Animated Solution for Mathematics - Definite Integration: Find all maxima and minima of the function $y=x(x-1{)}^2,0\le x\le 2$. Also determine the area bounded by the curve $y=x(x-1{)}^2$, the $y$-axis and the line $y=2$.

A Journey into Calculus

Welcome, future engineer! Today, we are not just solving a problem; we are exploring the anatomy of a cubic function. We are looking at $y=x(x-1{)}^2$ on the interval $[0,2]$.

This isn't just an equation; it is a path. Imagine yourself walking along this curve. It starts at the origin $(0,0)$, climbs to a peak, dips down to kiss the x-axis at $(1,0)$, and then surges upward toward the point $(2,2)$.

Our mission is twofold: to find the 'peaks and valleys' (the extrema) and to measure the 'shadow' (the area) trapped between this curve and the line $y=2$. Let us begin.

The Hunt for Extrema

To find the maxima and minima, we need to know where the curve stops climbing and starts falling. This happens when the slope of the tangent line is zero. We need the derivative, $\frac{dy}{dx}$.

Our function is a product: $u=x$ and $v=(x-1{)}^2$. The Product Rule is our best friend here:

Applying this, the derivative of $x$ is $1$, and the derivative of $(x-1{)}^2$ (using the Chain Rule) is $2(x-1)$. Putting it together, we get $y^{\prime }=1\cdot (x-1{)}^2+x\cdot 2(x-1)$.

Now, notice the beauty of algebra—we have a common factor of $(x-1)$! Factoring it out, we get $y^{\prime }=(x-1)[(x-1)+2x]$. Simplifying the bracket gives us $y^{\prime }=(x-1)(3x-1)$.

Setting $y^{\prime }=0$, we find our critical points: $x=1$ and $x=\frac{1}{3}$. Evaluating the function at these points, we find the local maximum at $x=\frac{1}{3}$ with $y=\frac{4}{27}$, and the local minimum at $x=1$ with $y=0$. We have mapped the terrain!

The Area Under the Curve

Now, let us tackle the area. We are looking for the region bounded by the line $y=2$, the $y$-axis ($x=0$), and our curve.

The area $A$ is defined by the integral of the 'top' function minus the 'bottom' function:

First, we must expand the integrand. $x(x-1{)}^2=x(x^2-2x+1)=x^3-2x^2+x$. Subtracting this from $2$, we get the polynomial $2-x^3+2x^2-x$.

Using the power rule $\int x^{n}dx=\frac{x^{n+1}}{n+1}$, the integration becomes a rhythmic process:

The Final Calculation

Now, we substitute the limits. At $x=2$, we have $2(2)-\frac{16}{4}+\frac{2(8)}{3}-\frac{4}{2}$.

This simplifies to $4-4+\frac{16}{3}-2$. The $4$ and $-4$ cancel out beautifully, leaving us with $\frac{16}{3}-2$.

Converting $2$ to $\frac{6}{3}$, we get $\frac{16-6}{3}=\frac{10}{3}$.

And there it is! The area is $\frac{10}{3}$ square units. You have navigated the derivative, identified the critical points, and mastered the definite integral. This is the essence of JEE Advanced preparation—not just finding the answer, but understanding the elegant dance of the mathematics behind it. Keep practicing, and keep that curiosity alive!