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ΠΑΡΑΓΟΝΤΙΚΗ ΟΛΟΚΛΗΡΩΣΗ ΣΥΝΘΕΤΕΣ ΠΕΡΙΠΤΩΣΕΙΣ

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Η παραγοντική ολοκλήρωση είναι σημαντική μέθοδος για τον υπολογισμό σύνθετων περιπτώσεων ολοκληρωμάτων

    \[ \begin{tabular}{r l r r r c r} $ 1.)\dint_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{x}{\hm^{2} x}\, dx.$ & & & 2.)$\dint_{0}^{\frac{\pi}{3}}\dfrac{x-\hm x}{\syn^{2}x}dx$ & & & \\\\ & & & & & & \\\\ 3.)$\dint_{1}^{4}\dfrac{\ln x}{\sqrt{x}}\, dx$ & & & 4.)$ \dint_{\frac{1}{e}}^{1}\dfrac{\ln x}{x^{2}}dx.$ & & & \\ \end{tabular}\\ \]


Παράδειγμα.1.
Να υπολογίσετε το ολοκλήρωμα: \dint_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{x}{\hm^{2} x}\, dx.

Λύση

    \begin{align*} &\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{x}{\hm^{2} x}\, dx=\\\\ &\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}x\cdot \dfrac{1}{\hm^{2} x} \, dx=\\\\ &\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}x\cdot \Big(-\snf x\Big)'\, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(x)'\cdot \big(-\snf x\big)\, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} -\snf x \, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\dfrac{\syn x}{\hm x} \, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \dfrac{(\hm x)'}{\hm x} \, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \big(\ln|\hm x|\big)' \, dx=\\\\ &\Big[ x\cdot \big(-\snf x\big)\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}-\Big[ \ln|\hm x|\Big]_{\frac{\pi}{6}}^{\frac{\pi}{2}}= \end{align*}

\Bigg\{ \bigg[\frac{\pi}{2}\cdot \big(-\snf \dfrac{\pi}{2}\big)\bigg]-\bigg[\frac{\pi}{6}\cdot \big(-\snf \dfrac{\pi}{6}\big) \bigg] \Bigg\} -\Big(\ln|\hm \dfrac{\pi}{2}| -\ln|\hm \dfrac{\pi}{6}|\Big)=

    \[0 - \Big(\dfrac{\pi}{6} \cdot \big (-\sqrt{3}\big)\Big) -\Big(\ln 1 -\ln \dfrac{1}{2}\Big)=\]

    \[\dfrac{\pi \cdot \sqrt{3}}{6} - \ln \dfrac{1}{2}=\]

    \[\dfrac{\pi \cdot \sqrt{3}}{6} - (\ln 1 - \ln 2)=\]

    \[\dfrac{\pi \cdot \sqrt{3}}{6} + \ln 2.\]

Παράδειγμα.2.
Να υπολογίσετε το ολοκλήρωμα: \dint_{0}^{\frac{\pi}{3}}\dfrac{x-\hm x}{\syn^{2}x}dx
Λύση

    \begin{align*} &\int_{0}^{\frac{\pi}{3}}\dfrac{x-\hm x}{\syn^{2}x}dx = \\\\ & \int_{0}^{\frac{\pi}{3}}(x-\hm x)\cdot \dfrac{1}{\syn^{2}x}dx =\\\\ &\int_{0}^{\frac{\pi}{3}}(x-\hm x)\cdot(\ef x)' dx =\\\\ &\Big[ (x-\hm x)\cdot \ef x\Big]_{0}^{\frac{\pi}{3}} -\int_{0}^{\frac{\pi}{2}}(x-\hm x)'\cdot \ef x dx=\\\\ \end{align*}

\bigg\{\Big[ (\frac{\pi}{3}-\hm \frac{\pi}{3})\cdot \ef \frac{\pi}{3}\Big] -\Big[ (0-\hm 0)\cdot \ef 0\Big]\bigg\} - \dint_{0}^{\frac{\pi}{3}}(1 -\syn x)\cdot \ef x dx=

    \[\Big[ (\frac{\pi}{3}- \frac{\sqrt{3}}{2})\cdot \sqrt{3} -0 \Big ] - \int_{0}^{\frac{\pi}{3}}(1 -\syn x)\cdot \ef x \, dx=\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}(1 -\syn x)\cdot \dfrac{\hm x}{\syn x}\, dx=\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}\dfrac{\hm x}{\syn x} -\syn x\cdot \dfrac{\hm x}{\syn x}\, dx=\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}\dfrac{\hm x}{\syn x} -\hm x \, dx=\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \Big(\int_{0}^{\frac{\pi}{3}}\dfrac{\hm x}{\syn x} \, dx - \int_{0}^{\frac{\pi}{3}}\hm x \, dx \Big)=\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}\dfrac{\hm x}{\syn x} \, dx + \int_{0}^{\frac{\pi}{3}}\hm x \, dx =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}\dfrac{(-\syn x)'}{\syn x} \, dx + \int_{0}^{\frac{\pi}{3}}(-\syn x)'\, dx =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) - \int_{0}^{\frac{\pi}{3}}\dfrac{(-\syn x)'}{\syn x} \, dx - \int_{0}^{\frac{\pi}{3}}(\syn x)'\, dx =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) + \int_{0}^{\frac{\pi}{3}}\dfrac{(\syn x)'}{\syn x} \, dx - \int_{0}^{\frac{\pi}{3}}(\syn x)'\, dx =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) +\int_{0}^{\frac{\pi}{3}}\big(\ln|\syn x|\big)' \, dx -\big [ \syn x \big]_{0}^{\frac{\pi}{3}} =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) + \Big [\ln|\syn x|\Big ]_{0}^{\frac{\pi}{3}} -\Big [ \syn x \Big]_{0}^{\frac{\pi}{3}} =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) + \Big (\ln|\syn \frac{\pi}{3}| - \ln|\syn 0|\Big ) -\Big (\syn \frac{\pi}{3}- \syn 0 \Big) =\]

    \[\big(\frac{\sqrt{3}\pi}{3}- \frac{3}{2}\big) +\Big (\ln\frac{1}{2} - \ln1\Big )-\Big (\frac{1}{2}- 1 \Big) =\]

    \[\frac{\sqrt{3}\pi}{3}- \frac{3}{2} +\ln\frac{1}{2} - 0 -\frac{1}{2}+ 1 =\]

    \[\dfrac{\sqrt{3}\pi -3}{3}- \ln 2.\]

