Animated Solution for Mathematics - Definite Integration: ${\int }_0^2[x^2]dx$ is

Mastering the Greatest Integer Function

My dear student, welcome to a beautiful exploration of one of the most misunderstood functions in the JEE syllabus: the Greatest Integer Function (GIF). Many students see the symbol $[x^2]$ and immediately feel a sense of dread, fearing complex calculus.

But I want you to shift your perspective. Think of the GIF not as a mathematical monster, but as a staircase. It is a function that stays flat, then suddenly jumps, stays flat again, and jumps once more.

Our goal today is to calculate the area under this staircase from $x=0$ to $x=2$.

Visualizing the Discontinuities

We are tasked with evaluating the integral:

The core of this problem lies in understanding that the value of $[x^2]$ is constant as long as $x^2$ does not cross an integer. So, the function only changes its value when $x^2$ hits an integer.

As $x$ travels from $0$ to $2$, $x^2$ travels from $0$ to $4$. Within this range, $x^2$ hits the integers $1,2,$ and $3$. These are the moments where our staircase takes a step up.

To find the exact $x$-coordinates of these steps, we solve the equations $x^2=1$, $x^2=2$, and $x^2=3$. This gives us $x=1$, $x=\sqrt{2}$, and $x=\sqrt{3}$. These are the points where we must pause and split our journey.

The Art of Splitting

Now that we have identified our critical points, we can break our integral into four distinct, manageable segments. This is the secret to solving any GIF integral: turn one big, scary problem into four small, easy ones.

Our integral becomes:

Within each of these intervals, the value of $[x^2]$ is constant. For instance, in the interval $[1,\sqrt{2})$, $x^2$ is between $1$ and $2$, so the greatest integer less than or equal to $x^2$ is simply $1$.

The Summation of Areas

Let us evaluate these pieces one by one:

1. In the interval $[0,1)$, $[x^2]=0$. The integral is ${\int }_0^{1}0dx=0$. 2. In the interval $[1,\sqrt{2})$, $[x^2]=1$. The integral is ${\int }_1^{\sqrt{2}}1dx=1\cdot (\sqrt{2}-1)=\sqrt{2}-1$. 3. In the interval $[\sqrt{2},\sqrt{3})$, $[x^2]=2$. The integral is ${\int }_{\sqrt{2}}^{\sqrt{3}}2dx=2\cdot (\sqrt{3}-\sqrt{2})=2\sqrt{3}-2\sqrt{2}$. 4. In the interval $[\sqrt{3},2)$, $[x^2]=3$. The integral is ${\int }_{\sqrt{3}}^{2}3dx=3\cdot (2-\sqrt{3})=6-3\sqrt{3}$.

Now, we simply add these results together:

Grouping the terms, we have constants: $-1+6=5$. The $\sqrt{2}$ terms: $\sqrt{2}-2\sqrt{2}=-\sqrt{2}$. The $\sqrt{3}$ terms: $2\sqrt{3}-3\sqrt{3}=-\sqrt{3}$.

Combining these, we get the final result:

This is the elegant result of our journey. By breaking the problem down, we didn't just find the answer; we understood the geometry of the function. Keep practicing this method, and you will find that even the most intimidating integrals become simple, logical steps.