Animated Solution for Mathematics - Definite Integration: Let $[t]$ denote the greatest integer function. If ${\int }_0^{2.4}[x^2]dx=\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$, then $\alpha +\beta +\gamma +\delta$ is equal to

Unlocking the Greatest Integer Function

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to demystify one of the most intimidating-looking problems in calculus: the integral of a Greatest Integer Function.

When you see ${\int }_0^{2.4}[x^2]dx$, your first instinct might be to panic. How do we integrate a function that isn't continuous? How do we handle those sharp, jagged steps?

The secret, my friend, is to stop seeing it as a single, terrifying expression and start seeing it as a series of simple, manageable rectangles.

Visualizing the Landscape

Imagine you are walking along the $x$-axis from $0$ to $2.4$. The function $f(x)=[x^2]$ is like a staircase. It stays flat for a while, then suddenly jumps to a new height.

It jumps exactly when $x^2$ hits an integer value. If $x^2$ is $0.9$, the floor is $0$. The moment $x^2$ becomes $1$, the floor jumps to $1$.

Our journey takes us from $x=0$ to $x=2.4$. This means $x^2$ travels from $0$ to $2.4^2=5.76$.

The integers we encounter on this path are $1,2,3,4,$ and $5$. These are our 'critical points.' By solving $x^2=k$ for $k\in \{1,2,3,4,5\}$, we find the exact locations where the staircase jumps: $x=1,\sqrt{2},\sqrt{3},2,$ and $\sqrt{5}$.

The Art of Partitioning

Now, we break the integral into pieces. Think of this as dividing a complex task into small, bite-sized chores. We are calculating the area under the curve for each interval:

1. For $x\in [0,1)$, $x^2<1$, so $[x^2]=0$. The area is ${\int }_0^{1}0dx=0$.

2. For $x\in [1,\sqrt{2})$, $1\le x^2<2$, so $[x^2]=1$. The area is:

3. For $x\in [\sqrt{2},\sqrt{3})$, $2\le x^2<3$, so $[x^2]=2$. The area is:

4. For $x\in [\sqrt{3},2)$, $3\le x^2<4$, so $[x^2]=3$. The area is:

5. For $x\in [2,\sqrt{5})$, $4\le x^2<5$, so $[x^2]=4$. The area is:

6. Finally, for $x\in [\sqrt{5},2.4]$, $5\le x^2<5.76$, so $[x^2]=5$. The area is:

The Grand Summation

This is where the magic happens. We add these areas together. We group the constants and the coefficients of $\sqrt{2},\sqrt{3},$ and $\sqrt{5}$:

Grouping the constants: $-1+6-8+12=9$.

Grouping the $\sqrt{2}$ terms: $1\sqrt{2}-2\sqrt{2}=-\sqrt{2}$.

Grouping the $\sqrt{3}$ terms: $2\sqrt{3}-3\sqrt{3}=-\sqrt{3}$.

Grouping the $\sqrt{5}$ terms: $4\sqrt{5}-5\sqrt{5}=-\sqrt{5}$.

We are left with $9-\sqrt{2}-\sqrt{3}-\sqrt{5}$. Comparing this to the form $\alpha +\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$, we identify $\alpha =9,\beta =-1,\gamma =-1,\delta =-1$.

The final sum is $9-1-1-1=6$.

Look at what we just achieved. We took a function that seemed impossible to integrate and dismantled it using nothing but logic and basic geometry.

The beauty of JEE Advanced problems isn't in the complexity of the formulas, but in the elegance of the process. You didn't need a supercomputer; you needed a clear mind and a steady hand.

Keep this confidence with you. Every 'impossible' problem is just a series of small, solvable steps waiting for you to find them.