Animated Solution for Mathematics - Definite Integration: Let $f(x)=$ Maximum $\left\{x^2,(1-x{)}^2,2x(1-x)\right\}$, where $0\le x\le 1$. Determine the area of the region bounded by the curves $y=f(x),x$-axis, $x=0$ and $x=1$.

A Journey into Calculus

Welcome, future engineer. Today, we are not just solving a calculus problem; we are embarking on a journey to understand the 'upper envelope' of functions.

Imagine you are standing on a landscape defined by three distinct paths: $g_1(x)=x^2$, $g_2(x)=(1-x{)}^2$, and $g_3(x)=2x(1-x)$. Your task is to walk from $x=0$ to $x=1$ while always staying at the highest possible altitude.

This path is the function $f(x)=\max \{x^2,(1-x{)}^2,2x(1-x)\}$. This is the essence of the problem—finding the area under this 'highest ridge'.

The Intersection Hunt

Before we can integrate, we must know where the path changes. We are looking for the transition points where one curve overtakes another.

Let's start by comparing the green curve, $g_2(x)=(1-x{)}^2$, and the red curve, $g_3(x)=2x(1-x)$. We set them equal:

Since we are in the interval $(0,1)$, we know $x eq 1$, so we can safely divide by $(1-x)$. This leaves us with $1-x=2x$, which simplifies beautifully to $3x=1$, or $x=1/3$. This is our first transition point.

Next, we look at the blue curve, $g_1(x)=x^2$, and the red curve, $g_3(x)=2x(1-x)$. We set $x^2=2x(1-x)$.

Again, since $x eq 0$ in this region, we divide by $x$ to get $x=2(1-x)$, which expands to $x=2-2x$. Solving for $x$, we find $3x=2$, or $x=2/3$. These two points, $x=1/3$ and $x=2/3$, are the keys to the kingdom. They partition our interval $[0,1]$ into three distinct zones.

Defining the Piecewise Landscape

With our transition points in hand, we can define our piecewise function $f(x)$.

In the first interval, $[0,1/3]$, the green curve $(1-x{)}^2$ is the highest. In the middle interval, $[1/3,2/3]$, the red curve $2x(1-x)$ takes the lead. Finally, in the last interval, $[2/3,1]$, the blue curve $x^2$ is the topmost.

We have successfully tamed the 'max' function into a manageable piecewise form.

The Integration

Now, we calculate the area $A$ by summing the integrals of these three pieces: $A=I_1+I_2+I_3$.

Let's tackle $I_1$:

The antiderivative is $-\frac{(1-x{)}^3}{3}$. Evaluating from $0$ to $1/3$, we get:

Next, we calculate $I_2$:

The antiderivative is $x^2-\frac{2x^3}{3}$. Evaluating this from $1/3$ to $2/3$ gives us:

Finally, we calculate $I_3$:

The antiderivative is $\frac{x^3}{3}$. Evaluating from $2/3$ to $1$ gives:

Notice the symmetry! $I_1$ and $I_3$ are identical because the entire system is symmetric about $x=1/2$.

Summing these up:

Simplifying by dividing by 3, we arrive at our final answer: $17/27$ square units.

You have successfully navigated the landscape of this function, identified the transitions, and calculated the area with precision. This is the power of calculus—breaking down complex, jagged shapes into simple, solvable parts. Keep this mindset, and no problem will ever be too daunting.