For any real number x, let [x] denote the largest integer less than or equal to x. Let f be a real v

Q: Wait, why can we just multiply the integral of one period by 10?

This is a fundamental property of periodic functions! If a function $g(x)$ has a period $T$, then the integral over any interval of length $nT$ is exactly $n$ times the integral over one period $T$. Since our interval is from $-10$ to $10$ (length 20) and our period is $2$, we have exactly $10$ full cycles. It is a powerful shortcut that saves us from tedious calculations!

Q: Why do the boundary terms vanish in the integration by parts?

Great observation! When you apply integration by parts, you get a term like $[u \cdot v]$. Here, $u$ is a linear function and $v$ involves $\sin(\pi x)$. At the limits $0, 1,$ and $2$, the $\sin(\pi x)$ term becomes $\sin(0), \sin(\pi),$ or $\sin(2\pi)$, all of which are $0$. Thus, the boundary terms disappear, leaving only the integral part. It is a beautiful simplification!

Q: Is $f(x)$ continuous? I am worried about the jump at $x=1$.

That is a very sharp question! Let's check the limit at $x=1$. As $x \to 1^-$, $f(x) = 1-x \to 0$. As $x \to 1^+$, $f(x) = x-1 \to 0$. Since the left-hand limit equals the right-hand limit, the function is indeed continuous at $x=1$. This continuity ensures that our integration is well-behaved and valid across the entire interval.

Question

For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. Let $f$ be a real valued function defined on the interval $[-10, 10]$ by $f(x) = \begin{cases} x - [x] & 	ext{if } [x] 	ext{ is odd,} \ 1 + [x] - x & 	ext{if } [x] 	ext{ is even.} \end{cases}$ Then the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos \pi x dx$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex integral, one that spans from $-10$ to $10$. At first glance, the function $f(x)$ seems erratic, shifting its definition based on the parity of the greatest integer function $[x]$.

It feels like a trap, a mountain of algebra waiting to crush your spirit. But take a breath. In the world of JEE Advanced, complexity is often a mask for hidden symmetry. Let us peel back that mask.

### Decoding the Sawtooth Wave

First, let us demystify $f(x)$. The definition tells us that $f(x)$ behaves differently depending on whether $[x]$ is even or odd.

For $x \in [0, 1)$, $[x] = 0$, which is even, so $f(x) = 1 + 0 - x = 1 - x$. For $x \in [1, 2)$, $[x] = 1$, which is odd, so $f(x) = x - 1$.

If you sketch this, you will see a beautiful, repeating triangle wave. It rises and falls with a period of $T = 2$. This periodicity is our greatest ally.

### The Power of Periodicity

Now, look at the integrand: $f(x) \cos(\pi x)$. We know $f(x)$ repeats every $2$ units.

The period of $\cos(\pi x)$ is $\frac{2\pi}{\pi} = 2$. Since both components of our product repeat every $2$ units, their product $h(x) = f(x) \cos(\pi x)$ must also repeat every $2$ units.

The entire interval from $-10$ to $10$ is $20$ units long, meaning we have exactly $10$ identical cycles. Instead of integrating over the whole range, we can use the property:

$$ \int_{-10}^{10} f(x) \cos(\pi x) dx = 10 \int_{0}^{2} f(x) \cos(\pi x) dx $$

### The Art of Integration

We must split our integral at $x = 1$ because the definition of $f(x)$ changes there. We are left with two integrals:

$$ I_1 = \int_{0}^{1} (1-x) \cos(\pi x) dx \quad 	ext{and} \quad I_2 = \int_{1}^{2} (x-1) \cos(\pi x) dx $$

Let us tackle $I_1$ using integration by parts. We set $u = 1-x$ and $dv = \cos(\pi x) dx$.

The boundary terms, involving $\sin(\pi x)$, vanish at both $0$ and $1$. We are left with the integral of $\frac{\sin(\pi x)}{\pi}$, which yields:

$$ \left[ \frac{(1-x)\sin(\pi x)}{\pi} ight]_0^1 + \int_0^1 \frac{\sin(\pi x)}{\pi} dx = 0 + \left[ -\frac{\cos(\pi x)}{\pi^2} ight]_0^1 = \frac{2}{\pi^2} $$

For $I_2$, we follow the same path with $u = x-1$. Again, the boundary terms vanish at $1$ and $2$. The integration leads us to the exact same result: $\frac{2}{\pi^2}$.

The sum of these two integrals is:

$$ I_1 + I_2 = \frac{2}{\pi^2} + \frac{2}{\pi^2} = \frac{4}{\pi^2} $$

### The Grand Finale

We are almost at the finish line. We have the integral for one period as $\frac{4}{\pi^2}$. Multiplying by our $10$ cycles, the total integral becomes:

$$ 10 	imes \frac{4}{\pi^2} = \frac{40}{\pi^2} $$

The question asks for the value of $\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos(\pi x) dx$. Substituting our result:

$$ \frac{\pi^2}{10} 	imes \frac{40}{\pi^2} = 4 $$

The $\pi^2$ terms cancel out, and $40$ divided by $10$ gives us a clean, satisfying **4**. You see? What seemed like a terrifying calculus problem was just a test of your ability to see the underlying rhythm of the function.

Enter Numerical Value:

Visualized Solution

Defining the Function $f (x)$

Analyzing Periodicity of $f (x)$

Periodicity of the Integrand

Simplifying the Integral Limits

Splitting the Integral

Evaluating the First Integral

Evaluating the Second Integral

Combining the Integrals

Final Answer

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Decoding the Sawtooth Wave

The Power of Periodicity

The Art of Integration

The Grand Finale

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