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 $\displaystyle{ \int \ln x \ dx } = \displaystyle { x \ln x - \int \Big(\frac{1}{x}\Big)(x) \ dx } = \displaystyle { x \ln x - \int 1 \ dx } = \displaystyle { x \ln x - x + C } . \Big)$ $$= \displaystyle { \Big( (x \ln x - x) - x \Big) \Big\vert_{e}^{e^{2}} }$$ $$= \displaystyle { \Big( x \ln x - 2x \Big) \Big\vert_{e}^{e^{2}} }$$ $$= \displaystyle { \Big( e^{2} \ln e^{2} - 2e^{2} \Big) - \Big( e \ln e - 2e \Big) }$$ $$= \displaystyle { \Big( e^{2}(2) - 2e^{2} \Big) - \Big( e(1) - 2e \Big) }$$ $$= \displaystyle { \Big( 0 \Big) - \Big( -e \Big) }$$ $$= \displaystyle { e }$$