Let [t] denote the largest integer less than or equal to t. If ∫_0^3 ([x²] + [x²/2]) dx = a + b√2 -

Q: Why can't I just integrate $[x^2]$ as $\frac{x^3}{3}$?

Because the Greatest Integer Function is a step function, not a continuous one. It is constant between integers, so its derivative is zero almost everywhere. You must treat it as a sum of areas, not as a continuous function with an antiderivative.

Q: Do I really have to write out every single interval? It takes so long!

In the heat of the JEE exam, precision is your best friend. While it feels tedious, this method is foolproof. The 'long' way is often the 'fast' way because it prevents silly errors. Trust the process; the algebra will simplify beautifully at the end.

Q: How do I know where to split the integral?

The jump points occur exactly where the argument of the GIF becomes an integer. For $[x^2]$, set $x^2 = k$ and solve for $x$. For $[\frac{x^2}{2}]$, set $\frac{x^2}{2} = k$ and solve for $x$. These are your critical points.

Question

Let $[t]$ denote the largest integer less than or equal to $t$. If $\int_0^3 ([x^2] + [\frac{x^2}{2}]) dx = a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$, where $a, b, c \in \mathbb{Z}$, then $a+b+c$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem requires us to evaluate the integral:

$$I = \int_0^3 ([x^2] + [\frac{x^2}{2}]) dx$$

The presence of the Greatest Integer Function (GIF) indicates that the integrand is a step function. We must avoid standard calculus rules and instead treat this as a summation of areas under a staircase.

### Phase 1: The Linearity Strategy

The beauty of integration lies in its linearity. We do not need to solve this as one giant, terrifying expression.

We can split the integral into two manageable parts:

$$I = I_1 + I_2$$

Where:

$$I_1 = \int_0^3 [x^2] dx$$

$$I_2 = \int_0^3 [\frac{x^2}{2}] dx$$

By separating them, we reduce the complexity. We are no longer fighting a dragon; we are fighting two smaller, manageable lizards.

### Phase 2: The Anatomy of $I_1$

Let us focus on $I_1 = \int_0^3 [x^2] dx$. The function $[x^2]$ is a staircase that jumps whenever $x^2$ hits an integer value.

The critical points occur at $x^2 \in \{1, 2, 3, 4, 5, 6, 7, 8\}$, which corresponds to:

$$x \in \{1, \sqrt{2}, \sqrt{3}, 2, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, 3\}$$

We break the integral at these points to sum the areas of the resulting rectangles:

$$I_1 = 0(1-0) + 1(\sqrt{2}-1) + 2(\sqrt{3}-\sqrt{2}) + 3(2-\sqrt{3}) + 4(\sqrt{5}-2) + 5(\sqrt{6}-\sqrt{5}) + 6(\sqrt{7}-\sqrt{6}) + 7(\sqrt{8}-\sqrt{7}) + 8(3-\sqrt{8})$$

After expanding and collecting terms, we find a beautiful cancellation pattern:

$$I_1 = 21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}$$

### Phase 3: The Anatomy of $I_2$

Now, we turn our attention to $I_2 = \int_0^3 [\frac{x^2}{2}] dx$. The jump occurs when $\frac{x^2}{2}$ is an integer, meaning $x^2$ must be an even number: $2, 4, 6, 8$.

Our critical points are $x \in \{\sqrt{2}, 2, \sqrt{6}, \sqrt{8}\}$. We partition the integral accordingly:

$$I_2 = 0(\sqrt{2}-0) + 1(2-\sqrt{2}) + 2(\sqrt{6}-2) + 3(\sqrt{8}-\sqrt{6}) + 4(3-\sqrt{8})$$

Simplifying this expression yields:

$$I_2 = 10 - 3\sqrt{2} - \sqrt{6}$$

### Phase 4: The Final Synthesis

Now, we bring the components together to find $I = I_1 + I_2$. Adding our results:

$$I = (21 - 3\sqrt{2} - \sqrt{3} - \sqrt{5} - \sqrt{6} - \sqrt{7}) + (10 - 3\sqrt{2} - \sqrt{6})$$

Combining the integers and the like radicals, we obtain:

$$I = 31 - 6\sqrt{2} - \sqrt{3} - \sqrt{5} - 2\sqrt{6} - \sqrt{7}$$

Comparing this to the form $a + b\sqrt{2} - \sqrt{3} - \sqrt{5} + c\sqrt{6} - \sqrt{7}$, we identify the constants:

$$a = 31, \quad b = -6, \quad c = -2$$

The final calculation is:

$$a + b + c = 31 - 6 - 2 = \mathbf{23}$$

You have mastered the staircase. The problem was not about complex calculus; it was about patience, precision, and the courage to break a complex problem into simple, solvable steps.

Animated Solution for Mathematics - Definite Integration: Let $[t]$ denote the largest integer less than or equal to $t$ . If $\int_{0}^{3} ([x^{2}] + [\frac{x ^{2}}{2}]) d x = a + b 2 - 3 - 5 + c 6 - 7$ , where $a, b, c \in Z$ , then $a + b + c$ is equal to

Enter Numerical Value:

Visualized Solution

Understanding the Problem

Splitting the Integral

Critical Points for $I_{1}$

Setting up $I_{1}$ Sum

Evaluating $I_{1}$

Simplifying $I_{1}$

Critical Points for $I_{2}$

Setting up and Evaluating $I_{2}$

Simplifying $I_{2}$

Combining $I_{1}$ and $I_{2}$

Comparing Coefficients

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

Understanding the Problem

Splitting the Integral

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