Animated Solution for Mathematics - Definite Integration: The area enclosed by the curves $y={\log }_e(x+e^2)$, $x={\log }_e(\frac{2}{y})$ and $x={\log }_{e}2$, above the line $y=1$ is

Welcome, future engineers! Today, we are not just solving a calculus problem; we are architects of a region. When you see a problem asking for the area enclosed by curves, your first instinct should always be to visualize.

We are given three boundaries: $y=\ln (x+e^2)$, $x=\ln (\frac{2}{y})$, and $x=\ln 2$. But wait, there is a fourth, silent boundary: the line $y=1$. This is our floor.

Everything we calculate must exist above this line. Imagine standing on a platform at $y=1$ and looking up at the curves. This perspective changes everything.

Analyzing the Curves

Let's look at our two primary curves. The first, $y=\ln (x+e^2)$, is a classic logarithmic function, shifted to the left.

To find where it meets our floor, we set $y=1$, which gives $1=\ln (x+e^2)$, leading to $x=e-e^2$. This is our starting point on the left.

Now, look at the second curve: $x=\ln (\frac{2}{y})$. It looks intimidating, but let's rewrite it. Exponentiating both sides gives $e^x=\frac{2}{y}$, which rearranges beautifully to $y=2e^{-x}$.

This is a decaying exponential. Where does it hit our floor? Setting $y=1$ gives $1=2e^{-x}$, so $e^x=2$, or $x=\ln 2$. We now have our horizontal boundaries: from $x=e-e^2$ to $x=\ln 2$.

The Aha! Moment

Now, where do these two curves meet? We need to find the peak of our region. Let's test $x=0$.

For the first curve, $y=\ln (0+e^2)=\ln (e^2)=2$. For the second curve, $y=2e^{-0}=2(1)=2$. They meet perfectly at $(0,2)$!

This is the pivot point of our problem. Because the 'ceiling' of our region changes from the logarithmic curve to the exponential curve at $x=0$, we cannot use a single integral. We must split our work into two distinct regions.

Breaking Down the Area

We are now ready to calculate. The total area is the sum of two integrals. For the first region, from $x=e-e^2$ to $x=0$, the upper boundary is $y=\ln (x+e^2)$.

Since our floor is $y=1$, the height of our vertical strips is $(\ln (x+e^2)-1)$. The integral is:

Using the standard integral $\int \ln (X)dX=X\ln X-X$, we find the antiderivative to be $(x+e^2)\ln (x+e^2)-(x+e^2)-x$. Evaluating this from $e-e^2$ to $0$ yields exactly $e$.

For the second region, from $x=0$ to $x=\ln 2$, the upper boundary is $y=2e^{-x}$. The height is $(2e^{-x}-1)$. The integral is:

This evaluates to $[-2e^{-x}-x]$ from $0$ to $\ln 2$. Plugging in the limits, we get:

Adding these two results together, we get the final area:

It is a clean, elegant result. As a final pro-tip, remember that in JEE Advanced, efficiency is key. If you had integrated with respect to $y$ from $y=1$ to $y=2$, you would have found the area in a single step. Always look for the path of least resistance.