Key Concepts and Formulas

• Area between curves with respect to $y$: If $x = f(y)$ and $x = g(y)$ are continuous functions and $f(y) \ge g(y)$ on the interval $[c, d]$, then the area between the curves is given by $A = \int_{c}^{d} [f(y) - g(y)] \, dy$.
• Logarithmic and exponential relationships: $y = \log_e x \Leftrightarrow x = e^y$.
• Integral of the exponential function: $\int e^y \, dy = e^y + C$.

Step-by-Step Solution

Step 1: Express the curves in terms of x as a function of y.

We are given the curves $y = \log_e(x + e^2)$ and $x = \log_e(\frac{2}{y})$. We need to express $x$ as a function of $y$ for both curves.

For the first curve, $y = \log_e(x + e^2)$, we have: $e^y = x + e^2$ $x = e^y - e^2$

For the second curve, $x = \log_e(\frac{2}{y})$, we already have $x$ as a function of $y$.

The third curve is $x = \log_e 2$.

Step 2: Determine the limits of integration.

We are given that we want the area above the line $y = 1$. We need to find the intersection points of the curves to determine the upper limit of integration.

First, let's find the intersection of $x = e^y - e^2$ and $x = \log_e 2$: $e^y - e^2 = \log_e 2$ $e^y = e^2 + \log_e 2$ $y = \log_e(e^2 + \log_e 2)$

Next, let's find the intersection of $x = \log_e(\frac{2}{y})$ and $x = \log_e 2$: $\log_e(\frac{2}{y}) = \log_e 2$ $\frac{2}{y} = 2$ $y = 1$

Since we are looking for the area above the line $y=1$, the lower limit of integration is $y=1$. The upper limit is found by equating $x=e^y - e^2$ and $x = \log_e(2/y)$ to find their intersection. However, notice that $y=2$ is an important value. When $y=2$, $x=\log_e(2/2)=\log_e(1)=0$. When $y=2$, $x = e^2 - e^2 = 0$. Thus, the upper limit of integration is $y=2$.

Step 3: Set up the integral for the area.

We need to find which function is to the right. At $y=1$, $x = e^1-e^2 = e-e^2 < 0$ and $x = \log_e(2/1) = \log_e 2 > 0$. So, $x=\log_e(2/y)$ is to the right of $x=e^y-e^2$ on the interval $[1,2]$. Thus, the area is given by: $A = \int_{1}^{2} \left[ \log_e\left(\frac{2}{y}\right) - (e^y - e^2) \right] \, dy$ $A = \int_{1}^{2} \left[ \log_e 2 - \log_e y - e^y + e^2 \right] \, dy$

Step 4: Evaluate the integral.

$A = \int_{1}^{2} \log_e 2 \, dy - \int_{1}^{2} \log_e y \, dy - \int_{1}^{2} e^y \, dy + \int_{1}^{2} e^2 \, dy$ $A = (\log_e 2) \int_{1}^{2} dy - \int_{1}^{2} \log_e y \, dy - [e^y]_{1}^{2} + e^2 \int_{1}^{2} dy$ $A = (\log_e 2) [y]_{1}^{2} - \int_{1}^{2} \log_e y \, dy - (e^2 - e) + e^2 [y]_{1}^{2}$ $A = (\log_e 2) (2-1) - \int_{1}^{2} \log_e y \, dy - e^2 + e + e^2 (2-1)$ $A = \log_e 2 - \int_{1}^{2} \log_e y \, dy - e^2 + e + e^2$ $A = \log_e 2 - \int_{1}^{2} \log_e y \, dy + e$

Now we need to evaluate $\int_{1}^{2} \log_e y \, dy$. We use integration by parts: let $u = \log_e y$ and $dv = dy$. Then $du = \frac{1}{y} dy$ and $v = y$. $\int_{1}^{2} \log_e y \, dy = [y \log_e y]_{1}^{2} - \int_{1}^{2} y \cdot \frac{1}{y} \, dy$ $= [y \log_e y]_{1}^{2} - \int_{1}^{2} 1 \, dy$ $= (2 \log_e 2 - 1 \log_e 1) - [y]_{1}^{2}$ $= 2 \log_e 2 - 0 - (2 - 1)$ $= 2 \log_e 2 - 1$

Substitute this back into the expression for $A$: $A = \log_e 2 - (2 \log_e 2 - 1) + e$ $A = \log_e 2 - 2 \log_e 2 + 1 + e$ $A = 1 + e - \log_e 2$

Step 5: Sanity Check

The area cannot be negative. $1 + e - \log_e 2$ is positive.

Common Mistakes & Tips

• Remember to determine which curve is to the right when integrating with respect to $y$.
• Don't forget the constant of integration when evaluating indefinite integrals.
• Integration by parts is a common technique for integrals involving logarithmic functions.

Summary

We found the area enclosed by the curves by expressing $x$ as a function of $y$ and integrating with respect to $y$. We determined the limits of integration by finding the intersection points of the curves and the line $y=1$. We then evaluated the definite integral to find the area. The final area is $1 + e - \log_e 2$.

Final Answer

The final answer is \boxed{1+e-\log _{e} 2}, which corresponds to option (B).