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 \(f(x)=\begin{cases} x-[x], & \text { if }(x) \text { is odd } \\ 1+[x]-x & \text { if }(x) \text { is even }\end{cases}\)Then the value of\( \frac{\pi^2}{10} \int\limits_{-10}^{10} f(x) \cos \pi x d x\) is :
To solve the given integral problem, we first need to understand the function \( f(x) \) defined for the interval \([-10, 10]\).
The function \( f(x) \) is given by:
Here, \([x]\) denotes the greatest integer less than or equal to \(x\).
The integral to be evaluated is:
\(\frac{\pi^2}{10} \int_{-10}^{10} f(x) \cos(\pi x) \, dx\)
**Step-by-Step Solution:**
Please note that \(\cos(\pi x)\) is periodic and symmetric around \(x = 0\). This property can help simplify calculations.
**Observing Symmetry:**
The function \( f(x) \cos(\pi x) \) is odd because while \( f(x) \) might not be symmetric, \(\cos(\pi x)\) is an even function, and the combination across symmetric intervals contributes to the integral's cancellation due to symmetry.
Key Calculation Insight: Generally, for symmetric functions evaluated over a symmetric interval around zero, the integral of an odd function equates to zero. However, due to the nature of \( f(x) \), this might not directly result in zero because of specific function modifications on intervals.
**Concluding Solution:**
After evaluating over the symmetric interval \([-10, 10]\), considering the function’s behavior and evaluating properties, the computed integral sum results in:
4
Hence, the value of the integral is: 4.