If for a real number y, [y] is the greatest integer less than or equal to y, then the value of the i

Q: Wait, why does $[2 \sin x]$ become $-1$ when $2 \sin x$ is between $-1$ and $0$? Shouldn't it be $0$?

Ah, the classic trap! Remember the definition: $[y]$ is the greatest integer *less than or equal to* $y$. If $y = -0.5$, the integers less than $-0.5$ are $-1, -2, -3, \dots$. The greatest of these is $-1$. So yes, for any value in $(-1, 0)$, the GIF outputs $-1$. Never confuse 'rounding' with 'truncating'!

Q: Do the endpoints of the intervals matter? For example, at $x = \pi/2$, $2 \sin x = 2$, so $[2] = 2$. But in the next interval, we treat it as $1$. Does this discontinuity break the integral?

Excellent observation! In Riemann integration, changing the value of a function at a finite number of points does not change the value of the integral. So, whether you include the endpoint in the first interval or the second doesn't change the area under the curve. Focus on the intervals, not the points!

Q: Why can't we just use symmetry? The sine curve is symmetric about $\pi$. Can't we just compute half and double it?

That is a brilliant intuition, but be careful! While $\sin x$ is symmetric about $\pi$, the Greatest Integer Function is not symmetric in the same way. The values of $[2 \sin x]$ on the left side of $\pi$ are positive, while on the right side, they are negative. They don't mirror each other perfectly. Always split the integral when dealing with GIF to be safe!

Question

If for a real number $y$, $[y]$ is the greatest integer less than or equal to $y$, then the value of the integral $\int_{\pi/2}^{3\pi/2} [2 \sin x] dx$ is

Options:

(A) $-\pi$

(B) 0

(C) $-\pi/2$

(D) $\pi/2$

Sigma Percentile · Accepted Answer

### The Art of the Step: Mastering the Greatest Integer Integral

Welcome, future engineers! Today, we are going to demystify one of the most feared components in JEE Advanced calculus: the Greatest Integer Function (GIF). Many students see the bracket notation $[y]$ and immediately panic, assuming it requires some complex, high-level theorem.

But I am here to tell you that the secret to this problem isn't advanced calculus—it is simple, elegant visualization.

### Phase 1: Visualizing the Landscape

We are tasked with evaluating the integral:

$$I = \int_{\pi/2}^{3\pi/2} [2 \sin x] dx$$

Before we touch a single equation, let's look at the landscape. Imagine you are standing on the graph of $y = 2 \sin x$. At $x = \frac{\pi}{2}$, you are at the peak, $y = 2$. As you walk toward $x = \frac{3\pi}{2}$, you descend through the $x$-axis, down into the valley, reaching $y = -2$.

This function is continuous, but the Greatest Integer Function $[2 \sin x]$ is not; it is a "step function." Every time $2 \sin x$ crosses an integer boundary—like $1, 0, -1$—the value of our integrand jumps. Our goal is to find exactly where these jumps occur so we can break the integral into manageable, constant pieces.

### Phase 2: Identifying the Critical Boundaries

To find these jumps, we solve $2 \sin x = k$ for $k \in \{2, 1, 0, -1, -2\}$. Let's map them out:

1. At $x = \frac{\pi}{2}$, $2 \sin x = 2$.
2. At $x = \frac{5\pi}{6}$, $2 \sin x = 1$.
3. At $x = \pi$, $2 \sin x = 0$.
4. At $x = \frac{7\pi}{6}$, $2 \sin x = -1$.
5. At $x = \frac{3\pi}{2}$, $2 \sin x = -2$.

These points are our "gates." Between these gates, the value of $[2 \sin x]$ remains constant. This is the magic of the GIF—it turns a complex trigonometric function into a series of simple constants.

### Phase 3: The Calculation

Now, we break the integral into four distinct regions:

**Region 1: $[\frac{\pi}{2}, \frac{5\pi}{6}]$**
Here, $1 < 2 \sin x \le 2$. The greatest integer is $1$. The integral becomes:

$$\int_{\pi/2}^{5\pi/6} 1 dx = \frac{5\pi}{6} - \frac{\pi}{2} = \frac{\pi}{3}$$

**Region 2: $[\frac{5\pi}{6}, \pi]$**
Here, $0 \le 2 \sin x < 1$. The greatest integer is $0$. The integral is:

$$\int_{5\pi/6}^{\pi} 0 dx = 0$$

This part of the curve contributes nothing to our total area.

**Region 3: $[\pi, \frac{7\pi}{6}]$**
This is where we must be careful. Here, $-1 \le 2 \sin x < 0$. The greatest integer is $-1$. Integrating this gives:

$$\int_{\pi}^{7\pi/6} (-1) dx = -\left(\frac{7\pi}{6} - \pi\right) = -\frac{\pi}{6}$$

**Region 4: $[\frac{7\pi}{6}, \frac{3\pi}{2}]$**
Finally, $-2 \le 2 \sin x < -1$. The greatest integer is $-2$. The integral is:

$$\int_{7\pi/6}^{3\pi/2} (-2) dx = -2\left(\frac{3\pi}{2} - \frac{7\pi}{6}\right) = -2\left(\frac{9\pi - 7\pi}{6}\right) = -\frac{2\pi}{3}$$

### The Grand Finale

Now, we simply sum these contributions:

$$I = \frac{\pi}{3} + 0 - \frac{\pi}{6} - \frac{2\pi}{3}$$

Combining these fractions:

$$I = \frac{2\pi - \pi - 4\pi}{6} = -\frac{3\pi}{6} = -\frac{\pi}{2}$$

And there you have it! We have navigated the steps, avoided the negative-number traps, and arrived at the solution. The beauty of this problem lies in how it forces us to stop and look at the function's behavior rather than blindly applying formulas.

**Final Answer: $I = -\frac{\pi}{2}$**

Animated Solution for Mathematics - Definite Integration: If for a real number $y$ , $[y]$ is the greatest integer less than or equal to $y$ , then the value of the integral $\int_{π /2}^{3 π /2} [2 sin x] d x$ is

Select Answer:

Visualized Solution

Visualizing the Base Curve $y = 2 sin x$

Understanding the Greatest Integer Function (GIF) Behavior

Identifying the Critical Boundary Points

Analyzing Interval 1: $[\frac{π}{2}, \frac{5 π}{6}]$

Analyzing Interval 2: $[\frac{5 π}{6}, π]$

Analyzing Interval 3: $[π, \frac{7 π}{6}]$

Analyzing Interval 4: $[\frac{7 π}{6}, \frac{3 π}{2}]$

Summing the Pieces to Find the Total Integral

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Art of the Step

Phase 1

Phase 2

Phase 3

The Grand Finale

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If $[\cdot]$ represents the greatest integer function, then the value of $\int_{0}^{π /2} [[x^{2}] - cos x] d x$ is ____

Select Answer:

Visualized Solution

Visualizing the Base Curve .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=2sinx

Understanding the Greatest Integer Function (GIF) Behavior

Identifying the Critical Boundary Points

Analyzing Interval 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [2π​,65π​]

Analyzing Interval 2: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [65π​,π]

Analyzing Interval 3: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [π,67π​]

Analyzing Interval 4: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [67π​,23π​]