Lösung 2.1:2c

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If we recall that
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If we recall that <math>\sqrt{x} = x^{1/2}</math>, the integral can be written as
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<math>\sqrt{x}=x^{{1}/{2}\;}</math>, the integral can be written as
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{{Displayed math||<math>\begin{align}
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\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx
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&= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt]
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&= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.}
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\end{align}</math>}}
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<math>\begin{align}
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This is a standard integral in which the integrand consists of two terms looking like <math>x^n</math>, where <math>n=1/2</math> and <math>n=-1/2\,</math>, respectively.
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& \int\limits_{4}^{9}{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}\,dx=\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-\frac{1}{x^{{1}/{2}\;}} \right)}\,dx \\
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& =\int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx \\
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\end{align}</math>
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This is a standard integral in which the integrand consists of two terms looking like
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<math>x^{n}</math>, where
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<math>n=\frac{1}{2}</math>
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and
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<math>n=-\frac{1}{2}</math>.
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We obtain
We obtain
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{{Displayed math||<math>\begin{align}
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<math>\begin{align}
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\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx
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& \int\limits_{4}^{9}{\left( x^{{1}/{2}\;}-x^{{-1}/{2}\;} \right)}\,dx=\left[ \frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right]_{4}^{9} \\
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&= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt]
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& =\left[ \frac{x^{1+\frac{1}{2}}}{\frac{3}{2}}-\frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right]_{4}^{9} \\
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&= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt]
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& =\left[ \frac{2}{3}x\sqrt{x}-2\sqrt{x} \right]_{4}^{9} \\
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&= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt]
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& =\frac{2}{3}\centerdot 9\centerdot \sqrt{9}-2\sqrt{9}-\left( \frac{2}{3}\centerdot 4\centerdot \sqrt{4}-2\sqrt{4} \right) \\
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&= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt]
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& =\frac{2}{3}93-2\centerdot 3-\left( \frac{2}{3}\centerdot 4\centerdot 2-2\centerdot 2 \right) \\
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&= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt]
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& =18-6-\frac{16}{3}+4=16-\frac{16}{3} \\
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&= 18-6-\frac{16}{3}+4\\[5pt]
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& =\frac{16\centerdot 3-16}{3}=\frac{32}{3} \\
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&= 16-\frac{16}{3}\\[5pt]
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\end{align}</math>
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&= \frac{16\cdot 3-16}{3}\\[5pt]
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&= \frac{32}{3}\,\textrm{.}
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\end{align}</math>}}

Version vom 12:59, 21. Okt. 2008

If we recall that \displaystyle \sqrt{x} = x^{1/2}, the integral can be written as

\displaystyle \begin{align}

\int\limits_{4}^{9} \bigl(\sqrt{x}-\frac{1}{\sqrt{x}}\Bigr)\,dx &= \int\limits_{4}^{9}\Bigl( x^{1/2}-\frac{1}{x^{1/2}}\Bigr)\,dx\\[5pt] &= \int\limits_{4}^{9}\bigl(x^{1/2} - x^{-1/2}\bigr)\,dx\,\textrm{.} \end{align}

This is a standard integral in which the integrand consists of two terms looking like \displaystyle x^n, where \displaystyle n=1/2 and \displaystyle n=-1/2\,, respectively.

We obtain

\displaystyle \begin{align}

\int\limits_{4}^{9} \bigl( x^{1/2}-x^{-1/2}\bigr)\,dx &= \Bigl[\ \frac{x^{1/2+1}}{1/2+1} - \frac{x^{-1/2+1}}{-1/2+1}\ \Bigr]_{4}^{9}\\[5pt] &= \Bigl[\ \frac{x^{1+1/2}}{3/2} - \frac{x^{1/2}}{1/2}\ \Bigr]_{4}^{9}\\[5pt] &= \Bigl[\ \frac{2}{3}x\sqrt{x} - 2\sqrt{x}\ \Bigr]_{4}^{9}\\[5pt] &= \frac{2}{3}\cdot 9\cdot\sqrt{9} - 2\sqrt{9} - \Bigl(\frac{2}{3}\cdot 4\cdot \sqrt{4}-2\sqrt{4} \Bigr)\\[5pt] &= \frac{2}{3}\cdot 9\cdot 3 - 2\cdot 3 - \Bigl( \frac{2}{3}\cdot 4\cdot 2 - 2\cdot 2 \Bigr)\\[5pt] &= 18-6-\frac{16}{3}+4\\[5pt] &= 16-\frac{16}{3}\\[5pt] &= \frac{16\cdot 3-16}{3}\\[5pt] &= \frac{32}{3}\,\textrm{.} \end{align}

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