Man berechne das Integral ∫23x2dx{\displaystyle \int _{2}^{3}x^{2}\,\mathrm {d} x} mit Hilfe von Obersummen bei äquidistanter Intervallteilung.
Hinweis: Man verwende ∑k=1nk=n(n+1)2{\displaystyle \sum _{k=1}^{n}k={\frac {n\left(n+1\right)}{2}}} und ∑k=1nk2=n(n+1)(2n+1)6{\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n\left(n+1\right)\left(2n+1\right)}{6}}}.
x2{\displaystyle x^{2}} ist streng monoton steigend, wodurch das Maximum eines Teilintervalls (Δx=b−an{\displaystyle \Delta x={\frac {b-a}{n}}}) immer "am Ende" ist. Daher kann man Folgendes schreiben:
A = ∫ a b f ( x ) d x = lim n → ∞ ∑ k = 1 n Δ x ⋅ f ( a + k ⋅ Δ x ) = lim n → ∞ ∑ k = 1 n Δ x ⋅ ( a + k ⋅ Δ x ) 2 = lim n → ∞ ∑ k = 1 n 1 n ( 2 + k n ) 2 = lim n → ∞ ∑ k = 1 n 1 n ( 4 + 4 k n + k 2 n 2 ) = lim n → ∞ ∑ k = 1 n 4 n + 4 k n 2 + k 2 n 3 = lim n → ∞ 4 ∑ k = 1 n 1 n + 4 n 2 ∑ k = 1 n k + 1 n 3 ∑ k = 1 n k 2 = lim n → ∞ 4 + 4 n 2 ⋅ n ( n + 1 ) 2 + 1 n 3 ⋅ n ( n + 1 ) ( 2 n + 1 ) 6 = lim n → ∞ 24 n 2 + 12 n ( n + 1 ) + ( n + 1 ) ( 2 n + 1 ) 6 n 2 = lim n → ∞ 24 n 2 + 12 n 2 + 12 n + 2 n 2 + 2 n + n + 1 6 n 2 = lim n → ∞ 38 n 2 + 15 n + 1 6 n 2 = lim n → ∞ 38 + 15 n + 1 n 2 6 = 38 6 = 19 3 = 6 + 1 3 {\displaystyle {\begin{aligned}A&=\int _{a}^{b}f(x)\,\mathrm {d} x\\&=\lim _{n\to \infty }\sum _{k=1}^{n}\Delta x\cdot f(a+k\cdot \Delta x)\\&=\lim _{n\to \infty }\sum _{k=1}^{n}\Delta x\cdot (a+k\cdot \Delta x)^{2}\\&=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{n}}\left(2+{\frac {k}{n}}\right)^{2}\\&=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{n}}\left(4+{\frac {4k}{n}}+{\frac {k^{2}}{n^{2}}}\right)\\&=\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {4}{n}}+{\frac {4k}{n^{2}}}+{\frac {k^{2}}{n^{3}}}\\&=\lim _{n\to \infty }4\sum _{k=1}^{n}{\frac {1}{n}}+{\frac {4}{n^{2}}}\sum _{k=1}^{n}k+{\frac {1}{n^{3}}}\sum _{k=1}^{n}k^{2}\\&=\lim _{n\to \infty }4+{\frac {4}{n^{2}}}\cdot {\frac {n(n+1)}{2}}+{\frac {1}{n^{3}}}\cdot {\frac {n(n+1)(2n+1)}{6}}\\&=\lim _{n\to \infty }{\frac {24n^{2}+12n(n+1)+(n+1)(2n+1)}{6n^{2}}}\\&=\lim _{n\to \infty }{\frac {24n^{2}+12n^{2}+12n+2n^{2}+2n+n+1}{6n^{2}}}\\&=\lim _{n\to \infty }{\frac {38n^{2}+15n+1}{6n^{2}}}\\&=\lim _{n\to \infty }{\frac {38+{\frac {15}{n}}+{\frac {1}{n^{2}}}}{6}}\\&={\frac {38}{6}}={\frac {19}{3}}=6+{\frac {1}{3}}\end{aligned}}}