The area enclosed by the curves y=ln(x+e^{2}) | vedclass

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(B) The curves are $y=\ln(x+e^{2})$ and $y=2e^{-x}$.For $y=\ln(x+e^{2})$,at $y=1$,$1=\ln(x+e^{2}) \implies x+e^{2}=e \implies x=e-e^{2}$.For $y=2e^{-x

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$1+e-\ln 2$

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(B) The curves are $y=\ln(x+e^{2})$ and $y=2e^{-x}$.For $y=\ln(x+e^{2})$,at $y=1$,$1=\ln(x+e^{2}) \implies x+e^{2}=e \implies x=e-e^{2}$.For $y=2e^{-x}$,at $y=1$,$1=2e^{-x} \implies e^{x}=2 \implies x=\ln 2$.At the intersection point,$\ln(x+e^{2})=2e^{-x}$. By inspection,at $x=0$,$y=\ln(e^{2})=2$ and $y=2e^{0}=2$. So they intersect at $(0, 2)$.The area is bounded above by the curves and below by $y=1$.Area $= \int_{e-e^{2}}^{0} (\ln(x+e^{2})-1) dx + \int_{0}^{\ln 2} (2e^{-x}-1) dx$.For the first integral,let $u=x+e^{2}$,$du=dx$. When $x=e-e^{2}$,$u=e$. When $x=0$,$u=e^{2}$.$\int_{e}^{e^{2}} (\ln u - 1) du = [u \ln u - u - u]_{e}^{e^{2}} = [u \ln u - 2u]_{e}^{e^{2}} = (e^{2} \cdot 2 - 2e^{2}) - (e \cdot 1 - 2e) = 0 - (-e) = e$.For the second integral,$\int_{0}^{\ln 2} (2e^{-x}-1) dx = [-2e^{-x}-x]_{0}^{\ln 2} = (-2e^{-\ln 2} - \ln 2) - (-2e^{0} - 0) = (-2(1/2) - \ln 2) + 2 = -1 - \ln 2 + 2 = 1 - \ln 2$.Total Area $= e + 1 - \ln 2$.

The area enclosed by the curves $y=\ln(x+e^{2})$,$x=\ln(2/y)$ (which is $y=2e^{-x}$) and $x=\ln 2$,above the line $y=1$ is:

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