= \displaystyle{ \int_{1}^{e} \ln x \ dx} - \displaystyle{ \int_{1}^{e} (\ln x)^{2} \ dx }$

$\Big($ Use Integration by Parts for

$\ \displaystyle \int (\ln x)^{2} \ dx$ . Recall that the Integration by Parts Formula is

$\ \int u \ dv = uv - \int v \ du$ . Let

$\ u = (\ln x)^{2} \$ and

$\ dv = dx \$ , so that

$\ du = 2 \ln x \cdot \displaystyle \frac{1}{x} \ dx \$ and

$\ v = x$ . Then

$\displaystyle \int (\ln x)^{2} \ dx = x (\ln x)^{2} - \displaystyle \int 2 x \ln x \cdot \frac{1}{x} \ dx = x (\ln x)^{2} - 2 \displaystyle \int \ln x \ dx$ Use Integration by Parts again. Let

$\ u = \ln x \$ and

$\ dv = dx \$ , so that

$\ du = \displaystyle \frac{1}{x} \ dx \$ and

$\ v = x$ . Then

$x (\ln x)^{2} - 2 \displaystyle \int \ln x \ dx = x (\ln x)^{2} - 2 \Big[ x \ln x - \int x \cdot \frac{1}{x} \ dx \Big] = x (\ln x)^{2} - 2 \Big[ x \ln x - \int 1 \ dx \Big] = x (\ln x)^{2} - 2 (x \ln x - x ) + C = x (\ln x)^{2} - 2 x \ln x + 2x + C \ \Big)$ Continuing with the definite integral, we get that

$\displaystyle{ \int_{1}^{e} \ln x \ dx} - \displaystyle{ \int_{1}^{e} (\ln x)^{2} \ dx } = \displaystyle{ \Big( \Big(x \ln x - x \Big) - \Big( x (\ln x)^{2} - 2 x \ln x + 2x \Big) \Big) \Big\vert_{1}^{e} }$

$= \displaystyle{ \Big( - x (\ln x)^{2} + 3 x \ln x - 3x ) \Big) \Big\vert_{1}^{e} }$

$= \displaystyle{ \Big( - e (\ln e)^{2} + 3 e \ln e - 3e \Big) - \Big( - (\ln 1)^{2} + 3 \ln 1 - 3 \Big) }$

$= \displaystyle{ \Big( - e (1)^{2} + 3 e (1) - 3e \Big) - \Big( - (0)^{2} + 3(0) - 3 \Big) }$

$= \displaystyle{ ( - e ) - ( - 3 ) }$

$= \displaystyle{ 3-e }$

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