Ricevo da Paola la seguente domanda: Gent.mo Professore, mi aiuta per favore a risolvere questo integrale? \[\int\limits_{2}^{3}{\frac{1}{\sqrt{\left( x-2 \right)\left( 3-x \right)}}\,}dx\quad .\] Grazie. Le rispondo così: Cara Paola, possiamo calcolare l’integrale improprio come limite di una funzione integrale in questo modo: \[\int\limits_{2}^{3}{\frac{1}{\sqrt{\left( x-2 \right)\left( 3-x \right)}}\,}dx=\underset{a\to {{2}^{+}}}{\mathop{\lim }}\,\left( \underset{b\to {{3}^{-}}}{\mathop{\lim }}\,\int\limits_{a}^{b}{\frac{1}{\sqrt{\left( x-2 \right)\left( 3-x \right)}}\,}dx \right)\] e poichè\[\frac{1}{\sqrt{\left( x-2 \right)\left( 3-x \right)}}=\frac{1}{\sqrt{-{{x}^{2}}+5x-6}}=\frac{1}{\sqrt{\frac{1}{4}-\left( {{x}^{2}}-5x+\frac{25}{4} \right)}}=\]\[=\frac{1}{\sqrt{\frac{1}{4}-\frac{{{\left( 2x-5 \right)}^{2}}}{4}}}=\frac{2}{\sqrt{1-{{\left( 2x-5 \right)}^{2}}}}\] si ha \[\int\limits_{2}^{3}{\frac{1}{\sqrt{\left( x-2 \right)\left( 3-x \right)}}\,}dx=\underset{a\to {{2}^{+}}}{\mathop{\lim }}\,\left( \underset{b\to {{3}^{-}}}{\mathop{\lim }}\,\int\limits_{a}^{b}{\frac{2}{\sqrt{1-{{\left( 2x-5 \right)}^{2}}}}\,}dx \right)=\]\[=\underset{a\to {{2}^{+}}}{\mathop{\lim }}\,\left( \underset{b\to {{3}^{-}}}{\mathop{\lim }}\,\left[ \arcsin \left( 2x-5 \right) \right]_{a}^{b} \right)=\underset{a\to {{2}^{+}}}{\mathop{\lim }}\,\left( \underset{b\to {{3}^{-}}}{\mathop{\lim }}\,\left[ \arcsin \left( 2b-5 \right)-\arcsin \left( 2a-5 \right) \right] \right)=\]\[=\underset{a\to {{2}^{+}}}{\mathop{\lim }}\,\left( \frac{\pi }{2}-\arcsin \left( 2a-5 \right) \right)=\frac{\pi }{2}+\frac{\pi }{2}=\pi \quad .\] Massimo Bergamini
