Question

The value of the integral $\int \limits^{2}_{-2} \frac{\sin^{2}x}{\left[\frac{x}{\pi}\right] + \frac{1}{2}} dx $ (where [x] denotes the greatest integer less than $^{20}C_r$ or equal to x) is :

• A. 4
• B. 4 - sin 4
• C. sin 4
• D. 0

Answer 1

The Correct Option is D

The given integral is:

\(I = \int \limits^{2}_{-2} \frac{\sin^{2}x}{\left[\frac{x}{\pi}\right] + \frac{1}{2}} \, dx\)

where \([x]\) denotes the greatest integer less than or equal to \(x\).

First, consider the function \(\left[\frac{x}{\pi}\right]\). As \(x\) moves from -2 to 2, the value \(\frac{x}{\pi}\) will move from approximately \(-\frac{2}{\pi}\) to \(\frac{2}{\pi}\).

Let's determine the value of \(\left[\frac{x}{\pi}\right]\) in this interval:

• For \(x = -2\), \(\frac{-2}{\pi} \approx -0.636\), so the greatest integer less than or equal to -0.636 is -1.
• For \(x = 2\), \(\frac{2}{\pi} \approx 0.636\), so the greatest integer less than or equal to 0.636 is 0.

Therefore, \([\frac{x}{\pi}] = -1\) or \(0\).

Now, observe that the function \(\sin^2 x\) is an even function: \(\sin^2(-x) = \sin^2 x\).

Additionally, since \({(x/\pi) + 1/2}\) changes sign over the symmetry of the interval \([-2, 2]\), the function:

\(f(x) = \frac{\sin^2 x}{[\frac{x}{\pi}] + \frac{1}{2}}\)

is an odd function because the denominator changes sign, i.e., \(f(-x) = -f(x)\).

The property of definite integrals of odd functions over symmetric limits \([-a, a]\) is that they evaluate to zero:

\(\int \limits^{a}_{-a} f(x) \, dx = 0\).

Therefore, the value of the integral is:

\(I = 0\).

Hence, the correct answer is 0.