Given the functional equation:
\[ f(x) + f(\pi - x) = \pi^2. \]Consider the given integral:
\[ I = \int_0^{\pi} f(x) \sin x \,dx. \]Using the substitution \( t = \pi - x \), we get:
\[ I = \int_0^{\pi} f(\pi - x) \sin (\pi - x) \,dx. \]Since \( \sin (\pi - x) = \sin x \), we obtain:
\[ I = \int_0^{\pi} f(\pi - x) \sin x \,dx. \]Adding both integrals:
\[ 2I = \int_0^{\pi} (f(x) + f(\pi - x)) \sin x \,dx. \]Substituting \( f(x) + f(\pi - x) = \pi^2 \):
\[ 2I = \pi^2 \int_0^{\pi} \sin x \,dx. \]Since \( \int_0^{\pi} \sin x \,dx = 2 \), we get:
\[ 2I = \pi^2 \times 0 = 0. \]Thus, \( I = 0 \).
Final Answer: \( \mathbf{0} \).
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
Consider an A.P. $a_1,a_2,\ldots,a_n$; $a_1>0$. If $a_2-a_1=-\dfrac{3}{4}$, $a_n=\dfrac{1}{4}a_1$, and \[ \sum_{i=1}^{n} a_i=\frac{525}{2}, \] then $\sum_{i=1}^{17} a_i$ is equal to