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

Q: Wait, why can we just change the limits from $[-10, 10]$ to $[0, 2]$? Is that always allowed?

Great question! We can only do this because both $f(x)$ and $\cos(\pi x)$ have a period of 2. When you integrate a periodic function over $n$ full periods, the integral is simply $n$ times the integral over a single period. Since the interval $[-10, 10]$ has a length of 20, it contains exactly 10 periods of length 2. Thus, the integral over $[-10, 10]$ is 10 times the integral over $[0, 2]$. The factor of $\frac{\pi^2}{10}$ in the original problem then cancels perfectly with this 10.

Q: How do I know for sure that $f(x)$ is periodic with period 2?

You can prove it algebraically! Check $f(x+2)$. If $[x]$ is even, then $[x+2] = [x]+2$, which is also even. So $f(x+2) = 1 + [x+2] - (x+2) = 1 + [x] + 2 - x - 2 = 1 + [x] - x = f(x)$. A similar logic holds if $[x]$ is odd. Since $f(x+2) = f(x)$, the function repeats every 2 units.

Q: Why does the boundary term in integration by parts vanish?

When you perform integration by parts on $\int (1-x) \cos(\pi x) dx$, the boundary term is $[(1-x) \frac{\sin(\pi x)}{\pi}]_0^1$. At $x=1$, $\sin(\pi) = 0$. At $x=0$, $\sin(0) = 0$. Since the sine function is zero at all integer multiples of $\pi$, the boundary term disappears at both limits. This is a common and beautiful feature in these types of trigonometric integrals!

Question

For any real number $x$, let $[x]$ denote the largest integer less than 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 :

Options:

(A) 4

(B) 2

(C) 1

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a path that stretches from $-10$ to $10$. You are looking at a function $f(x)$ that behaves like a rhythmic, repeating pulse.

This function is defined by the greatest integer function, $[x]$, which acts like a staircase. The rule for $f(x)$ flips depending on whether $[x]$ is odd or even, creating a symmetric zig-zag pattern that dances across the $x$-axis.

Our mission is to calculate the integral of this function multiplied by $\cos(\pi x)$ over the entire interval.

### Decoding the Rhythm

First, let us look at the function $f(x)$ in the interval $[0, 1)$. Here, $[x] = 0$, which is even. The definition tells us $f(x) = 1 + [x] - x$, so $f(x) = 1 - x$.

Now, look at the next interval, $[1, 2)$. Here, $[x] = 1$, which is odd. The definition changes to $f(x) = x - [x]$, so $f(x) = x - 1$.

If you sketch this, you see a line going down from $1$ to $0$, then a line going up from $0$ to $1$. By checking $f(x+2)$, we confirm that this pattern repeats every $2$ units. Thus, $f(x)$ is periodic with period $T = 2$.

### The Power of Periodicity

Now, consider the integrand $g(x) = f(x) \cos(\pi x)$. We know $f(x)$ has a period of $2$, and the function $\cos(\pi x)$ also has a period of:

$$\frac{2\pi}{\pi} = 2$$

When two periodic functions with the same period are multiplied, their product is also periodic with that same period. This is a massive shortcut.

The integral of a periodic function over multiple periods is just the number of periods multiplied by the integral over one period. Our interval $[-10, 10]$ has a length of $20$, which is exactly $10$ periods of length $2$.

The integral becomes:

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

When we multiply this by the $\frac{\pi^2}{10}$ factor from the original problem, the $10$s cancel out, leaving us with:

$$\pi^2 \int_{0}^{2} f(x) \cos(\pi x) \, dx$$

### The Calculus Journey

We have reduced the problem to integrating over just one period, $[0, 2]$. Because $f(x)$ changes its definition at $x=1$, we must split the integral:

$$\pi^2 \left[ \int_{0}^{1} (1-x) \cos(\pi x) \, dx + \int_{1}^{2} (x-1) \cos(\pi x) \, dx \right]$$

Let us call these $J_1$ and $J_2$. For $J_1 = \int_{0}^{1} (1-x) \cos(\pi x) \, dx$, we use integration by parts. We set $u = 1-x$ and $dv = \cos(\pi x) \, dx$.

The boundary term vanishes because $\sin(\pi) = 0$ and $\sin(0) = 0$. We are left with:

$$\int_{0}^{1} \frac{\sin(\pi x)}{\pi} \, dx = \left[ -\frac{\cos(\pi x)}{\pi^2} \right]_{0}^{1} = \frac{1}{\pi^2} - \left( -\frac{1}{\pi^2} \right) = \frac{2}{\pi^2}$$

For $J_2 = \int_{1}^{2} (x-1) \cos(\pi x) \, dx$, the process is identical. The boundary term vanishes again because $\sin(2\pi) = 0$ and $\sin(\pi) = 0$. The integral evaluates to $\frac{2}{\pi^2}$.

### The Grand Finale

We are almost there! We just need to sum our results:

$$\pi^2 (J_1 + J_2) = \pi^2 \left( \frac{2}{\pi^2} + \frac{2}{\pi^2} \right) = \pi^2 \left( \frac{4}{\pi^2} \right)$$

The $\pi^2$ terms cancel out, and we are left with the clean, beautiful integer **4**.

This problem teaches us that even the most intimidating integrals can be tamed by looking for symmetry and periodicity. Always pause, visualize the function, and let the properties of the math do the heavy lifting for you.

Select Answer:

Visualized Solution

Understanding the Function $f (x)$

Analyzing $f (x)$ for $x \in [0, 1)$

Analyzing $f (x)$ for $x \in [1, 2)$

Identifying Periodicity

Symmetry and the Integrand

Simplifying the Integral Range

Splitting the Integral

Integrating the First Part ( $J_{1}$ )

Integrating the Second Part ( $J_{2}$ )

Final Summation

Conclusion & Key Takeaways

The Sigma Insight: Fundamental Theorem & Properties of Definite Integrals

Analyzing the Setup

Decoding the Rhythm

The Power of Periodicity

The Calculus Journey

The Grand Finale

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Visualized Solution

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