Animated Solution for Mathematics - Definite Integration: The area of the region enclosed by the curves $y=x,x=e,y=1/x$ and the positive x-axis is

Imagine you are standing on the Cartesian plane, looking at a landscape defined by three distinct paths. We have the straight, rising path of $y=x$, the elegant, descending curve of the hyperbola $y=\frac{1}{x}$, and a solid, vertical wall at $x=e$.

Our goal is to find the area of the region trapped between these curves and the positive $x$-axis. This is not just a math problem; it is a study of boundaries.

When we look at the region, we see that it is not a simple, monolithic shape. It is a composite, a story of two halves.

Before we dive into the calculus, we must find the 'climax' of our region—the point where the upper boundary shifts. We set the two functions equal:

This simple algebraic step, $x^2=1$, reveals that the curves intersect at $x=1$ (since we are restricted to the positive $x$-axis). At this point, the line $y=x$ hands over the responsibility of being the 'ceiling' to the hyperbola $y=\frac{1}{x}$.

This is the transition point $(1,1)$. If we try to integrate from $0$ to $e$ without acknowledging this switch, we would be calculating the area under the wrong curve. We must respect the geometry and split our integral.

The Triangle

For the first part of our journey, from $x=0$ to $x=1$, the upper boundary is the line $y=x$. The area $A_1$ is the integral of this line:

This is the area of a simple right-angled triangle. Integrating $x$ gives us $\frac{x^2}{2}$.

Evaluating this from $0$ to $1$, we get:

It is elegant, simple, and confirms our geometric intuition.

The Logarithmic Curve

Now, we move to the second part, from $x=1$ to $x=e$. Here, the upper boundary is the hyperbola $y=\frac{1}{x}$. The area $A_2$ is the integral of this curve:

The antiderivative of $\frac{1}{x}$ is the natural logarithm, $\ln |x|$. Evaluating this from $1$ to $e$, we get $\ln (e)-\ln (1)$.

Since $\ln (e)=1$ and $\ln (1)=0$, the area $A_2$ is exactly $1$ square unit. The hyperbola, despite its infinite nature, creates a finite, beautiful area here.

Finally, we bring the two parts together. The total area $A$ is the sum of our two sub-regions:

We have successfully navigated the boundaries, respected the transition, and arrived at the solution.

The total area is $\frac{3}{2}$ square units. This problem teaches us that in calculus, as in life, the key is to identify the transition points and handle each phase with care. You have mastered the boundaries!