Animated Solution for Mathematics - Definite Integration: The value of the integral, ${\int }_1^3[x^2-2x-2]dx$, where $[x]$ denotes the greatest integer less than or equal to $x$, is:

My dear student, welcome to a beautiful problem. When you see the Greatest Integer Function (GIF) inside an integral, your first instinct might be to panic. It looks jagged, discontinuous, and frankly, quite intimidating.

But let us pause and breathe. The GIF is not a monster; it is simply a staircase. It stays flat, then jumps, then stays flat again. Our job is to find exactly where those jumps happen.

We are looking at the integral: To conquer this, we must first understand the landscape of the function inside the brackets, $f(x)=x^2-2x-2$.

Before we dive into calculus, let us use the power of algebra to visualize this parabola. By completing the square, we transform $f(x)=x^2-2x-2$ into:

This is a revelation! We now see that the vertex of our parabola lies at $x=1$. As we move from $x=1$ to $x=3$, the function is strictly increasing.

At the start, $f(1)=-3$. At the end, $f(3)=1$. This means our function sweeps through the values from $-3$ to $1$. Because the GIF jumps at every integer, we must identify exactly when $f(x)$ hits the integers $-2,-1,$ and $0$.

This is where the precision of the JEE Advanced exam comes into play. We need to solve $(x-1{)}^2-3=k$ for $k\in \{-2,-1,0\}$.

First, for $k=-2$: Since we are looking in the interval $[1,3]$, we take the positive root: $x-1=1$, so $x=2$.

Next, for $k=-1$: This gives us $x=1+\sqrt{2}$.

Finally, for $k=0$: This gives us $x=1+\sqrt{3}$. These points—$2,1+\sqrt{2},$ and $1+\sqrt{3}$—are the exact locations where our 'staircase' jumps.

Now, the integral becomes a sum of simple, constant-value integrals. We split the domain $[1,3]$ into four intervals based on our critical points:

1. For $x\in [1,2)$, $f(x)\in [-3,-2)$, so $[f(x)]=-3$. 2. For $x\in [2,1+\sqrt{2})$, $f(x)\in [-2,-1)$, so $[f(x)]=-2$. 3. For $x\in [1+\sqrt{2},1+\sqrt{3})$, $f(x)\in [-1,0)$, so $[f(x)]=-1$. 4. For $x\in [1+\sqrt{3},3]$, $f(x)\in [0,1)$, so $[f(x)]=0$.

Our integral $I$ is now the sum of four distinct parts:

Let us calculate these with care. The first part is $-3(2-1)=-3$.

The second part is $-2(1+\sqrt{2}-2)=-2(\sqrt{2}-1)=-2\sqrt{2}+2$.

The third part is $-1(1+\sqrt{3}-(1+\sqrt{2}))=-(\sqrt{3}-\sqrt{2})=-\sqrt{3}+\sqrt{2}$.

The final part is zero. Summing these up:

Grouping the rational and irrational terms, we get $(-3+2)+(-2\sqrt{2}+\sqrt{2})-\sqrt{3}$.

The final result is:

There it is! A complex problem reduced to simple, elegant arithmetic. Never fear the function; simply find where it jumps, and the rest will follow.