積分定義
請問這題怎麼寫成黎曼和，答案是\(\displaystyle \frac{26}{3}\)，謝謝解題\( \displaystyle \lim_{n \to \infty}\frac{1}{n^3}\left[n^2+(n+1)^2+(n+2)^2+\ldots+(3n-1)^2 \right]= \)
回復 1# chwjh32 的帖子
\(\displaystyle \lim_{n\to\infty} \frac{1}{n^3}\left[n^2+\left(n+1\right)^2+\cdots\left(3n-1\right)^2\right]\)\(\displaystyle =\lim_{n\to\infty} \sum_{k=0}^{2n-1} \left(1+\frac{k}{n}\right)^2\cdot\frac{1}{n}\)
\(\displaystyle =\int_1^3 x^2 dx\)
\(\displaystyle =\frac{x^3}{3}\Bigg|_{x=1}^{x=3}\)
\(\displaystyle =\frac{26}{3}\) 謝謝
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