If [x] is the greatest integer ≤ x, then π² ∫₀² (sin(πx/2)) (x - [x])^[x] dx is equal to : (1) 2(π -

Q: Why do we split the integral at $x=1$? Can't we just integrate it as one piece?

Great question! The Greatest Integer Function $[x]$ is discontinuous at every integer. If you try to integrate it as a single block, you are treating $[x]$ as a constant, which it is not. It jumps from $0$ to $1$ at $x=1$. Splitting is not just a technique; it is a necessity to respect the definition of the function.

Q: Wait, in the first interval, we get $(x-0)^0$. Is that always $1$?

Yes, for the interval $[0, 1)$, $x$ is not exactly $0$ (except at the very start). Even if $x=0$, the limit of the function as $x \to 0^+$ is $1$. In the context of Riemann integration, the value at a single point doesn't change the integral, so we treat $(x-0)^0$ as $1$.

Q: Why use Integration by Parts for the second integral?

Look at the structure: we have a polynomial term $(x-1)$ multiplied by a trigonometric term $\sin(\frac{\pi x}{2})$. This is the classic signature of Integration by Parts. The ILATE rule tells us to pick the algebraic function as $u$ and the trigonometric function as $dv$ to reduce the power of the polynomial.

Question

If $[x]$ is the greatest integer $\le x$, then $\pi^2 \int_{0}^{2} \left(\sin \frac{\pi x}{2}\right) (x - [x])^{[x]} dx$ is equal to :

Options:

(A) $2(\pi - 1)$

(B) $4(\pi - 1)$

(C) $4(\pi + 1)$

(D) $2(\pi + 1)$

Sigma Percentile · Accepted Answer

### The Staircase of Calculus: Unlocking the Greatest Integer Function

Imagine you are standing at the base of a staircase. Each step is a flat, horizontal surface, but to get from one to the next, you must make a distinct, vertical jump.

This is exactly how the Greatest Integer Function, $[x]$, behaves. It is the most common "trap" in JEE Advanced problems because it forces you to stop thinking about smooth, continuous curves and start thinking about discrete, piecewise behavior.

Today, we are going to dismantle this problem, $$\pi^2 \int_{0}^{2} \left(\sin \frac{\pi x}{2}ight) (x - [x])^{[x]} dx$$, piece by piece.

### Phase 1: The Anatomy of the Split

The first thing you must do when you see $[x]$ in an integral is to identify the "jump points." Since our limits of integration are from $0$ to $2$, the only integer between them is $1$.

This means the function behaves differently on $[0, 1)$ than it does on $[1, 2)$. We must split the integral:

$$I = \pi^2 \left[ \int_{0}^{1} f(x) dx + \int_{1}^{2} f(x) dx ight]$$

If you try to integrate this without splitting, you are essentially trying to walk up a staircase by treating it like a ramp—you will trip. By splitting, we respect the mathematical reality of the function.

### Phase 2: The First Interval - The Easy Win

Let's look at the first interval, $x \in [0, 1)$. Here, the greatest integer less than or equal to $x$ is always $0$.

So, our term $(x - [x])^{[x]}$ becomes $(x - 0)^0$. As long as $x$ is not exactly $0$, this is $1$.

The integral simplifies to:

$$I_1 = \int_{0}^{1} \sin\left(\frac{\pi x}{2}ight) dx$$

The anti-derivative of $\sin(\frac{\pi x}{2})$ is $-\frac{2}{\pi} \cos(\frac{\pi x}{2})$. Evaluating this from $0$ to $1$ gives us:

$$-\frac{2}{\pi} \left(\cos\left(\frac{\pi}{2}ight) - \cos(0)ight) = -\frac{2}{\pi} (0 - 1) = \frac{2}{\pi}$$

### Phase 3: The Second Interval - The Substitution Trick

Now, for the interval $x \in [1, 2)$, the greatest integer is $1$. Our expression becomes $(x - 1)^1$, which is simply $(x - 1)$.

The integral is:

$$I_2 = \int_{1}^{2} (x - 1) \sin\left(\frac{\pi x}{2}ight) dx$$

Let $t = x - 1$. Then $dx = dt$. When $x=1$, $t=0$; when $x=2$, $t=1$. The integral becomes:

$$\int_{0}^{1} t \sin\left(\frac{\pi (t+1)}{2}ight) dt$$

Using the trigonometric identity $\sin(	heta + \frac{\pi}{2}) = \cos(	heta)$, the integrand transforms into $t \cos(\frac{\pi t}{2})$.

