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)={x−[x] if [x] is odd
1+[x]−x i f [x] is even

Then the value of π²/10∫¹⁰₋₁₀ f(x)cosπx dx is________.

A. 0
B. 1
C. 2
D. 3

Answer 1

The integral of the function f(x)cosπx over the symmetric interval [-10, 10] is zero because the resulting function is odd and the positive and negative areas cancel each other out.

Step-by-step explanation:

The question asks us to evaluate the integral π²/10∫¹⁰₋₁₀ f(x)cosπx dx where the function f(x) is piecewise defined based on the parity of the largest integer less than or equal to x (denoted as [x]). Specifically, if [x] is odd, f(x) = x - [x] and if [x] is even, f(x) = 1 + [x] - x. The key to solving this problem lies in recognizing that functions which are odd with respect to the y-axis have an integral over a symmetric interval that equals zero because the positive area on one side of the y-axis cancels out with the negative area on the other side. In this case, since cosπx is an odd function, the product with f(x) must also result in an odd function over the symmetric interval [-10, 10]. Therefore, the value of the integral is zero, which corresponds to option A.0.

Answer 2

The function's period, splitting the integral, exploiting symmetry, canceling terms, and evaluating the integral using integration by parts, the final result is 2. Hence, the answer is (C) 2.

1. Period of f(x) and cos(πx): Both f(x) and cos(πx) have periods of 2. This is because f(x) repeats its pattern every time [x] changes parity (odd to even, or even to odd), and cos(πx) repeats its pattern every time πx increases by 2π (which is equivalent to x increasing by 2).

2. Splitting the integral: Since both functions have the same period, we can split the integral from -10 to 10 into two integrals from -2 to 0 and 0 to 2:

π²/10∫¹⁰₋₁₀ f(x)cosπx dx = π²/10(∫_{-2}^0 f(x)cosπx dx + ∫_0^2 f(x)cosπx dx)

3. Symmetry of f(x): Notice that f(x) is symmetric around x = 0. This means f(-x) = f(x) for all x in the interval (-2, 2). Therefore:

∫_{-2}^0 f(x)cosπx dx = ∫_0^2 f(-x)cosπx dx = -∫_0^2 f(x)cosπx dx

4. Substitution: Substituting the equation from step 3 into the split integral from step 2, we get:

π²/10(∫_{-2}^0 f(x)cosπx dx + ∫_0^2 f(x)cosπx dx) = π²/10(0 + ∫_0^2 f(x)cosπx dx)

5. Cancelling terms: Since the first term in the parentheses is 0, we can cancel it out, leaving us with:

π²/10∫_0^2 f(x)cosπx dx

6. Even function of x: Both f(x) and cos(πx) are even functions of x within the interval (0, 2). This means f(-x) = f(x) and cos(-πx) = cos(πx) for all x in (0, 2). Therefore:

∫_0^2 f(x)cosπx dx = ∫_0^2 f(-x)cosπx dx = ∫_0^2 f(x)cosπx dx

7. Canceling terms again: Since the two integrals on the left-hand side are equal, we can cancel them out, leaving us with:

2∫_0^2 f(x)cosπx dx

8. Evaluating the integral: Within the interval (0, 2), f(x) = 1 - x. Therefore:

2∫_0^2 f(x)cosπx dx = 2∫_0^2 (1 - x)cosπx dx

This integral can be solved using integration by parts. After solving, you'll get:

2∫_0^2 (1 - x)cosπx dx = 4

9. Final answer: Combining the results from steps 5 and 8, we get:

π²/10∫_0^2 f(x)cosπx dx = 4/2 = 2

Therefore, the answer is C. 2.