Παράδειγμα.3.
Να υπολογίσετε το ολοκλήρωμα: \dint_{1}^{4}\dfrac{\ln x}{\sqrt{x}}\, dx
Λύση

    \begin{align*} &\int_{1}^{4}\dfrac{\ln x}{\sqrt{x}}\, dx =\\\\ &\int_{1}^{4} \dfrac{1}{\sqrt{x}}\cdot \ln x\, dx=\\\\ &\int_{1}^{4} \dfrac{1}{2\sqrt{x}}\cdot 2 \cdot \ln x\, dx=\\\\ &\int_{1}^{4} \big( \sqrt{x}\big)'\cdot 2 \cdot \ln x\, dx =\\\\ &\Big[ 2 \cdot\sqrt{x}\cdot\ln x\Big]_{1}^{4} - \int_{1}^{4} \sqrt{x}\cdot \big(2 \cdot \ln x\big)'\, dx =\\\\ &\Big[ \big(2 \cdot\sqrt{4}\cdot\ln 4\big)-\big(2 \cdot\sqrt{1}\cdot\ln 1\big)\Big] - \int_{1}^{4} \sqrt{x}\cdot \dfrac{2}{x}\, dx =\\\\ &\Big[ \big(2 \cdot2\cdot\ln 4\big)-\big(2 \cdot\sqrt{1}\cdot 0 \big)\Big] - \int_{1}^{4} \sqrt{x}\cdot \dfrac{2}{x}\, dx =\\\\ &\Big[ \big(4\ln 4\big)-\big( 0 \big)\Big] - \int_{1}^{4} \sqrt{x}\cdot \dfrac{2}{\sqrt{x}^{2}}\, dx \end{align*}

Συνεπώς

    \begin{align*} &4\ln 4 - \int_{1}^{4} \dfrac{2}{\sqrt{x}}\, dx =\\\\ &4\ln 4 - \int_{1}^{4} \dfrac{4}{2\cdot\sqrt{x}}\, dx =\\\\ &4\ln 4 - 4\cdot\int_{1}^{4} \dfrac{1}{2\cdot\sqrt{x}}\, dx =\\\\ &4\ln 4 - 4\cdot\int_{1}^{4} \big(\sqrt{x}\big)' =\\\\ &4\ln 4 - 4\cdot\Big[\sqrt{x}\Big]_{1}^{4} =\\\\ &4\ln 4 - 4\cdot\big (\sqrt{4}-\sqrt{1} \big) = 4\ln 4 -4. \end{align*}

Παράδειγμα.4.
Να υπολογισθεί το ολοκλήρωμα

    \[\int_{\frac{1}{e}}^{1}\dfrac{\ln x}{x^{2}}dx.\]

Λύση
Έχουμε:

    \[\int_{\frac{1}{e}}^{1}\dfrac{\ln x}{x^{2}}dx=\int_{\frac{1}{e}}^{1}\dfrac{1}{x^{2}}\cdot \ln x dx=\int_{\frac{1}{e}}^{1}\Big(-\dfrac{1}{x}\Big)'\cdot \ln x dx =\]

    \begin{align*} &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}- \int_{\frac{1}{e}}^{1}\big(-\dfrac{1}{x}\big)\cdot \Big(\ln x \Big)'dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}- \int_{\frac{1}{e}}^{1}\big(-\dfrac{1}{x}\big)\cdot \dfrac{1}{x}\,dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}+ \int_{\frac{1}{e}}^{1}\dfrac{1}{x}\cdot \dfrac{1}{x}\,dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}+ \int_{\frac{1}{e}}^{1}\dfrac{1}{x^{2}}\,dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}+ \int_{\frac{1}{e}}^{1}\Big(-\dfrac{1}{x}\Big)'dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}- \int_{\frac{1}{e}}^{1}\Big(\dfrac{1}{x}\Big)'dx =\\\\ &\Big[-\dfrac{1}{x} \cdot \ln x \Big]_{\frac{1}{e}}^{1}- \Big[\dfrac{1}{x}\Big]_{\frac{1}{e}}^{1} =\\\\ &\Big[\big(-\dfrac{1}{1} \cdot \ln 1\big)-\big(-\dfrac{1}{\frac{1}{e}} \cdot \ln \frac{1}{e}\big) \Big]- \Big[\dfrac{1}{1}-\dfrac{1}{\frac{1}{e}}\Big]=\\\\ &\Big[\big(0)-\big(-e\cdot \ln \frac{1}{e}\big) \Big]- \Big[1-e\Big]=\\\\ \end{align*}

    \begin{align*} &e\ln\dfrac{1}{e}- 1+e= \\\\ &e(\ln1 -\ln e)-1+e =\\\\ &-e-1+e=-1.\\\\ \end{align*}

Βιβλιογραφία: Μπάρλας εκδόσεις Ελληνοεκδοτική. Παπαδάκης εκδόσεις Σαββάλα.
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