If the integral ∫₀¹⁰ [sin 2πx] / e^(x-[x]) dx = αe⁻¹ + βe⁻¹/² + γ, where α, β, γ are integers and [x

Q: Why can we just pull the 10 out? Does the periodicity property really apply here?

Yes! The property $\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$ is a powerful tool for periodic functions. Since the period is 1, integrating over 10 periods is exactly 10 times the integral over one period. It turns a massive problem into a bite-sized one.

Q: Wait, why is the numerator $-1$ on the interval $[1/2, 1)$? Shouldn't it be something else?

Great question. On $[1/2, 1)$, $2\pi x$ ranges from $\pi$ to $2\pi$. In this range, $\sin(2\pi x)$ is negative, specifically between $-1$ and $0$. The greatest integer function $[y]$ returns the largest integer less than or equal to $y$. Since our value is in $(-1, 0]$, the largest integer less than or equal to it is $-1$.

Q: Why do we reject the negative root or worry about the fractional part? Is $\{x\}$ always $x$?

In the interval $[0, 1)$, the fractional part $\{x\} = x - [x] = x - 0 = x$. So yes, in this specific window, the denominator simplifies beautifully to $e^x$. Always check your interval before simplifying functions!

Question

If the integral $\int_0^{10} \frac{[\sin 2\pi x]}{e^{x-[x]}} dx = \alpha e^{-1} + \beta e^{-1/2} + \gamma$, where $\alpha, \beta, \gamma$ are integers and $[x]$ denotes the greatest integer less than or equal to $x$, then the value of $\alpha + \beta + \gamma$ is equal to :

Options:

(A) 0

(B) 20

(C) 25

(D) 10

Sigma Percentile · Accepted Answer

### The Art of Taming the Mathematical Beast

When you first look at an integral like

$$I = \int_0^{10} \frac{[\sin 2\pi x]}{e^{x-[x]}} dx$$

it is natural to feel a surge of intimidation. It looks like a chaotic mess of greatest integer functions, exponential denominators, and a wide range of integration.

But in the world of JEE Advanced, intimidation is often just a mask for a hidden, elegant simplicity. Let us peel back the layers of this problem together.

### Phase 1: The Power of Periodicity

The first thing a seasoned problem solver looks for is symmetry or periodicity. Look at the integrand: $f(x) = \frac{[\sin 2\pi x]}{e^{\{x\}}}$.

We know that $x - [x]$ is the definition of the fractional part function, $\{x\}$. Both the fractional part function and the sine function have a period of $1$.

Because the entire function $f(x)$ repeats every single unit, we do not need to slog through the integration from $0$ to $10$. We can use the property:

$$\int_0^{nT} f(x) dx = n \int_0^T f(x) dx$$

By recognizing that $n=10$ and $T=1$, we transform our daunting integral into a much friendlier version:

$$I = 10 \int_0^1 \frac{[\sin 2\pi x]}{e^x} dx$$

Suddenly, the mountain has become a hill.

### Phase 2: Anatomy of the Numerator

Now, we must dissect the numerator: $[\sin 2\pi x]$. The greatest integer function is a step function; it only changes values at integers.

We need to see where $\sin 2\pi x$ hits these integers. On the interval $[0, 1]$, the argument $2\pi x$ ranges from $0$ to $2\pi$.

Imagine the sine wave. From $x = 0$ to $x = 1/2$, the argument $2\pi x$ goes from $0$ to $\pi$. In this region, $\sin 2\pi x$ is positive, living between $0$ and $1$.

The greatest integer of any value between $0$ and $1$ (exclusive of $1$) is $0$. Therefore, the entire first half of our integral vanishes into thin air!

But what happens from $x = 1/2$ to $x = 1$? Here, the argument $2\pi x$ goes from $\pi$ to $2\pi$.

The sine function dips below the x-axis, taking values between $-1$ and $0$. The greatest integer of any value in the interval $[-1, 0)$ is $-1$.

This is the crucial "trap" where many students lose points. We must be precise: the numerator becomes $-1$ in this second half.

### Phase 3: The Final Integration

With our interval split, the integral becomes:

$$I = 10 \left( \int_0^{1/2} 0 dx + \int_{1/2}^1 \frac{-1}{e^x} dx \right)$$

The first part is zero, leaving us with:

$$I = -10 \int_{1/2}^1 e^{-x} dx$$

This is a standard integral. The anti-derivative of $e^{-x}$ is $-e^{-x}$. Applying the Fundamental Theorem of Calculus, we get:

$$I = -10 \left[ -e^{-x} \right]_{1/2}^1 = 10 \left[ e^{-x} \right]_{1/2}^1$$

Substituting the limits, we find:

$$I = 10(e^{-1} - e^{-1/2})$$

This matches the form $\alpha e^{-1} + \beta e^{-1/2} + \gamma$. By comparing coefficients, we see $\alpha = 10$, $\beta = -10$, and $\gamma = 0$.

The sum $\alpha + \beta + \gamma$ is **$10 - 10 + 0 = 0$**.

### Conclusion

Look at that! The complexity dissolved because we took the time to understand the behavior of the functions involved.

Never rush into calculation. Visualize the graph, understand the periodicity, and let the math simplify itself. You have mastered this problem.

Select Answer:

Visualized Solution

Analyze the Integrand $f (x)$

Determine Periodicity

Apply Integral Property

Analyze Numerator on $[0, \frac{1}{2})$

Analyze Numerator on $[\frac{1}{2}, 1)$

Split the Integral

Evaluate the Integral

Substitute Limits

Compare Coefficients

Final Calculation

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

The Art of Taming the Mathematical Beast

Phase 1

Phase 2

Phase 3

Conclusion

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