7.
\(\displaystyle \lim_{n \to \infty}(\sqrt{\frac{4n^2-1^2}{36n^4}}+\sqrt{\frac{4n^2-2^2}{36n^4}}+\ldots+\sqrt{\frac{4n^2-n^2}{36n^4}})=\)？
[解答]
\(
\begin{array}{l}
\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\left( {\sqrt {\frac{1}{9} - (\frac{1}{{6n}})^2 } + \sqrt {\frac{1}{9} - (\frac{2}{{6n}})^2 } + ... + \sqrt {\frac{1}{9} - (\frac{n}{{6n}})^2 } } \right) \\
= \int_0^1 {\sqrt {\frac{1}{9} - (\frac{x}{6})^2 } dx} = \frac{1}{6}\int_0^1 {\sqrt {4 - x^2 } } dx = \frac{\pi }{{18}} + \frac{{\sqrt 3 }}{{12}} \\
\end{array}
\)
