Animated Solution for Mathematics - Definite Integration: Area of the region bounded by the curve $y=e^x$ and lines $x=0$ and $y=e$ is

A Journey Through the Exponential Plane

Welcome, future engineer. Today, we are not just solving a calculus problem; we are exploring the landscape of functions.

When you look at a problem involving $y=e^x$, you shouldn't just see symbols. You should see a curve that defines the very nature of growth—a curve that starts at $(0,1)$ and climbs with relentless, accelerating speed.

Our task is to find the area trapped between this curve, the vertical wall of the y-axis ($x=0$), and the horizontal ceiling of $y=e$. This is a classic JEE Advanced scenario where the math is simple, but the conceptual depth is profound.

The Intersection of Realities

Before we dive into integration, we must find our boundaries. We have the curve $y=e^x$ and the line $y=e$.

Where do they meet? We set them equal: $e^x=e^1$. Since the bases are identical, the exponents must be equal, leading us to $x=1$.

This tells us that our region of interest is confined within the horizontal interval $x\in [0,1]$. Imagine standing at the origin $(0,0)$. You walk to the right until you hit $x=1$. Above you is the line $y=e$, and below you is the curve $y=e^x$. This is the 'target area' we are about to conquer.

The Vertical Slicing Strategy ($dx$)

Let's use the most intuitive tool in our kit: the vertical strip. Imagine drawing an infinitely thin vertical line of width $dx$ at any point $x$ between $0$ and $1$.

The height of this strip is the difference between the top boundary and the bottom boundary. The top is the constant line $y=e$, and the bottom is the curve $y=e^x$. Therefore, the height of our strip is $(e-e^x)$.

To find the total area, we sum up all these strips from $x=0$ to $x=1$:

Using the linearity of the integral, we can split this into two parts: ${\int }_0^{1}edx-{\int }_0^1{e}^{x}dx$. The first part is simple: the integral of a constant $e$ over an interval of length $1$ is just $e$.

Thus, we arrive at the expression $e-{\int }_0^1{e}^{x}dx$. This is a beautiful, elegant form that matches one of our options. It reminds us that the area is essentially a rectangle of area $e$ minus the area under the exponential curve.

The Horizontal Slicing Strategy ($dy$)

But what if we turned our perspective sideways? In JEE Advanced, flexibility is your greatest asset. Let's slice the region horizontally with strips of thickness $dy$.

Now, the length of each strip is determined by the x-coordinate. Since $y=e^x$, we take the natural logarithm of both sides to get $x=\ln y$. As we stack these horizontal strips from the bottom of the region ($y=1$) to the top ($y=e$), the area becomes:

This is a powerful realization. We have transformed a problem about exponential growth into a problem about logarithmic accumulation. Both integrals represent the exact same physical area of $1$ square unit. Seeing this equivalence is the hallmark of a true mathematician.

The Magic of King's Rule

Finally, let's address the most elegant option of all. We have our integral $A={\int }_1^e\ln ydy$.

We can apply the property known as King's Rule: ${\int }_a^{b}f(y)dy={\int }_a^{b}f(a+b-y)dy$. Here, $a=1$ and $b=e$, so $a+b=e+1$. Substituting this into our integral, we get:

This is the same area, just viewed through a different algebraic lens. It is fascinating how the same geometric reality can be described by such diverse mathematical expressions.

Whether you use vertical strips, horizontal strips, or the symmetry of King's Rule, the truth remains constant: the area is 1. Keep this flexibility in your mind as you approach your exams. Don't just solve for the answer; solve for the connections.