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# Zweistellige Operationen/Durchschnittswerte/Aufgabe/Lösung

a) Nach Fubini ist

{\displaystyle {}{\begin{aligned}\int _{[a,b]\times [c,d]}x+yd\lambda ^{2}&=\int _{[a,b]\times [c,d]}xd\lambda ^{2}+\int _{[a,b]\times [c,d]}yd\lambda ^{2}\\&=\int _{a}^{b}xdx\int _{c}^{d}1dy+\int _{a}^{b}1dx\int _{c}^{d}ydy\\&={\frac {1}{2}}(b^{2}-a^{2})(d-c)+{\frac {1}{2}}(b-a)(d^{2}-c^{2}).\end{aligned}}}

Der Durchschnittswert ergibt sich, wenn man durch die Grundfläche dividiert, das ist also

${\displaystyle {}{\frac {{\frac {1}{2}}(b^{2}-a^{2})(d-c)+{\frac {1}{2}}(b-a)(d^{2}-c^{2})}{(b-a)(d-c)}}={\frac {1}{2}}{\left(b+a+c+d\right)}\,.}$

b) Es ist

{\displaystyle {}{\begin{aligned}\int _{[a,b]\times [c,d]}xyd\lambda ^{2}&=\int _{a}^{b}xdx\cdot \int _{c}^{d}ydy\\&={\frac {1}{2}}(b^{2}-a^{2})\cdot {\frac {1}{2}}(d^{2}-c^{2})\\&={\frac {1}{4}}(b^{2}-a^{2})(d^{2}-c^{2}).\end{aligned}}}

Der Durchschnittswert ist also

${\displaystyle {}{\frac {{\frac {1}{4}}(b^{2}-a^{2})(d^{2}-c^{2})}{(b-a)(d-c)}}={\frac {1}{4}}(b+a)(c+d)={\frac {1}{2}}(b+a)\cdot {\frac {1}{2}}(c+d)\,.}$

c) Es ist

{\displaystyle {}{\begin{aligned}\int _{[a,b]\times [c,d]}{\frac {x}{y}}d\lambda ^{2}&=\int _{a}^{b}xdx\cdot \int _{c}^{d}{\frac {1}{y}}dy\\&={\frac {1}{2}}(b^{2}-a^{2})\cdot {\left(\ln y|_{c}^{d}\right)}\\&={\frac {1}{2}}(b^{2}-a^{2}){\left(\ln d-\ln c\right)}.\end{aligned}}}

Der Durchschnittswert ist also

${\displaystyle {}{\frac {{\frac {1}{2}}(b^{2}-a^{2}){\left(\ln d-\ln c\right)}}{(b-a)(d-c)}}={\frac {1}{2}}(b+a){\frac {\ln d-\ln c}{d-c}}\,.}$