Berechnen Sie ∫12x2dx{\displaystyle \int _{1}^{2}x^{2}\,dx} mit Hilfe von Obersummen bei äquidistanter Teilung.
Riemannsches Integral:
A=∫abf=limn→∞∑k=0nΔx⋅f(k⋅Δx+a){\displaystyle A=\int _{a}^{b}f=\lim _{n\to \infty }\sum _{k=0}^{n}\Delta x\cdot f(k\cdot \Delta x+a)}
Δx=b−an{\displaystyle \Delta x={\frac {b-a}{n}}}
∑k=1nk2=n(n+1)(2n+1)6{\displaystyle \sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}}
∑k=1nk=n(n+1)2{\displaystyle \sum _{k=1}^{n}k={\frac {n(n+1)}{2}}}
∫12x2dx=limn→∞1n∑k=0n(kn+1)2=limn→∞1n∑k=0n(k2n2+2kn+1)=limn→∞(n+1)(14n+1)6n2=73{\displaystyle {\begin{aligned}\int _{1}^{2}x^{2}\,dx&=\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=0}^{n}({\frac {k}{n}}+1)^{2}\\&=\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=0}^{n}({\frac {k^{2}}{n^{2}}}+{\frac {2k}{n}}+1)\\&=\lim _{n\to \infty }{\frac {(n+1)(14n+1)}{6n^{2}}}\\&={\frac {7}{3}}\end{aligned}}}
1n∑k=0nk2n2+2kn+1=1n(1n2∑k=0nk2+2n∑k=0nk+n+1){\displaystyle {\frac {1}{n}}\sum _{k=0}^{n}{\frac {k^{2}}{n^{2}}}+{\frac {2k}{n}}+1={\frac {1}{n}}{\Bigl (}{\frac {1}{n^{2}}}\sum _{k=0}^{n}k^{2}+{\frac {2}{n}}\sum _{k=0}^{n}k+n+1{\Bigr )}}
∑k=1n+1k2=(n+1)(n+2)(2n+3)6=(n2+3n+2)(2n+3)6=2n3+3n2+6n2+9n+4n+66=2n3+9n2+13n+66{\displaystyle {\begin{aligned}\sum _{k=1}^{n+1}k^{2}&={\frac {(n+1)(n+2)(2n+3)}{6}}={\frac {(n^{2}+3n+2)(2n+3)}{6}}\\&={\frac {2n^{3}+3n^{2}+6n^{2}+9n+4n+6}{6}}={\frac {2n^{3}+9n^{2}+13n+6}{6}}\end{aligned}}}
1n2∑k=1n+1k2=2n3+9n2+13n+66=a{\displaystyle {\frac {1}{n^{2}}}\sum _{k=1}^{n+1}k^{2}={\frac {2n^{3}+9n^{2}+13n+6}{6}}=a}
∑k=1n+1k=(n+1)(n+2)2=(n2+3n+2)2{\displaystyle \sum _{k=1}^{n+1}k={\frac {(n+1)(n+2)}{2}}={\frac {(n^{2}+3n+2)}{2}}}
2n∑k=1n+1k=(n2+3n+2)n=b{\displaystyle {\frac {2}{n}}\sum _{k=1}^{n+1}k={\frac {(n^{2}+3n+2)}{n}}=b}
1n(a+b)=1n(2n3+9n2+13n+6+(6n(n2+3n+2))6n2)=8n3+27n2+25n+66n3=14n3+33n2+25n+66n3⇒f(n){\displaystyle {\begin{aligned}{\frac {1}{n}}(a+b)&={\frac {1}{n}}{\Bigl (}{\frac {2n^{3}+9n^{2}+13n+6+(6n(n^{2}+3n+2))}{6n^{2}}}{\Bigr )}\\&={\frac {8n^{3}+27n^{2}+25n+6}{6n^{3}}}\\&={\frac {14n^{3}+33n^{2}+25n+6}{6n^{3}}}\Rightarrow f(n)\end{aligned}}}
das n+1 ist einfach nur durch erweitern der brüche in eine andere form gebracht, so dass man es addieren kann
limn→∞f(n)=146=73{\displaystyle \lim _{n\to \infty }f(n)={\frac {14}{6}}={\frac {7}{3}}}
ge-TeX-ed von [1]