\[{{x}_{B}}=\frac{2}{ab}\int\limits_{0}^{a}{\left( \int\limits_{0}^{-(b/a)x+b}{x\,dy} \right)\,}dx=\frac{2}{ab}\int\limits_{0}^{a}{\left( -\frac{b}{a}{{x}^{2}}+bx \right)dx=}\]\[=\frac{2}{ab}\left[ -\frac{b}{3a}{{x}^{3}}+\frac{b}{2}{{x}^{2}} \right]_{0}^{a}=\frac{2}{ab}\left( \frac{b{{a}^{2}}}{6} \right)=\frac{a}{3}\]\[{{y}_{B}}=\frac{2}{ab}\int\limits_{0}^{a}{\left( \int\limits_{0}^{-(b/a)x+b}{y\,dy} \right)\,}dx=\frac{1}{ab}\int\limits_{0}^{a}{{{\left( -\frac{b}{a}x+b \right)}^{2}}dx=}\]\[=\frac{b}{a}\left[ \frac{{{x}^{3}}}{3{{a}^{2}}}+x-\frac{{{x}^{2}}}{a} \right]_{0}^{a}=\frac{b}{a}\left( \frac{a}{3}+a-a \right)=\frac{b}{3}\quad .\]
2) trapezio rettangolo. Area: \(S=\frac{\left( a+b \right)c}{2}\), retta \(AB\): \(y=-\frac{c}{a-b}\left( x-a \right)\)
\[{{x}_{B}}=\frac{2}{\left( a+b \right)c}\left[ \int\limits_{0}^{b}{\left( \int\limits_{0}^{c}{x\,dy} \right)\,}dx+\int\limits_{b}^{a}{\left( \int\limits_{0}^{c(a-x)/(a-b)}{x\,dy} \right)\,}dx \right]=\]\[\frac{2}{\left( a+b \right)c}\left[ \int\limits_{0}^{b}{cx\,dx}+\int\limits_{b}^{a}{\frac{cx\left( a-x \right)}{a-b}dx} \right]=\]\[=\frac{2}{\left( a+b \right)c}\left[ \frac{c{{b}^{2}}}{2}+\frac{c}{a-b}\left[ \frac{a{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{b}^{a} \right]=\frac{{{a}^{2}}+ab+{{b}^{2}}}{3\left( a+b \right)}\]
\[{{y}_{B}}=\frac{2}{\left( a+b \right)c}\left[ \int\limits_{0}^{b}{\left( \int\limits_{0}^{c}{y\,dy} \right)\,}dx+\int\limits_{b}^{a}{\left( \int\limits_{0}^{-c(x-a)/(a-b)}{y\,dy} \right)\,}dx \right]=\]\[=\frac{2}{\left( a+b \right)c}\left[ \frac{{{c}^{2}}}{2}\int\limits_{0}^{b}{dx}+\frac{{{c}^{2}}}{2{{\left( a-b \right)}^{2}}}\int\limits_{b}^{a}{{{\left( x-a \right)}^{2}}dx} \right]=\frac{2}{\left( a+b \right)c}\left[ \frac{{{c}^{2}}b}{2}+\frac{{{c}^{2}}}{2{{\left( a-b \right)}^{2}}}\left[ \frac{{{\left( x-a \right)}^{3}}}{3} \right]_{b}^{a} \right]=\frac{\left( 2b+a \right)c}{3\left( a+b \right)}\quad .\]
3) semiellisse. Area: \(S=\frac{\pi ab}{2}\), ordinata: \(y=b\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}\)
\[{{x}_{B}}=\frac{2}{\pi ab}\int\limits_{-a}^{a}{\left( \int\limits_{0}^{b\sqrt{1-{{x}^{2}}/{{a}^{2}}}}{x\,dy} \right)\,}dx=\frac{2}{\pi ab}\int\limits_{-a}^{a}{bx\sqrt{1-\frac{{{x}^{2}}}{{{a}^{2}}}}}\,dx=\]\[=\frac{2a}{\pi }\int\limits_{-1}^{1}{t\sqrt{1-{{t}^{2}}}}\,dt=0\quad (simmetria\;rispetto\;a\;y\;...)\]
\[{{y}_{B}}=\frac{2}{\pi ab}\int\limits_{-a}^{a}{\left( \int\limits_{0}^{b\sqrt{1-{{x}^{2}}/{{a}^{2}}}}{y\,dy} \right)\,}dx=\frac{b}{\pi a}\int\limits_{-a}^{a}{\left( 1-\frac{{{x}^{2}}}{{{a}^{2}}} \right)}\,dx=\frac{b}{\pi a}\left( 2a-\frac{2}{3}a \right)=\frac{4b}{3\pi }\quad .\]
3) segmento parabolico. Area: \(S=\frac{4ab}{3}\), ordinata: \(y=-\frac{b}{{{a}^{2}}}{{x}^{2}}+b\)
\[{{x}_{B}}=0\quad (simmetria\;rispetto\;a\;y\;...)\]
\[{{y}_{B}}=\frac{3}{4ab}\int\limits_{-a}^{a}{\left( \int\limits_{0}^{-\left( b/{{a}^{2}} \right){{x}^{2}}+b}{y\,dy} \right)\,}dx=\frac{3b}{8a}\int\limits_{-a}^{a}{\left( 1+\frac{{{x}^{4}}}{{{a}^{4}}}-\frac{2{{x}^{2}}}{{{a}^{2}}} \right)}\,dx=\]\[=\frac{3b}{8a}\left( 2a+\frac{2}{5}a-\frac{4}{3}a \right)=\frac{2}{5}b\quad .\]
Massimo Bergamini
