Animated Solution for Mathematics - Definite Integration: Let $[x]$ denote the greatest integer $\le x$. Consider the function $f(x)=\max \{x^2,1+[x]\}$. Then the value of the integral ${\int }_0^{2}f(x)dx$ is :

Imagine you are standing on a coordinate plane, looking at two distinct paths. One is a smooth, accelerating parabola, $y=x^2$, which starts at the origin and climbs steadily. The other is a rigid, stepped staircase, $y=1+[x]$, which jumps to a new height every time $x$ crosses an integer.

Our function, $f(x)=\max \{x^2,1+[x]\}$, is like a traveler who always chooses the higher path. At any point $x$, the traveler looks at both the parabola and the staircase and decides to walk on whichever one is currently higher.

This is what we call the 'upper envelope' of the two functions. Our goal is to find the area under this path from $x=0$ to $x=2$.

To calculate this area, we cannot simply integrate the whole thing at once because the 'upper' path changes its identity. We need to break our journey into segments where the identity of the higher function is clear.

First, consider the interval $x\in [0,1)$. In this region, the greatest integer function $[x]$ is $0$. So, our staircase is at height $1+[0]=1$.

Meanwhile, the parabola $x^2$ starts at $0$ and reaches $1$ only at $x=1$. Since $x^2<1$ for all $x<1$, the staircase is clearly the winner. Our function here is simply $f(x)=1$.

Now, let's step into the interval $x\in [1,2)$. Here, the greatest integer function $[x]$ becomes $1$. Our staircase is now at height $1+1=2$.

But wait—the parabola is also climbing. It starts at $1$ (when $x=1$) and climbs toward $4$ (when $x=2$). They are going to cross! We need to find exactly where $x^2=2$. Solving this, we find the intersection point $x=\sqrt{2}$.

This gives us two sub-intervals in the range $[1,2)$:

1. For $x\in [1,\sqrt{2})$, the parabola $x^2$ is less than $2$, so the staircase ($y=2$) is the higher path.

2. For $x\in [\sqrt{2},2)$, the parabola $x^2$ has surpassed $2$, so the parabola is now the higher path.

Now that we have our map, the integration becomes a simple matter of summing the areas of these three distinct regions:

Let's evaluate these one by one. The first part is the area of a rectangle with height $1$ and width $1$, which is $1$.

The second part is the area of a rectangle with height $2$ and width $(\sqrt{2}-1)$, which gives us $2\sqrt{2}-2$.

Finally, the third part is the integral of $x^2$, which is $\frac{x^3}{3}$. Evaluating this from $\sqrt{2}$ to $2$, we get:

Adding these pieces together, we have:

Simplifying this, we get $2\sqrt{2}-1+\frac{8-2\sqrt{2}}{3}$. Combining the terms under a common denominator of $3$, we reach the final result:

The final answer is $\frac{5+4\sqrt{2}}{3}$.