Animated Solution for Mathematics - Definite Integration: In what ratio does the $x$-axis divide the area of the region bounded by the parabolas $y=4x-x^2$ and $y=x^2-x$ ?

Welcome, fellow explorers of the mathematical universe! Today, we are embarking on a journey to solve a classic JEE Advanced problem. It is not just about finding a number; it is about understanding the geometry of curves and how they interact with the fundamental axes of our coordinate system.

Imagine the Cartesian plane as a canvas, and our two parabolas, $y=4x-x^2$ and $y=x^2-x$, as two dancers moving across it. Our goal is to find the ratio in which the x-axis divides the area they enclose together.

First, let us get to know our dancers. The first parabola, $y=4x-x^2$, is a downward-opening curve. Its roots are at $x=0$ and $x=4$, and its vertex sits proudly at $(2,4)$.

The second parabola, $y=x^2-x$, is an upward-opening curve with roots at $x=0$ and $x=1$, and its vertex dips just below the x-axis at $(\frac{1}{2},-\frac{1}{4})$.

To find the boundaries of the region they enclose, we must find where they meet. We set their equations equal to each other:

Rearranging this, we get $2x^2-5x=0$, which factors beautifully into $x(2x-5)=0$. Thus, our dancers meet at $x=0$ and $x=\frac{5}{2}$. These are our limits of integration for the total area.

Now, let us calculate the total area $A_{\text{total}}$ enclosed between these two curves. In the interval $[0,\frac{5}{2}]$, the downward-opening parabola $y=4x-x^2$ is the upper boundary, and $y=x^2-x$ is the lower boundary. The area is given by the integral:

Simplifying the integrand, we get ${\int }_0^{\frac{5}{2}}(5x-2x^2)dx$. Integrating term by term, we have:

Substituting the upper limit $x=\frac{5}{2}$, we find $A_{\text{total}}=\frac{125}{8}-\frac{125}{12}=\frac{125}{24}$. This is our total area, the entire space enclosed by the two parabolas.

Here is where the problem gets interesting. The x-axis, which is the line $y=0$, acts as a silent judge, splitting this total area into two parts. Looking at our curves, we see that the upward-opening parabola $y=x^2-x$ dips below the x-axis between its roots, $x=0$ and $x=1$.

This creates a region below the x-axis. To find the area below the x-axis, $A_{\text{below}}$, we integrate the difference between the upper boundary (the x-axis, $y=0$) and the lower boundary (the parabola $y=x^2-x$):

Evaluating this integral, we get:

To make our comparison easier, let us express this with a denominator of 24: $A_{\text{below}}=\frac{4}{24}$.

The area above the x-axis, $A_{\text{above}}$, is simply the total area minus the area below the x-axis:

Finally, the ratio in which the x-axis divides the area is $A_{\text{above}}:A_{\text{below}}$, which is $\frac{121}{24}:\frac{4}{24}$. The denominators cancel out, leaving us with the elegant result of $121:4$.