リーマン和
等差数列 x k = 1 + k n ( k = 0 , 1 , 2 , ... , n ) {\displaystyle x_{k}=1+{\frac {k}{n}}\;(k=0,1,2,\dots ,n)} をとると、 左リーマン和と右リーマン和は、それぞれ、
∑ k = 1 n x k − 1 2 Δ x k = ∑ k = 1 n ( 1 + k − 1 n ) 2 1 n = 7 3 − 3 2 n + 1 6 n 2 {\displaystyle \sum _{k=1}^{n}x_{k-1}^{2}\Delta x_{k}=\sum _{k=1}^{n}\left(1+{\frac {k-1}{n}}\right)^{2}{\frac {1}{n}}={\frac {7}{3}}-{\frac {3}{2n}}+{\frac {1}{6n^{2}}}}
∑ k = 1 n x k 2 Δ x k = ∑ k = 1 n ( 1 + k n ) 2 1 n = 7 3 + 3 2 n + 1 6 n 2 {\displaystyle \sum _{k=1}^{n}x_{k}^{2}\Delta x_{k}=\sum _{k=1}^{n}\left(1+{\frac {k}{n}}\right)^{2}{\frac {1}{n}}={\frac {7}{3}}+{\frac {3}{2n}}+{\frac {1}{6n^{2}}}}
となる。
例えば、 [ 1 , 2 ] {\displaystyle [1,2]} で f ( x ) = x 2 {\displaystyle f(x)=x^{2}} のとき