Animated Solution for Mathematics - Definite Integration: Compute the area of the region bounded by the curves $y=ex\ln x$ and $y=\frac{\ln x}{ex}$ where $\ln e=1$.

A Journey into Area

Welcome, future engineers! Today, we are not just solving a math problem; we are embarking on a journey to visualize the hidden behavior of functions. We are tasked with finding the area bounded by two curves: $y_1=ex\ln x$ and $y_2=\frac{\ln x}{ex}$.

At first glance, these functions might look intimidating, but they are actually elegant partners in a mathematical dance. Let us break this down step-by-step.

The Hunt for Intersections

Before we can calculate the area, we must know where these curves meet. We set $y_1=y_2$, which gives us the equation:

If we bring everything to one side, we get $(e^2{x}^2-1)\ln x=0$. This is a beautiful moment of factorization!

We have two possibilities: either $\ln x=0$ or $e^2{x}^2-1=0$. Solving these, we find our boundaries: $x=1$ and $x=1/e$. These are the walls of our region.

The Trap of the Interval

Now, here is where many students stumble. We are integrating between $x=1/e$ and $x=1$. In this interval, $\ln x$ is negative.

This changes everything! Because $y_1$ is multiplied by $ex$ and $y_2$ is multiplied by $1/ex$, the curve $y_1$ becomes 'more negative' than $y_2$. Therefore, $y_2$ is actually the upper curve.

If you do not verify this, you might end up with a negative area, which is physically impossible for a region. Always check your signs!

The Calculus Battle

With our boundaries set and our upper curve identified, we set up the integral:

We can split this into two manageable parts. For the first part, $\frac{1}{e}\int \frac{\ln x}{x}dx$, we use the substitution $u=\ln x$. This transforms the integral into $\frac{1}{e}\int udu$, which is simply:

For the second part, $e\int x\ln xdx$, we must use Integration by Parts. Recall the formula $\int udv=uv-\int vdu$. By choosing $u=\ln x$ and $dv=xdx$, we get:

This simplifies beautifully to:

The Final Victory

Now, we apply the limits from $1/e$ to $1$. Substituting $x=1$ gives us zero for the logarithmic terms, which simplifies our arithmetic significantly.

Substituting $x=1/e$ requires careful attention to the negative signs. After the dust settles and we combine our two parts, we arrive at the final result:

Take a moment to appreciate this result. It is not just a number; it is the culmination of understanding intersection, interval analysis, and the power of integration techniques. You have successfully navigated the trap and emerged victorious. Keep this confidence for your next challenge!