Animated Solution for Mathematics - Definite Integration: If $[t]$ denotes the greatest integer $\le t$, then the value of ${\int }_0^1[2x-|3x^2-5x+2|+1]dx$ is:

Welcome, fellow traveler of the mathematical landscape. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of brackets and absolute values.

We are looking at the integral:

It is intimidating, isn't it? But here is the secret: every complex problem is just a series of simple, logical steps waiting to be uncovered. Let us peel back the layers together.

Taming the Modulus

First, let us define our function $g(x)=2x-|3x^2-5x+2|+1$. The greatest integer function $[g(x)]$ is a step function; it only changes its value when $g(x)$ crosses an integer.

Before we worry about the integers, we must deal with that modulus. The expression inside, $3x^2-5x+2$, factors into $(3x-2)(x-1)$. The roots are $x=\frac{2}{3}$ and $x=1$.

This is our first critical boundary. For $x\in [0,\frac{2}{3}]$, the quadratic is positive, so the modulus is redundant. For $x\in [\frac{2}{3},1]$, the quadratic is negative, so the modulus flips the sign.

Mapping the Terrain

Now, we define $g(x)$ piecewise. In the first region, $x\in [0,\frac{2}{3}]$, $g(x)=-3x^2+7x-1$. In the second region, $x\in [\frac{2}{3},1]$, $g(x)=3x^2-3x+3$.

For the first piece, we solve $g(x)=0$, $g(x)=1$, and $g(x)=2$ to find where the floor function jumps. Solving $g(x)=0$ gives $x=\frac{7\pm \sqrt{37}}{6}$.

Since we are in the interval $[0,\frac{2}{3}]$, we take the root $\alpha =\frac{7-\sqrt{37}}{6}\approx 0.15$. This is where the function crosses from $-1$ to $0$.

Solving $g(x)=1$ gives $x=\frac{1}{3}$. This is where the function crosses from $0$ to $1$. Finally, solving $g(x)=2$ gives $\beta =\frac{7-\sqrt{13}}{6}\approx 0.56$. This is where the function crosses from $1$ to $2$.

The Integration Journey

Now, look at the second piece, $x\in [\frac{2}{3},1]$. Here, $g(x)=3x^2-3x+3$.

At $x=\frac{2}{3}$, $g(x)=\frac{7}{3}\approx 2.33$. At $x=1$, $g(x)=3$. Since the function is strictly increasing, the value of $g(x)$ stays between $2$ and $3$ for the entire interval. Thus, $[g(x)]=2$ throughout this entire segment.

We have mapped our path! The integral is now a simple sum of areas:

Evaluating these is straightforward. We get $I=-\alpha +0+(\beta -\frac{1}{3})+2(1-\beta )$. Simplifying this, we arrive at $I=\frac{5}{3}-\alpha -\beta$.

Substituting our values for $\alpha$ and $\beta$ back into the equation:

After a moment of careful arithmetic, the terms align perfectly to give us the final answer:

Look at that result. It is elegant, precise, and entirely derived from your logical steps. You didn't just solve a problem; you navigated a complex system and emerged victorious.