### Phase 4: Integration by Parts - The Heavy Lifting

We are left with $\int_{0}^{1} t \cos(\frac{\pi t}{2}) dt$. We use the formula $\int u \, dv = uv - \int v \, du$.

Following the ILATE rule, we set $u = t$ and $dv = \cos(\frac{\pi t}{2}) dt$. Then $du = dt$ and $v = \frac{2}{\pi} \sin(\frac{\pi t}{2})$.

Applying the formula:

$$\left[t \cdot \frac{2}{\pi} \sin\left(\frac{\pi t}{2}ight)ight]_{0}^{1} - \int_{0}^{1} \frac{2}{\pi} \sin\left(\frac{\pi t}{2}ight) dt$$

The first part evaluates to $\frac{2}{\pi}$. The second part, after integrating, becomes:

$$\left[ \frac{4}{\pi^2} \cos\left(\frac{\pi t}{2}ight) ight]_{0}^{1} = \frac{4}{\pi^2} (0 - 1) = -\frac{4}{\pi^2}$$

Subtracting this from the first part, we get $I_2 = \frac{2}{\pi} - \frac{4}{\pi^2}$.

### The Final Synthesis

We have $I_1 = \frac{2}{\pi}$ and $I_2 = \frac{2}{\pi} - \frac{4}{\pi^2}$. Adding them together, the total integral inside the brackets is:

$$\frac{4}{\pi} - \frac{4}{\pi^2}$$

Finally, we multiply by the constant $\pi^2$ that was waiting outside the integral from the very beginning:

$$\pi^2 \left(\frac{4}{\pi} - \frac{4}{\pi^2}ight) = 4\pi - 4 = 4(\pi - 1)$$

The final answer is **$4(\pi - 1)$**.

Animated Solution for Mathematics - Definite Integration: If $[x]$ is the greatest integer $\leq x$ , then $π^{2} \int_{0}^{2} (sin \frac{π x}{2}) (x - [x])^{[x]} d x$ is equal to :

Select Answer:

Visualized Solution

Understanding the Integral

Splitting the Interval

Simplifying Interval $0 \leq x < 1$

Simplifying Interval $1 \leq x < 2$

Evaluating the First Integral

Substitution for the Second Integral

Using Trigonometric Identity

Integration by Parts - Setup

Integration by Parts - Execution

Combining the Results

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Staircase of Calculus

Phase 1

Phase 2

Phase 3

Phase 4

The Final Synthesis

Similar Questions

The value of the integral $\int_{- 2}^{2} \frac{s i n ^{2} x}{[ x / π ] + 1/2} d x$ (where $[x]$ denotes the greatest integer less than or equal to $x$ ) is :

The value of $\int_{π}^{2 π} [2 sin x] d x$ where $[.]$ represents the greatest integer function is

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

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

Let $[x]$ denote the greatest integer $\leq x$ . Consider the function $f (x) = max {x^{2}, 1 + [x]}$ . Then the value of the integral $\int_{0}^{2} f (x) d x$ is :

The integral $\int_{0}^{1.5} [x^{2}] d x$ , where $[]$ denotes the greatest integer function, equals $\dots\dots\dots$

$\int_{0}^{5} cos (π (x - [\frac{x}{2}])) d x$ , where $[t]$ denotes greatest integer less than or equal to $t$ , is equal to :

$\int_{0}^{π} [cot x] d x$ , where $[.]$ denotes the greatest integer function, is equal to:

If the integral $\int_{0}^{10} \frac{[ s i n 2 π x ]}{e ^{x - [x]}} d x = α e^{- 1} + β e^{- 1/2} + γ$ , where $α, β, γ$ are integers and $[x]$ denotes the greatest integer less than or equal to $x$ , then the value of $α + β + γ$ is equal to :

Let $f (x)$ be a function satisfying $f (x) + f (π - x) = π^{2}, \forall x \in R$ . Then $\int_{0}^{π} f (x) sin x d x$ is equal to

Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If [x] is the greatest integer ≤x, then π2∫02​(sin2πx​)(x−[x])[x]dx is equal to :